text:
\section{Introduction and main results}

Let $\H(\mathbb{D})$ denote the space of analytic functions in the unit disc $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. Each $g\in\H(\mathbb{D})$ induces the integral operator defined by
	$$
	T_g(f)(z)=\int_0^zf(\zeta)g'(\zeta)\,d\zeta,\quad z\in\mathbb{D}.
	$$
The question of when this operator is either bounded or compact has been extensively studied in a large variety of spaces of analytic functions since the appearance of the seminal works, related to the Hardy and Bergman spaces, due to Aleman, Cima, Pommerenke and Siskakis~\cite{AC,AS,Pom}. Getting neat conditions on the symbol $g$ which describe the bounded and compact operators $T_g$ acting from a Banach space $X\subset\H(\mathbb{D})$ to the Hardy space $H^\infty$ is known to be a tough problem~\cite{AJS,CPPR,SSV2017}. However, recently an abstract approach to the study of this question was given in \cite{CPPR}. One of the basic results there is the reproducing kernel dual testing condition provided in \cite[Theorem~2.2]{CPPR}.
It states that, if $X^\star\simeq Y$ via the $H(\beta)$-pairing
	$$
	\langle f,g\rangle_{H(\beta)}=\lim_{r\to1^-}\sum_{n=0}^\infty\widehat{f}(n)\overline{\widehat{g}(n)}\beta_nr^n,
	\quad f(z)=\sum_{n=0}^\infty\widehat{f}(n)z^n,\quad g(z)=\sum_{n=0}^\infty\widehat{g}(n)z^n,
	$$
where $\lim_{n\to\infty}\sqrt[n]{\beta_n}=1$, then $T_g:X\to H^\infty$ is bounded if and only if $\sup_{z\in\mathbb{D}}\left\|G^{H(\beta)}_{g,z}\right\|_Y<\infty$, where
		$$
		\overline{G^{H(\beta)}_{g,z}(w)}
		=\int_0^zg'(\zeta)\overline{K_\zeta^{H(\beta)}(w)}\,d\zeta
		=\overline{T_g^*(K_z^{H(\beta)})(w)},\quad z\in\mathbb{D},\quad w\in\overline{\mathbb{D}},
		$$
and $K_\zeta^{H(\beta)}$ are the reproducing kernels of the Hilbert space $H(\beta)$.

Theoretically, the above relatively simple result offers a characterization of the boundedness in the case of most of the natural spaces one can think of. However, if one tries to apply this in praxis one observes that it looks like a reformulation rather than a solution of the problem. This is due to the fact that treating the function $G^{H(\beta)}_{g,z}$ in the dual space of $X$ is often laborious if not even frustrating. Because of these reasons, in this study we restrict ourselves to the case in which the symbol $g$ has non-negative Maclaurin coefficients, and search for neat and useful conditions in terms of the Maclaurin coefficients of $g$ that can be used to test if $T_g$ is either bounded or compact from $X$ to $H^\infty$. The starting point is the characterization \cite[Theorem~2.2]{CPPR} given above, and the choices for $X$ that will be considered in the sequel contain the Hardy and the Hardy-Littlewood spaces, and certain Dirichlet-type spaces, as well as the classical Bloch space and $\mathord{\rm BMOA}$. Next, the main findings of this study along with necessary definitions are stated.

For $0<p\le\infty$, the Hardy space $H^p$ consists of $f\in\H(\mathbb{D})$ for which
    \begin{equation*}\label{normi}
    \|f\|_{H^p}=\sup_{0<r<1}M_p(r,f)<\infty,
    \end{equation*}
where
    $$
    M_p(r,f)=\left (\frac{1}{2\pi}\int_0^{2\pi}
    |f(re^{i\theta})|^p\,d\theta\right )^{\frac{1}{p}},\quad 0<p<\infty,
    $$
and
    $$
    M_\infty(r,f)=\max_{0\le\theta\le2\pi}|f(re^{i\theta})|.
    $$
Further, $f\in\H(\mathbb{D})$ belongs to the Dirichlet-type space $D^p_{p-1}$ if
	$$
	\|f\|^p_{D^p_{p-1}}=\int_\mathbb{D}|f'(z)|^p(1-|z|)^{p-1}\,dA(z)+|f(0)|^p<\infty,
	$$
where $dA(z)=\frac{dx\,dy}{\pi}$ is the normalized area measure on $\mathbb{D}$. The closely related Hardy-Littlewood space $\mathord{\rm HL}_p$ contains those $f\in\H(\mathbb{D})$ whose Maclaurin coefficients $\{\widehat{f}(n)\}_{n=0}^\infty$ satisfy
	$$
	\nm{f}^p_{\mathord{\rm HL}_p}=\sum_{n=0}^{\infty}\abs{\widehat{f}(n)}^p (n+1)^{p-2}<\infty.
	$$
These spaces satisfy the well-known inclusions
	\begin{equation}\label{Hardy1}
	D^p_{p-1}\subset H^p\subset\mathord{\rm HL}_p,\quad 0<p\le 2,
	\end{equation}
and
	\begin{equation}\label{Hardy2}
	\mathord{\rm HL}_p\subset H^p\subset D^p_{p-1} ,\quad 2\le p<\infty,
	\end{equation}
by \cite{D,Flett,LP}. Each of these inclusions is strict unless $p=2$, in which case all the spaces are the same by direct calculations or straightforward applications of Parseval's formula and Green's theorem.

Our first main result reveals that $T_g$ does not distinguish $H^p$, $\mathord{\rm HL}_p$ and $D^p_{p-1}$ when it acts boundedly or compactly from one of these spaces to $H^\infty$, provided $1<p<\infty$ and the symbol $g$ has non-negative Maclaurin coefficients. Here, as usual, the conjugate index of $1<p<\infty$ is the number $p'$ such that $\frac1p+\frac1{p'}=1$.

\begin{theorem}\label{Theorem:Hardy-p<2}
Let $1<p<\infty$ and $g\in H^\infty$ such that $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$. Further, let $X_p\in\{H^p,D^p_{p-1}\mathord{\rm HL}_p\}$. Then $T_g:~X_p\rightarrow H^\infty$ is bounded (equivalently compact) if and only if
	\begin{equation}\label{pipeli}
	\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}<\infty.
	\end{equation}
Moreover,
	\begin{equation}\label{eq:j11}
	\|T_g\|_{X_p\rightarrow H^\infty}^{p'}
	\asymp\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}.
	\end{equation}
\end{theorem}

On the way to Theorem~\ref{Theorem:Hardy-p<2} we show in Theorem~\ref{th:dpsuf} below that for each $g\in\H(\mathbb{D})$ we have
	\begin{equation}\label{poi}
	\|T_g\|_{X_p\rightarrow H^\infty}^{p'}
  \lesssim
  \sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)|\widehat{g}(n+1)|}{n+k+1}\right)^{p'}
	\end{equation}
for each $X_p\in\{H^p,D^p_{p-1}\mathord{\rm HL}_p\}$. This observation offers a relatively easy way to see if a given general symbol $g$ induces a bounded operator on the Hardy space $H^p$ with $1<p<\infty$.

The proof of Theorem~\ref{Theorem:Hardy-p<2} relies on \cite[Theorem~2.2]{CPPR}, and duality relations for $H^p$, $D^p_{p-1}$ and $\mathord{\rm HL}_p$ with $1<p<\infty$. The dual of $H^p$ is of course isomorphic to $H^{p'}$ via the $H^2$-pairing (the Cauchy-pairing), and certainly many experts working on the field now that $(D^p_{p-1})^\star\simeq D^{p'}_{p'-1}$ and $(\mathord{\rm HL}_p)^\star\simeq\mathord{\rm HL}_{p'}$ via the same pairing. Since we do not know exact references for the last-mentioned two dualities, we give proofs in Section~\ref{Sec:dualities} where also other less obvious duality relations are treated. Another tool that we will employ in the proof of Theorem~\ref{Theorem:Hardy-p<2} is of technical nature, and concerns smooth universal Ces\'aro basis of polynomials~\cite[Section 5.2]{Pavlovic}. The proof of Theorem~\ref{Theorem:Hardy-p<2} is presented in Section~\ref{Sec:pili}.

If $0<p<1$, then $T_g:H^p\to H^\infty$ is bounded if and only if $g$ is a constant by~\cite[Theorem~2.5(vi)]{CPPR}. Further, by~\cite[Theorem~4.2(ii)]{CPPR}, $T_g:H^1\to H^\infty$ is compact if and only if $g$ is a constant. Therefore the same conclusions are valid for $T_g$ acting on $\mathord{\rm HL}_p$ by \eqref{Hardy1}. The following result clarifies the situation with regard to the Dirichlet-type spaces $D^p_{p-1}$. Here and from now on $T(X,H^\infty)$ (resp. $T_c(X,H^\infty)$) denotes the set of $g\in\H(\mathbb{D})$ such that $T_g:X\to H^\infty$ is bounded (resp. compact).

\begin{theorem}\label{Theorem:p=1}
Let $g\in H^\infty$ and $X_p\in\{H^p,D^p_{p-1},\mathord{\rm HL}_p\}$. Then the following assertions hold:
\begin{itemize}
\item[\rm(i)] $T(X_1,H^\infty)$ contains all polynomials;
\item[\rm(ii)] $T(X_p,H^\infty)$ consists of constant functions only if $0<p<1$;
\item[\rm(iii)] $T_c(X_1,H^\infty)$ consists of constant functions only.
\end{itemize}
\end{theorem}

In the proof of Theorem~\ref{Theorem:p=1} we use identifications of the duals of $\mathord{\rm HL}_1$ and $D^p_{p-1}$ with $0<p\le1$.
Since many dual spaces $X^\star$ can be described, via the $H^2$-pairing,  as  the space of coefficient multipliers from $X$ to the disc algebra  \cite[Proposition~1.3]{P}, a natural characterization of the dual of $\mathord{\rm HL}_1$ is easy to find by using the relation $(\ell^1)^\star\simeq\ell^\infty$. We do this in Section~\ref{Sec:dualities} when we prove Lemma~\ref{le:dualHL1} which states that $(\mathord{\rm HL}_1)^\star\simeq\mathord{\rm HL}_{\infty}$ via the $H^2$-pairing with equivalence of norms. The space $\mathord{\rm HL}_{\infty}$ consists of $f\in\H(\mathbb{D})$ such that its Maclaurin coefficients $\{\widehat{f}(n)\}_{n=0}^{\infty}$ satisfy
	$$
	\|f\|_{\mathord{\rm HL}_{\infty}}=\sup_{n\in\mathbb{N}\cup\{0\}}\left(|\widehat{f}(n)|(n+1)\right)<\infty.
	$$
To find a suitable dual of $D^p_{p-1}$ with $0<p\le1$ is not that straightforward. Lemma~\ref{lemma:dual-Dirichlet-2} in Section~\ref{Sec:dualities} states that $(D^p_{p-1})^\star\simeq\mathcal{B}^2$ via the $A^2_{\frac1p-1}$-pairing.	Here $A^2_\beta$ refers to the Bergman Hilbert space induced by the standard weight $(\beta+1)(1-|z|^2)^\beta$. Further, for $0<\alpha<\infty$ and $f\in\H(\mathbb{D})$, the $\alpha$-Bloch space $\mathcal{B}^\alpha$ consists of $f\in\H(\mathbb{D})$ such that
	$$
	\|f\|_{\mathcal{B}^\alpha}=\sup_{z\in\mathbb{D}}|f'(z)|(1-|z|^2)^\alpha+|f(0)|<\infty.
	$$
The proof of the duality relation $(D^p_{p-1})^\star\simeq\mathcal{B}^2$ is lengthy, and apart from standard tools, such as Green's theorem and continuous embeddings between different weighted Bergman spaces, it also relies on a use of  smooth universal Ces\'aro basis of polynomials. The last-mentioned creatures are used to show that a certain function, dependent of $p$, is a coefficient multiplier of $\mathcal{B}^2$.

The next result is the counterpart of Theorem~\ref{Theorem:Hardy-p<2} in the case $p=1$. Observe that \eqref{Eq:p=1} is the limit case $p'=\infty$ of \eqref{eq:j11}, and that the supremum there is in fact the limit as $k\to\infty$ since the quantity over which the supremum is taken is increasing in $k$.

\begin{theorem}\label{Theorem:p=1'}
Let $g\in H^\infty$ such that $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$, and $X_1\in\{H^1,D^1_{0},\mathord{\rm HL}_1\}$. Then $T_g:X_1\rightarrow H^\infty$ is bounded if and only if
	\begin{equation}\label{Eq:p=1}
	\sup_{k\in\mathbb{N}\cup\{0\}}\left((k+1)\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)<\infty.
	\end{equation}
Moreover,
	\begin{equation}\label{eq:j11'}
	\|T_g\|_{X_1\rightarrow H^\infty}
	\asymp\sup_{k\in\mathbb{N}\cup\{0\}}\left((k+1)\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right).
	\end{equation}
\end{theorem}

Theorem~\ref{Theorem:p=1'} is relatively straightforward to establish once the tools needed for Theorem~\ref{Theorem:p=1} are on the table.
Both of these theorems are proved in Section~\ref{Sec:2-3}.

Our last result concerns the case when $T_g$ acts from the Bloch space or $\mathord{\rm BMOA}$ to $H^\infty$. Recall that the classical Bloch space $\mathcal{B}$ is just the space $\mathcal{B}^1$ defined before Theorem~\ref{Theorem:p=1'}. Further, let
	$$
	\|f\|_{H^\infty_{\log}}=\sup_{z\in\mathbb{D}}|f(z)|\left(\log\frac{e}{1-|z|}\right)^{-1},
	$$
and recall that $\mathord{\rm BMOA}$ consists of the functions in the Hardy space $H^1$ that have
bounded mean oscillation on the boundary $\mathbb{T}$. The space $\mathord{\rm BMOA}$ can be equipped with several different norms~\cite{Daniel}. We will use the one given by
    $$
    \|g\|^2_{\mathord{\rm BMOA}}=\sup_{a\in\mathbb{D}}\frac{\int_{S(a)}|g'(z)|^2(1-|z|^2)\,dA(z)}{1-|a|}+|g(0)|^2,
    $$
where $S(a)=\{\z:1-|a|<|\z|<1,\,|\arg\z-\arg a|<(1-|a|)/2\}$ is the Carleson square induced by $a\in\mathbb{D}\setminus\{0\}$ and $S(0)=\mathbb{D}$.
It is well-known that
	\begin{equation}\label{Eq:embedding}
	\|f\|_{H^\infty_{\log}}\lesssim\|f\|_\mathcal{B}\lesssim\|f\|_{\mathord{\rm BMOA}}\lesssim\|f\|_{H^\infty},\quad f\in\H(\mathbb{D}).
	\end{equation}

Our last main result says that $T_g$ does not distinguish $\mathord{\rm BMOA}$, $\mathcal{B}$ and $H^\infty_{\log}$ when it acts boundedly or compactly from one of these spaces to $H^\infty$, if the symbol $g$ has non-negative Maclaurin coefficients.

\begin{theorem}\label{Bounded}
Let $X\subset\H(\mathbb{D})$ be a Banach space such that $\mathord{\rm BMOA}\subset X\subset H^\infty_{\log}$ and
let $g\in H^{\infty}$ with $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$. Then the following statements are equivalent:
\begin{itemize}
\item[(i)] $T_g:~H_{\log}^\infty\to H^\infty$ is bounded (equivalently compact);
\item[(ii)] $T_g:~X\to H^\infty$ is bounded (equivalently compact);
\item[(iii)] $T_g:~\mathord{\rm BMOA}\to H^\infty$ is bounded (equivalently compact);
\item[(iv)] $\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2)<\infty$;
\item[(v)] $\int_0^1M_{\infty}(r,g')\log\frac{e}{1-r}dr<\infty$.
\end{itemize}
Moreover,
	\begin{equation}\label{Eq:norm-estimates}
	\begin{split}
	\|T_g\|_{\mathord{\rm BMOA}\rightarrow H^\infty}
	&\asymp\|T_g\|_{X\rightarrow H^\infty}
	\asymp\|T_g\|_{H_{\log}^\infty\rightarrow H^\infty}
	\asymp\int_0^1M_\infty(r,g')\log\frac{e}{1-r}dr\\
	&\asymp\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2).
	\end{split}
	\end{equation}
\end{theorem}

The proof of Theorem~\ref{Bounded}, given in Section~\ref{Sec:Bloch}, reveals that
	\begin{equation*}
	\|T_g\|_{H_{\log}^\infty\rightarrow H^\infty}
	\lesssim\int_0^1M_\infty(r,g')\log\frac{e}{1-r}dr
	\lesssim\sum_{n=0}^\infty|\widehat{g}(n+1)|\log(n+2)
	\end{equation*}
for each $g\in\H(\mathbb{D})$. The hypothesis on the coefficients is only used when the right most quantity above is dominated by the operator norm.

Probably the most obvious election for the space $X$ in the statement of Theorem~\ref{Bounded} is the classical Bloch space $\mathcal{B}$. However, there are other choices for $X$ which arise naturally in the theory of  integral operators, see Section~\ref{Sec:Bloch} for further details.

It is worth mentioning that the hypotheses $g\in H^\infty$ in  Theorems~\ref{Theorem:Hardy-p<2}-\ref{Bounded} is not a restriction, because it is an obvious necessary condition so that $T_g: X\to H^\infty$ is bounded.

To this end, couple of words about the notation used in this paper. The letter $C=C(\cdot)$ will denote an absolute constant whose value depends on the parameters indicated
in the parenthesis, and may change from one occurrence to another. If there exists a constant
$C=C(\cdot)>0$ such that $a\le Cb$, then we write either $a\lesssim b$ or $b\gtrsim a$. In particular, if $a\lesssim b$ and
$a\gtrsim b$, then we denote $a\asymp b$ and say that $a$ and $b$ are comparable.

\section{Dualities}\label{Sec:dualities}

In this section we will discuss the duality relations employed to prove the main results of the paper. Apart from the well-known relation $(H^p)^\star\simeq H^{p'}$, $1<p<\infty$, we will need to know the dual spaces, with respect to appropriate pairings, of $D^p_{p-1}$ and $\mathord{\rm HL}_p$ for $0<p\le 1$ and $1<p<\infty$, respectively.

The following lemma describes the dual of the Dirichlet-type space $D^p_{p-1}$ when $1<p<\infty$, and it will be needed in the proof of Theorem~\ref{Theorem:Hardy-p<2}. We believe that the result itself must be known at least by experts working on the field, but since we do not know an exact reference, we include a proof here.

\begin{lemma}\label{le:dualdp}
Let $1<p<\infty$. Then $(D^p_{p-1})^\star\simeq D^{p'}_{p'-1}$ via the $H^2$-pairing with equivalence of norms.
\end{lemma}

\begin{proof}
Let us first show that each $g\in D^{p'}_{p'-1}$ induces a bounded linear functional on $D^p_{p-1}$. Green's theorem implies
	\begin{equation}\label{Green}
	\langle f,g\rangle_{H^2}=2\int_{\mathbb{D}}f'(\z)\overline{g'(\z)}\log\frac{1}{|\z|}\,dA(\z)+f(0)\overline{g(0)},
	\end{equation}
from which H\"{o}lder's inequality yields
	\begin{equation*}\begin{split}
|\langle f,g\rangle_{H^2}|
 &\lesssim \int_{\mathbb{D}}  |f'(\z)||g'(\z)|(1-|\z|)\,dA(\z)+|f(0)||g(0)|
\lesssim\|f\|_{D^p_{p-1}}\|g\|_{D^{p'}_{p'-1}},\quad f,g\in\H(\mathbb{D}),
\end{split}
	\end{equation*}
where the first step is an easy consequence of the inequality $-\log t\le\frac1t(1-t)$, valid for all $0<t\le1$, and the monotonicity of $M_p(r,h)$ for each $0<p<\infty$ and $h\in\H(\mathbb{D})$.
Thus each $g\in D^{p'}_{p'-1}$ induces a bounded linear functional on $D^p_{p-1}$ via the $H^2$-pairing.

Let now $L$ be a bounded linear functional on $D^p_{p-1}$. Consider the weights $\om(z)=-2\log|z|$ and $\nu(z)=\left(-2\log|z|\right)^\frac1{p-1}$, defined in the punctured unit disc. The proof of \cite[Theorem~3]{PR2014} now shows that the Bergman projection $P_\om$, induced by $\om$, is bounded from $L^{p'}_\nu$ into itself because the weight $\left(\frac{\om}{\nu}\right)^p\nu=\om^{p-1}$ is sufficiently smooth. It then follows from the proof of \cite[Theorem~6]{PR2014} and standard arguments that $(A^p_{p-1})^\star\simeq A^{p'}_{p'-1}$ under the pairing
	$$
	\langle f,g\rangle_{A^2_{\omega}}=2\int_{\mathbb{D}}f(z)\overline{g(z)}\log\frac{1}{|z|}\,dA(z).
	$$
We note that this duality relation of the weighted Bergman spaces is essentially contained in \cite[Theorem~2.1]{L} as a special case, but with respect to a slightly different pairing. The method there would certainly work also in our setting and therefore offers an alternative way to deduce this duality. Getting back to the proof of the lemma, we observe that, for each $f\in D^p_{p-1}$, there exists $F=F_f\in A^p_{p-1}$ such that $\mathcal{I}(F)=f-f(0)$, where $\mathcal{I}(F)(z)=\int_0^zF(\zeta)\,d\zeta$. Further, $\mathcal{I}$ is an isometric mapping from $A^p_{p-1}$ to $D^p_{p-1}$, in particular, it is bounded. Therefore the composition $L\circ\mathcal{I}$ is a bounded linear functional on $A^p_{p-1}$, and hence there exists a unique $G\in A^{p'}_{p'-1}$ such that $\|G\|_{A^{p'}_{p'-1}}\lesssim\|L\circ\mathcal{I}\|\lesssim\|L\|$ and
	\begin{equation*}
	\begin{split}
	L(f)=L(f-f(0)+f(0))&=(L\circ\mathcal{I})(F)+f(0)L(1)\\
	&=2\int_{\mathbb{D}}F(z)\overline{G(z)}\log\frac{1}{|z|}dA(z)+f(0)L(1)\\
	&=2\int_{\mathbb{D}}f'(z)\overline{G(z)}\log\frac{1}{|z|}dA(z)+f(0)L(1).
	\end{split}
	\end{equation*}
Further, since $G\in A^{p'}_{p'-1}$, there exists a unique $g\in D^{p'}_{p'-1}$ such that $g'=G$ and $\overline{g(0)}=L(1)$. Consequently, there exists a unique $g\in D^{p'}_{p'-1}$ such that
	$$
	L(f)=2\int_{\mathbb{D}}f'(z)\overline{g'(z)}\log\frac{1}{|z|}dA(z)+f(0)\overline{g(0)}=\langle f,g\rangle_{H^2},
	$$
where the last identity follows from \eqref{Green}. Moreover, $\|g\|^{p'}_{D^{p'}_{p'-1}}=\|G\|^{p'}_{A^{p'}_{p'-1}}+|L(1)|^{p'}\lesssim\|L\|^{p'}$, and the assertion is proved.
\end{proof}

To prove Theorems~\ref{Theorem:p=1} and~\ref{Theorem:p=1'} we need to know the dual of $D^p_{p-1}$ with $0<p\le1$. In order to do that, some more notation is needed. For $0<\alpha<\infty$, the space $H^\infty_\alpha$ consists of $f\in\H(\mathbb{D})$ such that
	$$
	\|f\|_{H^\infty_\alpha}=\sup_{z\in\mathbb{D}}|f(z)|(1-|z|^2)^\alpha<\infty.
	$$
It is well-known that
	\begin{equation}\label{Eq-alpha-bloch}
	\|f\|_{H^\infty_\alpha}\asymp\|f\|_{\mathcal{B}^{\alpha+1}},\quad f\in\H(\mathbb{D}),
	\end{equation}
for each $0<\alpha<\infty$.

We will also need background on certain smooth polynomials defined in terms of Hadamard products. Recall that the Hadamard product of $f\in\H(\mathbb{D})$ and $g\in\H(\mathbb{D})$ is formally defined as
    $$
    (f\ast g)(z)=\sum_{k=0}^\infty\widehat{f}(k)\widehat{g}(k)z^k,\quad z\in\mathbb{D}.
    $$
A direct calculation shows that
    \begin{equation}\label{eq:hadprod}
    (f\ast g)(r^2e^{it})
    =\frac{1}{2\pi}\int_{-\pi}^\pi f(re^{i(t+\theta)})g(re^{-i\t})\,d\t.
    \end{equation}
If  $W(z)=\sum_{k\in J}b_kz^k$ is a polynomial and $f\in\H(\mathbb{D})$, then the Hadamard product
    $$
    (W\ast f)(z)=\sum_{k\in J}b_k\widehat{f}(k)z^k
    $$
is well defined. Further, if $\Phi:\mathbb{R}\to\mathbb{C}$ is a $C^\infty$-function with compact support $\supp(\Phi)$ in $(0,\infty)$, set
    $$
    A_{\Phi,m}=\max_{s\in\mathbb{R}}|\Phi(s)|+\max_{s\in\mathbb{R}}|\Phi^{(m)}(s)|,
    $$
and consider the polynomials
   \begin{equation}\label{eq:polynomials}
    W_n^\Phi(z)=\sum_{k\in\mathbb
    Z}\Phi\left(\frac{k}{n}\right)z^{k},\quad n\in\mathbb{N}.
    \end{equation}
With this notation we can state the next auxiliary result that follows by \cite[p.~111--113]{Pavlovic}.

\begin{lettertheorem}\label{th:cesaro}
Let $\Phi:\mathbb{R}\to\mathbb{C}$ be a $C^\infty$-function such that
$\supp(\Phi)\subset(0, \infty)$ is compact. Then for each $p\in(0,\infty)$ and $m\in\mathbb{N}$ with $mp>1$, there exists a constant
$C=C(p)>0$ such that
    $$
    \|W_N^\Phi\ast f\|_{H^p}\le C A_{\Phi,m}\|f\|_{H^p}
    $$
for all $f\in H^p$ and $N\in\mathbb{N}$.
\end{lettertheorem}

Theorem~\ref{th:cesaro} shows that the polynomials $\{W_n^\Phi\}_{n\in\mathbb{N}}$ can be seen as a universal C\'esaro basis for $H^p$ for any $0<p<\infty$. A particular case of the previous construction is useful for our purposes. By following \cite[Section~2]{JevPac98}, let $\Psi:\mathbb{R}\to\mathbb{R}$ be a $C^\infty$-function such that
    \begin{enumerate}
    \item $\Psi\equiv1$ on $(-\infty,1]$,
    \item $\Psi\equiv0$ on $[2,\infty)$,
    \item $\Psi$ is decreasing and positive on $(1,2)$,
    \end{enumerate}
and set $\psi(t)=\Psi\left(\frac{t}{2}\right)-\Psi(t)$ for all $t\in\mathbb{R}$. Let $V_0(z)=1+z$
and
    \begin{equation}\label{vn}
    V_n(z)=W^\psi_{2^{n-1}}(z)=\sum_{k=0}^\infty
    \psi\left(\frac{k}{2^{n-1}}\right)z^k=\sum_{k=2^{n-1}}^{2^{n+1}-1}
    \psi\left(\frac{k}{2^{n-1}}\right)z^k,\quad n\in\mathbb{N}.
    \end{equation}
These polynomials have the following properties with regard to
smooth partial sums, see \cite[p. 175--177]{JevPac98} or \cite[p. 143--144]{Pabook2} for details:
    \begin{equation}
    \begin{split}\label{propervn}
    f(z)&=\sum_{n=0}^\infty (V_n\ast f)(z),\quad f\in\H(\mathbb{D}),\\
    \|V_n\ast f\|_{H^p}&\le C\|f\|_{H^p},\quad f\in H^p,\quad 0<p<\infty,\\
    \|V_n\|_{H^p}&\asymp 2^{n(1-1/p)}, \quad 0< p<\infty.
    \end{split}
    \end{equation}

With these preparations we can describe the dual of $D^p_{p-1}$ with $0<p\le1$.

\begin{lemma}\label{lemma:dual-Dirichlet-2}
Let $0<p\le1$. Then $(D^p_{p-1})^\star\simeq\mathcal{B}^2$ via the $A^2_{\frac1p-1}$-pairing with equivalence of norms.
\end{lemma}

\begin{proof}
Let $f,g\in\H(\mathbb{D})$. Then Green's formula and Fubini's theorem yield
	\begin{equation*}
	\begin{split}
	|\langle f,g\rangle_{A^2_{\frac1p-1}}|
	&=\left|\int_\mathbb{D} f(z)\overline{g(z)}(1-|z|^2)^{\frac1p-1}\,dA(z)\right|\\
	&= \left| \int_0^1 \left(2\int_\mathbb{D} f'(rz)r\overline{g'(rz)r}\log\frac1{|z|}\,dA(z)+f(0)\overline{g(0)}\right)
	(1-r^2)^{\frac1p-1}r\,dr\right|
    \\
	&\le2\int_\mathbb{D}|f'(\z)|g'(\z)|\left(\int_{|\z|}^1\log\frac{r}{|\z|}(1-r^2)^{\frac1p-1}r\,dr\right)\,dA(\z)+|f(0)||g(0)|.
	\end{split}
	\end{equation*}
The inequality $-\log t\le\frac1t(1-t)$, valid for all $0<t\le1$, now gives
	$$
	\int_{|\z|}^1\log\frac{r}{|\z|}(1-r^2)^{\frac1p-1}r\,dr
	\le2^\frac1p\log\frac{1}{|\z|}\int_{|\z|}^1(1-r)^{\frac1p-1}\,dr
	\le p2^\frac1p\frac{(1-|\z|)^{1+\frac1p}}{|\z|},\quad \z\in\mathbb{D}\setminus\{0\}.
	$$
By using this and the continuous embedding $A^{p}_{p-1}\subset A^1_{\frac1p-1}$, valid for $0<p\le1$ by \cite[Theorem~1]{L2}, we deduce
	\begin{equation*}
	\begin{split}
	|\langle f,g\rangle_{A^2_{\frac1p-1}}|
	&\le p2^\frac1p\int_\mathbb{D}|f'(\z)||g'(\z)|\frac{(1-|\z|)^{1+\frac1p}}{|\z|}\,dA(\z)+|f(0)||g(0)|\\
	&\lesssim\|g\|_{\mathcal{B}^2}\int_\mathbb{D}|f'(\z)|(1-|\z|)^{\frac1p-1}\,dA(\z)+|f(0)||g(0)|
	\lesssim\|g\|_{\mathcal{B}^2}\|f\|_{D^p_{p-1}},
	\end{split}
	\end{equation*}
and hence each $g\in\mathcal{B}^2$ induces a bounded linear functional on $D^p_{p-1}$ via the $A^2_{\frac1p-1}$-pairing.

Let $L\in(D^p_{p-1})^\star$, and recall that $\mathcal{I}(F)(z)=\int_0^zF(\zeta)\,d\zeta$. Then $|(L\circ\mathcal{I})(F)|\lesssim\|\mathcal{I}(F)\|_{D^p_{p-1}}=\|F\|_{A^p_{p-1}}$ for all $F\in A^p_{p-1}$. Therefore $L\circ\mathcal{I}\in(A^p_{p-1})^\star$. Since $(A^p_{p-1})^\star$ is isomorphic to the Bloch space via the $A^2_{\frac1p-1}$-pairing by \cite[Theorem~A]{Zhu2}, there exists a unique $G\in\mathcal{B}$ such that $\|G\|_\mathcal{B}\lesssim\|L\circ\mathcal{I}\|\lesssim\|L\|$ and $(L\circ\mathcal{I})(F)=\langle F,G\rangle_{A^2_{\frac1p}-1}$ for all $F\in A^p_{p-1}$. For each $f\in D^p_{p-1}$, there exists $F=F_f\in A^p_{p-1}$ such that $\mathcal{I}(F)=f-f(0)$. Therefore
	\begin{equation*}
	\begin{split}
	L(f)&=L(f-f(0)+f(0))=L(f-f(0))+f(0)L(1)\\
	&=L(\mathcal{I}(F))+f(0)L(1)
	=\langle F,G\rangle_{A^2_{\frac1p-1}}+f(0)L(1)\\
	&=\langle f',G\rangle_{A^2_{\frac1p-1}}+f(0)L(1),\quad f\in D^p_{p-1}.
	\end{split}
	\end{equation*}
By denoting
	$$
	w_{n,p}=\int_0^1r^{2n+1}(1-r^2)^{\frac1p-1}\,dr,\quad n\in\mathbb{N}\cup\{0\},
	$$
we deduce
	\begin{equation*}
	\begin{split}
	\langle f',G\rangle_{A^2_{\frac1p-1}}
	&=\lim_{r\to 1^-}\int_\mathbb{D} rf'(rz)\overline{G(z)}(1-|z|^2)^{\frac1p-1}\,dA(z)\\
	&= \lim_{r\to 1^-}\int_\mathbb{D} \left(\sum_{n=0}^\infty\widehat{f}(n+1)(n+1)r(rz)^n \right) \left(\sum_{k=0}^\infty\overline{\widehat{G}(k)}\overline{z}^k
    \right)(1-|z|^2)^{\frac1p-1}\,dA(z)\\
	&= \lim_{r\to 1^-}2\pi\sum_{n=0}^\infty\widehat{f}(n+1)r^{n+1}\left(\frac{w_{n,p}}{w_{n+1,p}}(n+1)\overline{\widehat{G}(n)}\right)w_{n+1,p}\\
	&=\lim_{r\to 1^-}\int_\mathbb{D} \sum_{n=0}^\infty\widehat{f}(n+1)(rz)^{n+1}
	\sum_{k=0}^\infty\left(\frac{w_{k,p}}{w_{k+1,p}}(k+1)\overline{\widehat{G}(k)}\right)\overline{z}^{k+1}(1-|z|^2)^{\frac1p-1}\,dA(z)\\
	&=\langle f-f(0),K_p\rangle_{A^2_{\frac1p-1}},
	\end{split}
	\end{equation*}
where
	$$
	K_p(z)=\sum_{k=0}^\infty\frac{w_{k,p}}{w_{k+1,p}}(k+1)\widehat{G}(k) z^{k+1},\quad z\in\mathbb{D}.
	$$
In the case $p=1$ we have
	$$
	K_1(z)=\sum_{k=0}^\infty(k+2)\widehat{G}(k) z^{k+1}
	=\frac{d}{dz}\left(z^2G(z)\right),\quad z\in\mathbb{D},
	$$
and hence $K_1\in\mathcal{B}^2$ by \eqref{Eq-alpha-bloch}. To obtain the same conclusion for each $0<p<1$ we first observe that
$J(z)=\sum_{k=0}^\infty(k+1)\widehat{G}(k)z^{k+1}=\frac{d}{dz}\left(zG(z)\right)$, and thus $J\in\mathcal{B}^2$. Therefore it suffices to show that
$\lambda_p(z)=\sum_{n=0}^\infty\frac{w_{n,p}}{w_{n+1,p}}z^{n+1}$ is a coefficient multiplier of $\mathcal{B}^2$ for each $0<p\le 1$.

To see this, for each $\beta\in\mathbb{N}$, denote $D^\beta f(z)=\sum_{n=0}^\infty (n+1)^\beta\widehat{f}(n)z^n$ for all
 $f\in\H(\mathbb{D})$, and for simplicity write $Df$ instead of $D^1f$. We claim that
\begin{equation}\label{eq:M_1}
M_1(r,D\lambda_p)\lesssim\frac1{1-r},\quad 0\le r<1,
\end{equation}
the proof of which is postponed for a moment. Direct calculations show that
	$$
	\|f\|_{\mathcal{B}^2}\asymp\sup_{z\in\mathbb{D}}|Df(z)|(1-|z|)^2
	\asymp\sup_{z\in\mathbb{D}}|D^2f(z)|(1-|z|)^3,\quad f\in\H(\mathbb{D}),
	$$
and hence \eqref{eq:M_1} yields
	\begin{equation*}
	\begin{split}
	|D^2(f*\lambda_p)(r^2e^{it})|
	&=|(Df*D\lambda_p)(r^2e^{it})|
	=\left|\frac1{2\pi}\int_0^{2\pi}Df(re^{i(t+\theta)})D\lambda_p(re^{-i\theta})\,d\theta\right|\\
	&\le M_\infty(r,Df)M_1(r,D\lambda_p)\lesssim\frac{M_1(r,D\lambda_p)}{(1-r)^2}\lesssim\frac{1}{(1-r)^3},\quad f\in\mathcal{B}^2.
	\end{split}
	\end{equation*}
It follows that $f*\lambda_p\in\mathcal{B}^2$ for all $f\in\mathcal{B}^2$ and $0<p\le1$. Thus $K_p\in\mathcal{B}^2$ for each $0<p\le1$. By choosing $H_p=K_p+\frac{\overline{L(1)}}{\omega_{0,p}}\in\mathcal{B}^2$, we deduce $L(f)=\langle f,H_p\rangle_{A^2_{\frac1p-1}}$ for all $f\in D^p_{p-1}$.

To complete the proof, it remains to establish~\eqref{eq:M_1}. To do this, we will use the families of polynomials defined by \eqref{eq:polynomials} and \eqref{vn}. It follows from \eqref{propervn} that
    \begin{equation}\label{fbnorm}
    M_1(r,D\lambda_p)=\|(D\lambda_p)_r\|_{H^1}\le C(p)+\sum_{n=2}^\infty\|V_n\ast(D\lambda_p)_r\|_{H^1},
    \end{equation}
where $(D\lambda_p)_r(z)=\sum_{n=1}^\infty n\frac{w_{n-1,p}}{w_{n,p}}r^{n}z^{n}$. Next, for each $n\in\mathbb{N}\setminus\{1\}$ and
$r\in\left[\frac12,1\right)$, consider
    $$
    F_n(x)=x\frac{w_{x-1,p}}{w_{x,p}}r^{x}\chi_{[2^{n-1},2^{n+1}]}(x),\quad x\in\mathbb{R}.
    $$
Since for each radial weight $\nu$ there exists a constant $C=C(\nu)>0$ such that
    \begin{equation*}
    \int_0^1 s^x\left(\log\frac1s\right)^n\nu(s)\,ds\le C\int_0^1s^x\nu(s)\,ds, \quad n\in\{1,2\},\quad x\ge 2,
    \end{equation*}
it follows by a direct calculation that
    $$
    |F_n''(x)|\le C|F_n(x)|, \quad n\in\mathbb{N}\setminus\{1\}, \quad r\in\left[\frac12,1\right),\quad x\ge2,
    $$
for some constant $C=C(\om)>0$. Therefore,
    \begin{equation*}
    \begin{split}
    A_{F_n,2}&=\max_{x\in[2^{n-1},2^{n+1}]}|F_n(x)|+\max_{x\in[2^{n-1},2^{n+1}]}|F_n''(x)|
		\lesssim \max_{x\in[2^{n-1},2^{n+1}]}|F_n(x)|\\
    &
    \lesssim \max_{x\in[2^{n-1},2^{n+1}]} (x+1) r^{x+1}
    \lesssim 2^{n} r^{2^{n-1}},\quad n\in\mathbb{N}\setminus\{1\}.
    \end{split}
    \end{equation*}
For each $n\in\mathbb{N}\setminus\{1\}$, choose a $C^\infty$-function $\Phi_n$ with compact support contained in $[2^{n-2},2^{n+2}]$ such that
$\Phi_n=F_n$ on $[2^{n-1},2^{n+1}]$ and
    \begin{equation}\label{phin}
    A_{\Phi_n,2}=\max_{x\in\mathbb{R}}|\Phi_n(x)|+\max_{x\in\mathbb{R}}|\Phi''_n(x)|
    \lesssim 2^{n} r^{2^{n-1}},\quad n\in\mathbb{N}\setminus\{1\}.
    \end{equation}
Since
    $$
    W_1^{\Phi_n}(z)
    =\sum_{k\in\mathbb{Z}}\Phi_n\left(k\right)z^k
    =\sum_{k\in\mathbb{Z}\cap[2^{n-2},2^{n+2}]}\Phi_n\left(k\right)z^k,
    $$
the identity \eqref{vn} yields
    \begin{equation*}
    \begin{split}
    \left(V_n\ast(D\lambda_p)_r\right)(z)
    &=\sum_{k=2^{n-1}}^{2^{n+1}-1}\psi\left(\frac{k}{2^{n-1}}\right)k\frac{w_{k-1,p}}{w_{k,p}}r^{k}z^k
    =\sum_{k=2^{n-1}}^{2^{n+1}-1}\psi\left(\frac{k}{2^{n-1}}\right)\Phi_n(k)z^k\\
    &=\left(W_1^{\Phi_n}\ast V_n\right)(z),\quad n\in\mathbb{N}\setminus\{1\}.
    \end{split}
    \end{equation*}
This together with Theorem~\ref{th:cesaro}, \eqref{phin} and \eqref{propervn} implies
    \begin{equation*}
    \begin{split}
    \|V_n\ast(D\lambda)_r\|_{H^1}
    &=\|W_1^{\Phi_n}\ast V_n\|_{H^1}
    \lesssim A_{\Phi_n,2}\|V_n\|_{H^1}
    \lesssim  2^{n} r^{2^{n-1}}\|V_n\|_{H^1}\\
    &\lesssim 2^{n} r^{2^{n-1}},\quad r\in \left[\frac12,1\right),\quad n\in\mathbb{N}\setminus\{1\},
    \end{split}
    \end{equation*}
which combined with \eqref{fbnorm} gives
    \begin{equation}\label{endstep1new1}
    M_1(r,D\lambda_p)

    \lesssim\sum_{n=2}^\infty2^{ n}r^{2^{n-1}}
    \lesssim\frac{1}{(1-r)}, \quad  r\in\left[\frac12,1\right).
    \end{equation}
This implies \eqref{eq:M_1}, and finishes the proof.
\end{proof}

For two Banach spaces $X,Y\subset\H(\mathbb{D})$, let $[X,Y]=\{g\in\H(\mathbb{D}): f*g\in Y \textrm{ for every }f\in X\}$ denote the space of coefficient multipliers from $X$ to $Y$. The dual space of the Banach space $\mathord{\rm HL}_p$, with $1<p<\infty$, with respect to the $H^2$-pairing is given in the next lemma. This result is needed in the proof of Theorem~\ref{Theorem:Hardy-p<2}.

\begin{lemma}\label{le:dualHLp}
Let $1<p<\infty$. Then $(\mathord{\rm HL}_p)^\star\simeq\mathord{\rm HL}_{p'}$ via the $H^2$-pairing with equivalence of norms.
\end{lemma}

\begin{proof}
Let $\mathcal{A}$ denote the disc algebra. By \cite[Proposition~1.3]{P}, the dual of $\mathord{\rm HL}_p$ can be identified with $[\mathord{\rm HL}_p, \mathcal{A}]$ under the $H^2$-pairing. We claim that $[\mathord{\rm HL}_p,\mathcal{A}]=\mathord{\rm HL}_{p'}$ for each $1<p<\infty$.

To see this, first observe that for all $g\in\mathord{\rm HL}_{p'}$ and $f\in\mathord{\rm HL}_p$, H\"{o}lder's inequality yields
	$$
	\sup_{z\in\overline{\mathbb{D}}}|(f*g)(z)|
	\le\sum_{n=0}^{\infty}|\widehat{f}(n)||\widehat{g}(n)|
	\le\|f\|_{\mathord{\rm HL}_p}\|g\|_{\mathord{\rm HL}_{p'}}<\infty,
	$$
which shows that $\mathord{\rm HL}_{p'}\subset[\mathord{\rm HL}_p, \mathcal{A}]$.

Conversely, if $g(z)=\sum_{n=0}^{\infty}\widehat{g}(n)z^n\in[\mathord{\rm HL}_p,\mathcal{A}]$, then
	$$\infty>\sup_{z\in\overline{\mathbb{D}}}\left|\sum_{n=0}^{\infty}\widehat{f}(n)\widehat{g}(n)z^n \right|
	\ge
	\left|\sum_{n=0}^{\infty}\widehat{f}(n)(n+1)^{\frac{p-2}{p}}\widehat{g}(n)(n+1)^{\frac{p'-2}{p'}}\right|,\quad f\in\mathord{\rm HL}_p.
	$$
Since $f\in\mathord{\rm HL}_p$, that is, $\{\widehat{f}(n)(n+1)^{\frac{p-2}{p}}\}\in\ell^p$, the duality $(\ell^p)^\star\simeq\ell^{p'}$ yields $\{\widehat{g}(n)(n+1)^{\frac{p'-2}{p'}}\}\in\ell^{p'}$. Thus $g\in\mathord{\rm HL}_{p'}$, and the proof is finished.
\end{proof}

Recall that the space $\mathord{\rm HL}_{\infty}$ consists of $f\in\H(\mathbb{D})$ such that
	$$
	\|f\|_{\mathord{\rm HL}_{\infty}}=\sup_{n\in\mathbb{N}\cup\{0\}}\left(|\widehat{f}(n)|(n+1)\right)<\infty.
	$$
The last lemma of the section describes $(\mathord{\rm HL}_1)^\star$. It will be used in the proof of Theorem~\ref{Theorem:p=1}.

\begin{lemma}\label{le:dualHL1}
$(\mathord{\rm HL}_1)^\star\simeq\mathord{\rm HL}_{\infty}$ via the $H^2$-pairing with equivalence of norms.
\end{lemma}

\begin{proof}
By \cite[Proposition~1.3]{P}, the dual of $\mathord{\rm HL}_1$ can be identified with $[\mathord{\rm HL}_1,\mathcal{A}]$ under the $H^2$-pairing. We claim that $[\mathord{\rm HL}_1,\mathcal{A}]=\mathord{\rm HL}_{\infty}$.

On one hand,
	$$
	\sup_{z\in\overline{\mathbb{D}}}|(f*g)(z)|
	\le\sum_{n=0}^{\infty}\frac{|\widehat{f}(n)|}{n+1}\left(|\widehat{g}(n)|(n+1)\right)
	\le\|f\|_{\mathord{\rm HL}_1}\|g\|_{\mathord{\rm HL}_{\infty}},
	$$
and hence $\mathord{\rm HL}_{\infty}\subset[\mathord{\rm HL}_1,\mathcal{A}]$.

On the other hand, let $g(z)=\sum_{n=0}^{\infty}\widehat{g}(n)z^n$ be a member of $[\mathord{\rm HL}_1,\mathcal{A}]$. Then
	$$
	\infty>\sup_{z\in\overline{\mathbb{D}}}\left|\sum_{n=0}^{\infty}\widehat{f}(n)\widehat{g}(n)z^n\right|
	\ge\left|\sum_{n=0}^{\infty}\frac{\widehat{f}(n)}{n+1}\left(\widehat{g}(n)(n+1)\right)\right|,\quad f\in\mathord{\rm HL}_1.
	$$
Since $f\in\mathord{\rm HL}_1$ if and only if $\{\frac{\widehat{f}(n)}{n+1}\}_{n=0}^\infty\in\ell^1$, the duality $(\ell^1)^\star\simeq\ell^{\infty}$ yields $g\in\mathord{\rm HL}_{\infty}$. Thus $\mathord{\rm HL}_{\infty}=[\mathord{\rm HL}_1,\mathcal{A}]$.
\end{proof}

\section{Hardy, Hardy-Littlewood and Dirichlet-type spaces with $1<p<\infty$}\label{Sec:pili}

The main  aim of this section is to prove Theorem~\ref{Theorem:Hardy-p<2}. To do that some notation and auxiliary results are needed. For each $g\in\H(\mathbb{D})$, with Maclaurin series expansion $g(z)=\sum_{k=0}^\infty \widehat{g}(k)z^k$, consider the dyadic polynomials defined by $\Delta_0 g (z)=g(0)$ and $\Delta_n g (z)=\sum_{k=2^{n}}^{2^{n+1}-1}\widehat{g}(k)z^k$ for all $n\in\mathbb{N}$ and $z\in\mathbb{D}$. Then, obviously, $g=\sum_{n=0}^\infty\Delta_n g$. Further, write $\Delta_0=1$ and $\Delta_n(z)= \sum_{k=2^{n}}^{2^{n+1}-1}z^k$ for all $n\in\mathbb{N}$ and $z\in\mathbb{D}$. Then \cite[Lemma~2.7]{CPPR} shows that
	\begin{equation}\label{Deltanorm}
	\|\Delta_n\|_{H^p}\asymp 2^{\frac{n}{p'}},\quad 1<p<\infty,\quad n\in \mathbb{N}\cup\{0\}.
	\end{equation}
For $a\in\mathbb{D}$, denote $f_a(z)=f(az)$ for all $z\in\mathbb{D}$. With these preparations we can state the first auxiliary result.

\begin{proposition}\label{pr:j1}
Let $1<q<\infty$ and $g\in \H(\mathbb{D})$ such that $\sum_{k=0}^\infty |\widehat{g}(k)|<\infty$. Then there exists a constant $C=C(q)>0$ such that
	\begin{equation}
	\begin{split}\label{eq:j1}
	\left\| \Delta_j f\ast G^{H^2}_{g,z} \right\|_{H^q}
	&=\left\|\sum_{k=2^j}^{2^{j+1}-1}\widehat{f}(k)\sum_{n=0}^\infty\frac{(n+1)\overline{\widehat{g}(n+1)}\overline{z}^{n+k+1} }{n+k+1}\z^k	\right\|_{H^q}\\
	&\le C\left\| \Delta_j f_{\overline{z}}  \right\|_{H^q}  \sum_{n=0}^\infty \frac{(n+1) |\widehat{g}(n+1)| |z|^{n+1} }{n+ 2^{j-1}+1},\quad z\in\mathbb{D},
	\end{split}
	\end{equation}
for all $f\in\H(\mathbb{D})$ and $j\in\mathbb{N}$.
\end{proposition}

\begin{proof}
For each $j\in\mathbb{N}$ and $z\in\mathbb{D}$, let us consider the $C^\infty$-function
	$$
	\Psi_{2^j,z}(s)=\sum_{n=0}^\infty \frac{(n+1)\overline{\widehat{g}(n+1)}\overline{z}^{n+1} }{n+1+2^{j}s},\quad s>0.
	$$
Then
\begin{equation}\label{eq:j2}
|\Psi_{2^j,z}(s)|\le \sum_{n=0}^\infty \frac{(n+1)|\widehat{g}(n+1)z^{n+1}| }{n+1+2^{j-1}},\quad s\ge \frac{1}{2}.
\end{equation}
Further,
	$$
	(\Psi_{2^j,z})'(s)=-2^j\sum_{n=0}^\infty \frac{(n+1)\overline{\widehat{g}(n+1)}\bar{z}^{n+1} }{(n+1+2^{j}s)^2},\quad s>0,$$
and hence
	\begin{equation}
	\begin{split}\label{eq:j3}
	|(\Psi_{2^j,z})'(s)|
	&\le2^j\sum_{n=0}^\infty\frac{(n+1)|\widehat{g}(n+1)z^{n+1}|}{(n+1+2^{j-1})^2}\\
	&\le2\sum_{n=0}^\infty\frac{(n+1)|\widehat{g}(n+1)z^{n+1}|}{n+1+2^{j-1}},\quad s\ge\frac{1}{2}.
	\end{split}
	\end{equation}
Therefore, by using \eqref{eq:j2} and \eqref{eq:j3}, we can find a $C^\infty$-function $\Phi_{2^j,z}$ and an absolute constant $C>0$ such that $\supp\Phi_{2^j,z}\subset\left( \frac{1}{2},4\right)$, $\Phi_{2^j,z}(s)=\Psi_{2^j,z}(s)$ for all $s\in [1,2]$ and
	$$
  A_{\Phi_{2^j,z},1}
	=\max_{s\in\mathbb{R}}|\Phi_{2^j,s}(s)|
	+\max_{s\in\mathbb{R}}|\Phi_{2^j,z}'(s)|
	\le C\sum_{n=0}^\infty \frac{(n+1)|\widehat{g}(n+1)z^{n+1}|}{n+1+2^{j-1}}.
  $$
Hence
	\begin{equation*}
	\begin{split}
	\Delta_j f\ast G^{H^2}_{g,z}(\z)
	&=\sum_{k=2^j}^{2^{j+1}-1}\widehat{f}(k)\overline{z}^{k}
	\sum_{n=0}^\infty \frac{(n+1)\overline{\widehat{g}(n+1)}\bar{z}^{n+1} }{n+k+1} \z^k\\
	&=\sum_{k=2^j}^{2^{j+1}-1}\widehat{f}(k)\overline{z}^{k}\Phi_{2^j,z}\left(\frac{k}{2^j}\right)\z^k
	=\left(\Delta_jf_{\overline{z}}\ast W^{\Phi_{2^j,z}}_{2^j}\right)(\z),\quad j\in\mathbb{N}.
	\end{split}
	\end{equation*}
Using now Theorem~\ref{th:cesaro} we find a constant $C=C(q)>0$ such that
	\begin{equation*}
	\begin{split}
	\left\|\Delta_j f\ast G^{H^2}_{g,z}(\z) \right\|_{H^q}
	&=\left\| \Delta_j f_{\bar{z}}\ast W^{\Phi_{2^j,z}}_{2^j}\right\|_{H^q}
	\le CA_{\Phi_{2^j,z},1}\left\|\Delta_j f_{\overline{z}}\right\|_{H^q}\\
	&\le C\left\| \Delta_j f_{\overline{z}}\right\|_{H^q} \sum_{n=0}^\infty \frac{(n+1)|\widehat{g}(n+1)z^{n+1}|}{n+1+2^{j-1}}.
	\end{split}
	\end{equation*}
This finishes the proof.
\end{proof}

The next result gives a sufficient condition for $T_g:~H^p\rightarrow H^\infty$ to be bounded and establishes the operator norm estimate \eqref{poi} announced in the introduction.

\begin{theorem}\label{th:dpsuf}
Let $1<p<\infty$ and $g\in\H(\mathbb{D})$ such that $$\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)|\widehat{g}(n+1)|}{n+k+1}\right)^{p'}<\infty.$$
If $X_p\in\{H^p,D^p_{p-1},\mathord{\rm HL}_p\}$,  then $T_g:~X_p\rightarrow H^\infty$ is bounded and
	$$
	\|T_g\|_{ X_p \rightarrow H^\infty}^{p'}
  \lesssim
  \sum_{k=0}^{\infty}(k+1)^{p'-2} \left(\sum_{n=0}^{\infty}\frac{(n+1)|\widehat{g}(n+1)|}{n+k+1}\right)^{p'}.
	$$
\end{theorem}

\begin{proof}
We begin with the case $X_p=D^p_{p-1}$. By Lemma~\ref{le:dualdp} and \cite[Theorem~2.2]{CPPR}, $T_g:~D^p_{p-1}\rightarrow H^\infty$ is bounded if and only if $\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{ D^{p'}_{p'-1}}<\infty$, and moreover,
	\begin{equation}\label{eq:j4}
	\|T_g\|_{D^p_{p-1}\rightarrow H^\infty}^{p'}
	\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|^{p'}_{D^{p'}_{p'-1}}.
  \end{equation}
Further, for each $1<q<\infty$, \cite[Theorem~2.1]{MatPav} yields
	\begin{equation}\label{eq:j5}
	\|F\|^q_{D^q_{q-1}}\asymp\sum_{j=0}^\infty\| \Delta_j\ast F\|^q_{H^q},\quad F\in\H(\mathbb{D}).
	\end{equation}
Therefore, by combining \eqref{eq:j4}, \eqref{eq:j5}, Proposition~\ref{pr:j1} and \eqref{Deltanorm}, we deduce
	\begin{equation}
	\begin{split}\label{eq:TgDp}
	\|T_g\|_{D^p_{p-1}\rightarrow H^\infty}^{p'}
	&\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|^{p'}_{D^{p'}_{p'-1}}
	\asymp\sup_{z\in\mathbb{D}}\sum_{j=0}^\infty \|\Delta_j \ast  G^{H^2}_{g,z}\|_{H^{p'}}^{p'}\\
	&\lesssim\sup_{z\in\mathbb{D}}\sum_{j=0}^\infty
	\|\Delta_j\|_{H^{p'}}^{p'}\left(\sum_{n=0}^\infty\frac{(n+1)|\widehat{g}(n+1)| |z|^{n+1} }{n+ 2^{j-1}+1}\right)^{p'}\\
	&\lesssim\sum_{j=0}^\infty 2^{j(p'-1)} \left( \sum_{n=0}^\infty \frac{(n+1) |\widehat{g}(n+1)|  }{n+ 2^{j-1}+1}\right)^{p'}
	\\
	&\lesssim \sum_{k=0}^{\infty}(k+1)^{p'-2} \left( \sum_{n=0}^\infty \frac{(n+1) |\widehat{g}(n+1)|  }{n+ k+1}  \right)^{p'}.
	\end{split}
	\end{equation}
Thus the assertion is proved for $X_p=D^p_{p-1}$.

Next we deal with the case $X_p=\mathord{\rm HL}_p$. By Lemma~\ref{le:dualHLp} and \cite[Theorem~1.1]{CPPR}, $T_g:\mathord{\rm HL}_p\rightarrow H^\infty$ is bounded if and only if $\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{\mathord{\rm HL}_{p'}}<\infty$, and moreover,
	\begin{equation}\label{eq:j4+}
	\|T_g\|_{\mathord{\rm HL}_p\rightarrow H^\infty}^{p'}
	\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|^{p'}_{\mathord{\rm HL}_{p'}}.
  \end{equation}
But
	\begin{equation*}
	\begin{split}
	\|G^{H^2}_{g,z}\|^{p'}_{\mathord{\rm HL}_{p'}}
	&=\sum_{k=0}^{\infty}(k+1)^{p'-2}\left|\sum_{n=0}^\infty\frac{(n+1)\overline{\widehat{g}(n+1)}\bar{z}^{n+k+1} }{n+k+1}\right|^{p'}\\
  &\le \sum_{k=0}^{\infty}(k+1)^{p'-2} \left( \sum_{n=0}^\infty \frac{(n+1) |\widehat{g}(n+1)|  }{n+ k+1}  \right)^{p'},
	\end{split}
	\end{equation*}
and thus $T_g:\mathord{\rm HL}_p\rightarrow H^\infty$ is bounded and
	$$
	\|T_g\|_{\mathord{\rm HL}_{p}\rightarrow H^\infty}^{p'}\le\sum_{k=0}^{\infty}
   (k+1)^{p'-2} \left( \sum_{n=0}^\infty \frac{(n+1) |\widehat{g}(n+1)|  }{n+ k+1}  \right)^{p'}.
	$$
Bearing in mind \eqref{Hardy1} and \eqref{Hardy2}, the remaining case $X_p=H^p$ follows from \cite[Theorem~1.1]{CPPR}, the well-known identification $(H^p)^\star\simeq H^{p'}$ via the $H^2$-pairing and the two cases already proven.
\end{proof}

Despite the inclusions in \eqref{Hardy1} and \eqref{Hardy2} are strict unless $p=2$, if one restricts to the class of power series with non-negative decreasing coefficients, then the following statements hold by \cite{HD}, \cite{P2} and \cite[Chapter~XII, Lemma~6.6]{Z}.

\begin{letterlemma}\label{Eq1}
Let $1\le p<\infty$, then there exist constants $C_1=C_1(p)>0$, $C_2=C_2(p)>0$ and $C_3=C_3(p)>0$ such that
	$$
	\|f\|^p_{H^p}
	\le C_1\|f\|^p_{D^p_{p-1}}
	\le C_2 \|f\|^p_{\mathord{\rm HL}_p}
	\le C_3 \|f\|^p_{H^p},
	$$
for all $f\in\H(\mathbb{D})$ such that its Maclaurin coefficients $\{\widehat{f}(n)\}_{n=0}^\infty$ form a sequence of non-negative numbers decreasing to zero. In particular,
	$$
	f\in H^p\quad\Longleftrightarrow\quad
	f\in D^p_{p-1}\quad\Longleftrightarrow\quad
	f\in\mathord{\rm HL}_p
	$$
for every such $f$.
\end{letterlemma}

We are now ready to prove Theorem~\ref{Theorem:Hardy-p<2}.

\medskip

\noindent{\emph {Proof of}} Theorem~\ref{Theorem:Hardy-p<2}. Assume first that $T_g:~X_p\rightarrow H^\infty$ is bounded. Then $g\in H^\infty$, and hence
	\begin{equation}\label{eq:j10}
	\sum_{n=0}^\infty |\widehat{g}(n)|=\sum_{n=0}^\infty\widehat{g}(n)<\infty.
	\end{equation}
Lemmas~\ref{le:dualdp} and~\ref{le:dualHLp} together with the well-known identification of $(H^p)^\star$ as $H^{p'}$ via the $H^2$-pairing imply $(X_p)^\star\simeq X_{p'}$. Therefore \cite[Theorem~1.1]{CPPR} yields
	\begin{equation}\label{q1}
	\begin{split}
	\|T_g\|_{X_p\rightarrow H^\infty}^{p'}
	&\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{X_{p'}}^{p'}
     \gtrsim \sup_{x\in (0,1)}\|G^{H^2}_{g,x}\|_{X_{p'}}^{p'}.
	\end{split}
	\end{equation}
Since $G^{H^2}_{g,x}(\z)=\sum_{k=0}^\infty \left(\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)x^{n+k+1}}{n+k+1} \right)\z^k$, for each $x\in(0,1)$, the Maclaurin coefficients
	$$
	\widehat{G^{H^2}_{g,x}}(k)=\sum_{n=0}^\infty \frac{(n+1)\widehat{g}(n+1)x^{n+k+1}}{n+k+1}, \quad k\in\mathbb{N}\cup\{0\},
	$$
form a sequence of non-negative and decreasing numbers. Therefore \eqref{q1}, Lemma~\ref{Eq1} and \eqref{eq:j10} imply
	\begin{equation*}
	\begin{split}
	\|T_g\|_{X_p\rightarrow H^\infty}^{p'}&\gtrsim
    \sup_{x\in (0,1)}\|G^{H^2}_{g,x}\|_{\mathord{\rm HL}_{p'}}^{p'}
	\asymp\sup_{x\in(0,1)}\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}x^{n+k+1}\right)^{p'}\\
	&\asymp\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}.
	\end{split}
	\end{equation*}
Thus \eqref{pipeli} holds.

Conversely, if \eqref{pipeli} is satisfied, then $T_g:~X_p\rightarrow H^\infty$ is bounded and
	$$
	\|T_g\|_{X_p\rightarrow H^\infty}^{p'}
	\lesssim \sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}
	$$
by Theorem~\ref{th:dpsuf}. The norm estimate \eqref{eq:j11} follows from the above inequalities.

To complete the proof we still need to show that $T_g:~X_p\rightarrow H^\infty$ is in fact compact if \eqref{pipeli} is satisfied. To see this, let first $X_p=\mathord{\rm HL}_p$. Further, let $\{f_n\}$ such that $\sup_{n}\|f_n\|_{\mathord{\rm HL}_p}<\infty$
and $f_n\to0$ uniformly on compact subsets of $\mathbb{D}$ as $n\to\infty$.
For each $\varepsilon>0$ there exists $k_0=k_0(\varepsilon)\in\mathbb{N}$ such that
	\begin{equation*}\label{eq:j16}
	\sum_{k=k_0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}<\varepsilon^{p'}.
	\end{equation*}
Moreover, by the uniform convergence we may pick up an $n_0=n_0(\varepsilon)\in\mathbb{N}$ such that
	\begin{equation*}
	\sup_{n\ge n_0}\sum_{k=0}^{k_0-1} (k+1)^{p-2}|\widehat{f_n}(k)|^p<\varepsilon^{p}.
	\end{equation*}
Then, H\"older's inequality yields
	\begin{equation*}
	\begin{split}
	\|T_g(f_n)\|_{H^\infty}
	&=\sup_{z\in\mathbb{D}}|\langle f_n,G^{H^2}_{g,z}\rangle_{H^2}|
	\le\sum_{k=0}^\infty |\widehat{f_n}(k)|\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)\\
	&=\sum_{k=0}^{k_0-1} |\widehat{f_n}(k)|\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)
	+\sum_{k=k_0}^\infty |\widehat{f_n}(k)|\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)\\
	&\le\left(\sum_{k=0}^{k_0-1}(k+1)^{p-2} |\widehat{f_n}(k)|^p \right)^{\frac1p}
 \left(\sum_{k=0}^{k_0-1}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'} \right)^{\frac1{p'}}\\
	&\quad+
 \left(\sum_{k=k_0}^{\infty}(k+1)^{p-2} |\widehat{f_n}(k)|^p  \right)^{\frac1p} \left(\sum_{k=k_0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'} \right)^{\frac1{p'}}\\
	&\le\varepsilon\left(\left(\sum_{k=0}^{\infty}(k+1)^{p'-2}
	\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}\right)^{1/p'}+ \sup_{n}\|f_n\|_{\mathord{\rm HL}_p}\right)\\
	&\lesssim \varepsilon,\quad n\ge n_0,
	\end{split}
	\end{equation*}
and hence $\lim_{n\to \infty}\|T_g(f_n)\|_{H^\infty}=0$. Therefore $T_g:\mathord{\rm HL}_p\to H^\infty$ is compact by \cite[Lemma~3.6]{Tjani}.

Let now $X_p=D^p_{p-1}$. We first show that
	\begin{equation}\label{eq:j16}
	\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_{\mathbb{D}\backslash\overline{\mathbb{D}(0,R)}}|(G^{H^2}_{g,z})'(w)|^{p'}(1-|w|)^{p'-1}\,dA(w)=0,
	\end{equation}
and then we use this fact to prove the compactness of $T_g:D^p_{p-1} \to H^\infty$.

If $2<p'<\infty$, then \eqref{Hardy2} and Fubini's theorem yield
	\begin{equation*}
	\begin{split}
	&\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_{\mathbb{D}\backslash\overline{\mathbb{D}(0,R)}}|(G^{H^2}_{g,z})'(w)|^{p'}(1-|w|)^{p'-1}dA(w)\\
	&=\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_R^1\int_0^{2\pi}\left|\sum_{k=0}^{\infty}(k+1)
	\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)z^{n+k+2}}{n+k+2}r^ke^{ik\theta}\right|^{p'} d\theta (1-r)^{p'-1}r\,dr\\
	&=2\pi\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_R^1 \| (G^{H^2}_{g,z})'_r\|^{p'}_{H^{p'}} (1-r)^{p'-1}r\,dr\\
	&\lesssim\lim_{R\rightarrow1^-}\int_R^1\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}
	(k+1)^{p'}r^{kp'+1}(1-r)^{p'-1}dr\\
	&=\lim_{R\rightarrow1^-}\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}
	(k+1)^{p'}\int_R^1r^{kp'+1}(1-r)^{p'-1}dr,
	\end{split}
	\end{equation*}
where
	$$
	(k+1)^{p'}\int_R^1r^{kp'+1}(1-r)^{p'-1}dr\leq(k+1)^{p'}\int_0^1r^{kp'+1}(1-r)^{p'-1}dr \asymp 1,\quad k\in\mathbb{N}\cup\{0\}.
	$$
The dominated convergence theorem now implies \eqref{eq:j16}.

If $1<p'\le2$, then Proposition~\ref{pr:j1} and an argument similar to that used in the proof of \eqref{eq:TgDp} allows us to find a constant $C=C(p)>0$ such that
	$$
	\| (G^{H^2}_{g,z})'_r\|^{p'}_{D^{p'}_{p'-1}}\le C\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}
	(k+1)^{p'}r^{kp'},\quad 0\le r<1.
	$$
This together with \eqref{Hardy1} implies
  \begin{equation*}
	\begin{split}
	&\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_{\mathbb{D}\backslash\overline{\mathbb{D}(0,R)}}|(G^{H^2}_{g,z})'(w)|^{p'}(1-|w|)^{p'-1}dA(w)\\

     & =2\pi\lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_R^1 \| (G^{H^2}_{g,z})'_r\|^{p'}_{H^{p'}} (1-r)^{p'-1}r\,dr
     \\ & \lesssim  \lim_{R\rightarrow1^-}\sup_{z\in\mathbb{D}}\int_R^1\| (G^{H^2}_{g,z})'_r\|^{p'}_{D^{p'}_{p'-1}}(1-r)^{p'-1}r\,dr
    \\ &\lesssim\lim_{R\rightarrow1^-}\int_R^1\sum_{k=0}^{\infty}(k+1)^{p'-2}\left(\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)^{p'}
	(k+1)^{p'}r^{kp'+1}(1-r)^{p'-1}dr=0.
	\end{split}
	\end{equation*}
Consequently, \eqref{eq:j16} holds for each $1<p<\infty$.

Let now $\{f_n\}$ such that $\sup_{n}\|f_n\|_{D^p_{p-1}}<\infty$
and $f_n\to0$ uniformly on compact subsets of $\mathbb{D}$. By \eqref{eq:j16}, for each $\varepsilon>0$, there exists $R=R(\varepsilon)\in(0,1)$  such that
	$$
	\sup_{z\in\mathbb{D}}\int_{\mathbb{D}\backslash\overline{\mathbb{D}(0,R)}}|(G^{H^2}_{g,z})'(w)|^{p'}(1-|w|)^{p'-1}dA(w)<\varepsilon^{p'}.
	$$
Further, by the uniform convergence we may choose $N=N(\varepsilon,R)\in\mathbb{N}$ such that $\max\{|f_n(0)|,|f_n'(\xi)|\}<\varepsilon$ for all $n\geq N$ and $\xi\in\overline\mathbb{D}(0,R)$. Therefore \cite[(2.4) and (4.4)]{CPPR} and H\"{o}lder's inequality yield
	\begin{equation*}
	\begin{split}
	\|T_g(f_n)\|_{H^\infty}
	&=\sup_{z\in\mathbb{D}}|\langle f_n,G^{H^2}_{g,z}\rangle_{H^2}|\\
	&\lesssim\sup_{z\in\mathbb{D}}\left|\int_{\mathbb{D}}f_n'(w)\overline{(G^{H^2}_{g,z})'(w)}\log\frac1{|w|}\,dA(w)\right|+|f_n(0)| \|g\|_{H^\infty}\\
	&\lesssim\sup_{z\in\mathbb{D}}\left(\left(\int_{\overline{\mathbb{D}(0,R)}}+\int_{\mathbb{D}\backslash\overline{\mathbb{D}(0,R)}}\right)
	|f_n'(w)||(G^{H^2}_{g,z})'(w)|(1-|w|)^{1-\frac1p+1-\frac1{p'}}dA(w)\right)\\
	&\quad+|f_n(0)| \|g\|_{H^\infty}\\
	&\lesssim\varepsilon\left(\|G^{H^2}_{g,z})\|^{p'}_{D^{p'}_{p'-1}}+ \sup_{n}\|f_n\|_{D^p_{p-1}}\right)\lesssim\varepsilon,\quad n\ge N.
	\end{split}
	\end{equation*}
Therefore $T_g:D^p_{p-1}\to H^\infty$ is compact by \cite[Lemma~3.6]{Tjani}.

Finally, let $X_p=H^p$. If $1<p<2$, then we may use the fact already proven that $T_g:\mathord{\rm HL}_p\to H^\infty$ is compact, and \eqref{Hardy1} to deduce the compactness of $T_g: H^p\to H^\infty$. In the case $2<p<\infty$ the same conclusion follows from  \eqref{Hardy2} and the compactness of $T_g:D^p_{p-1}\to H^\infty$. This finishes the proof of the theorem.
\hfill$\Box$

\section{Hardy, Hardy-Littlewood and Dirichlet-type spaces with $0<p\le1$}\label{Sec:2-3}

In this section we prove Theorems~\ref{Theorem:p=1} and \ref{Theorem:p=1'} in the said order. Since all the necessary auxiliary results are already stated in the previous sections, we can directly embark on the proofs.

\medskip

\noindent{\emph{Proof of}} Theorem~\ref{Theorem:p=1}.
(i). By \eqref{Hardy1} it suffices to show that $T(\mathord{\rm HL}_1,H^\infty)$ contains all polynomials. By Lemma~\ref{le:dualHL1} and \cite[Theorem~2.2]{CPPR}, we know that
	$$
	\|T_g\|_{\mathord{\rm HL}_1\rightarrow H^{\infty}}\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{\mathord{\rm HL}_{\infty}}.
	$$
Since
	$$
	\overline{G^{H^2}_{g,z}(w)}
	=\int_0^z\frac{g'(\z)}{1-\overline{w}\z}\,d\zeta
	=\sum_{k=0}^{\infty}\left(\sum_{n=0}^{\infty}(n+1)\widehat{g}(n+1)\frac{z^{n+k+1}}{n+k+1}\right)\overline{w}^k,
	$$
it is easy to show that
	\begin{equation}\label{pillo}
	\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{\mathord{\rm HL}_{\infty}}\lesssim
	\sup_{k\in\mathbb{N}\cup\{0\}}\left((k+1)\sum_{n=0}^{\infty}\frac{(n+1)|\widehat{g}(n+1)|}{n+k+1}\right).
	\end{equation}
If $g$ is a monomial, then $T_g:\mathord{\rm HL}_1\to H^\infty$ is bounded by \eqref{pillo}. It follows that $T(\mathord{\rm HL}_1,H^\infty)$ contains all polynomials.

(ii). Let $0<p<1$. Observe that
	$$
	\frac1{(1-\overline{w}\zeta)^{1+\frac1p}}=\sum_{k=0}^\infty c_p(k)(\overline{w}\zeta)^k,
	$$
where $c_p(k)\asymp(k+1)^\frac1p$ for all $k\in\mathbb{N}\cup\{0\}$. Hence
	$$
	\int_0^z\frac{g'(\zeta)}{(1-\overline{w}\zeta)^{1+\frac1p}}d\zeta
	=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{(n+1)\widehat{g}(n+1)c_p(k)}{n+k+1}\overline{w}^kz^{n+k+1}.
	$$
If $g$ is not a constant, then Lemma~\ref{lemma:dual-Dirichlet-2}, \cite[Theorem~2.2]{CPPR}, which can be applied to quasi-Banach spaces,  \eqref{Eq-alpha-bloch} and \eqref{Hardy1} yield
	\begin{equation*}
	\begin{split}
	\|T_g\|_{X_p\to H^\infty}
	&\gtrsim \|T_g\|_{D^p_{p-1}\to H^\infty}
	\asymp\sup_{z\in\mathbb{D}}\left(\left(\sup_{w\in\mathbb{D}}\left|\overline{\int_0^z\frac{g'(\zeta)\zeta}{(1-\overline{w}\zeta)^{2+\frac1p}}d\zeta}\right|(1-|w|)^2\right)+|g(z)-g(0)|\right)\\
	&\asymp\sup_{z\in\mathbb{D}}\left(\sup_{w\in\mathbb{D}}\left|\overline{\int_0^z\frac{g'(\zeta)}{(1-\overline{w}\zeta)^{1+\frac1p}}d\zeta}\right|(1-|w|)\right)\\
	&\gtrsim\sup_{z\in\mathbb{D}}\left|\int_0^z\frac{g'(\zeta)}{(1-\overline{z}\zeta)^{1+\frac1p}}d\zeta\right|(1-|z|)\\
	&=\sup_{z\in\mathbb{D}}\left|\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{(n+1)\widehat{g}(n+1)c_p(k)}{n+k+1}|z|^{2k}z^{n+1}\right|(1-|z|)\\
	&\ge\sup_{0<r<1}\frac1{2\pi}\int_0^{2\pi}\left|\sum_{n=0}^\infty
	\left(\sum_{k=0}^\infty\frac{c_p(k)}{n+k+1}r^{2k+n+1}\right)(n+1)\widehat{g}(n+1)e^{i\theta(n+1)}\,d\theta\right|(1-r)\\
	&\gtrsim\sup_{0<r<1}\sum_{n=0}^\infty\frac{\sum_{k=0}^\infty\frac{c_p(k)}{n+k+1}r^{2k+n+1}(n+1)|\widehat{g}(n+1)|}{n+1}(1-r)\\
	&\gtrsim\limsup_{r\to1^-}\sum_{k=0}^\infty\frac{c_p(k)}{k+1}r^{2k}(1-r)
	\asymp\lim_{r\to1^-}\frac1{(1-r)^{\frac1p-1}}=\infty
	\end{split}
	\end{equation*}
because $0<p<1$. Therefore $T_g:X_p\rightarrow H^\infty$ is bounded if and only if $g$ is a constant.

(iii). By \eqref{Hardy1}, it suffices to consider the case of $X_1=D^1_0$, so assume that $T_g: D^1_0\rightarrow H^\infty$ is compact. Let $(H^{\infty})^\star$ denote the identification of the dual space of $H^{\infty}$ via the $A^2$-pairing. Then $T_g^*: (H^{\infty})^\star\rightarrow\mathcal{B}^2$ is compact by Lemma~\ref{lemma:dual-Dirichlet-2}. Let $K^{A^2}_{z}$ denote the reproducing kernel of the Hilbert space $A^2$, associated to the point $z\in\mathbb{D}$. Then $T_g^*(K^{A^2}_{z})=G^{A^2}_{g,z}$ for all $z\in\mathbb{D}$, and
	\begin{equation*}
	\begin{split}
	\|K^{A^2}_{z}\|_{(H^{\infty})^\star}
	&=\sup_{\|f\|_{H^{\infty}}\le1}|\langle f, K^{A^2}_{z}\rangle|
	=\sup_{\|f\|_{H^{\infty}}\le1}\lim_{r\rightarrow1^-}\left|\sum_{n=0}^{\infty}\widehat{f}(n)\frac{z^n}{n+1}(n+1)r^n\right|\\
	&\le\sup_{\|f\|_{H^{\infty}}\le1}\|f\|_{H^{\infty}}
	\le1,\quad z\in\mathbb{D}.
	\end{split}
\end{equation*}
Therefore $\{G^{A^2}_{g,z}:z\in\mathbb{D}\}$ is relatively compact in $\mathcal{B}^2$. Hence, for given $\varepsilon>0$, there exist $z_1,\ldots,z_N\in\mathbb{D}$ such that for each $z\in\mathbb{D}$, we have $\|G^{A^2}_{g,z}-G^{A^2}_{g,z_j}\|_{\mathcal{B}^2}<\varepsilon$ for some $j=j(z)\in\{1,\ldots,N\}$. By using this and \eqref{Eq-alpha-bloch} we deduce
	\begin{equation*}
	\begin{split}
	&\sup_{a\in\mathbb{D}}\frac1{(1-|a|)^2}\int_{S(a)\setminus D(0,R)}|G^{A^2}_{g,z}(w)|^2(1-|w|)^2\,dA(w)\\
	&\quad\lesssim\|G^{A^2}_{g,z}-G^{A^2}_{g,z_j}\|_{H^\infty_1}^2+\sup_{a\in\mathbb{D}}\frac1{(1-|a|)^2}\int_{S(a)\setminus D(0,R)}|G^{A^2}_{g,z_j}(w)|^2(1-|w|)^2\,dA(w)\\
	&\quad \lesssim\varepsilon^2+\sup_{|a|\ge R}\frac1{(1-|a|)^2}\int_{S(a)}|G^{A^2}_{g,z_j}(w)|^2(1-|w|)^2\,dA(w).
	\end{split}
	\end{equation*}
Since $G^{A^2}_{g,z_j}\in\mathcal{A}\subset\mathcal{B}^2_0$ for each $j\in\{1,\ldots,N\}$, we obtain
	$$
	\lim_{R\rightarrow1^-}\sup_{a,z\in\mathbb{D}}\frac1{(1-|a|)^2}\int_{S(a)\setminus D(0,R)}|G^{A^2}_{g,z}(w)|^2(1-|w|)^2\,dA(w)=0,
	$$
which is equivalent to
	\begin{equation}\label{eq:compactdirichlet}
	\lim_{R\rightarrow1^-}\sup_{a,z\in\mathbb{D}}\int_{\mathbb{D}\setminus D(0,R)}|G^{A^2}_{g,z}(w)|^2(1-|\varphi_a(w)|^2)^2\,dA(w)=0
	\end{equation}
by the reasoning in the proof of \cite[Lemma 3.3]{G}, see \cite[Lemma 5.3]{PR} for further details. However, if $g$ is not a constant, then there exists an $N\in\mathbb{N}\cup\{0\}$ such that $\widehat{g}(N+1)\ne0$. Therefore,
	\begin{equation*}
	\begin{split}
	&\sup_{a,z\in\mathbb{D}}\int_{\mathbb{D}\setminus D(0,R)}|G^{A^2}_{g,z}(w)|^2(1-|\varphi_a(w)|^2)^2\,dA(w)\\
	&\quad\ge\sup_{z\in\mathbb{D}}\int_{\mathbb{D}\setminus D(0,R)}|G^{A^2}_{g,z}(w)|^2(1-|\varphi_z(w)|^2)^2\,dA(w)\\
	&\quad=\sup_{z\in\mathbb{D}}(1-|z|)^2\left(\int_{\mathbb{D}\setminus D(0,R)}\left|\frac{G^{A^2}_{g,z}(w)}{(1-z\overline{w})^2}\right|^2
    (1-|w|)^2
    \,dA_2(w)\right)\\
	&\quad\asymp\sup_{z\in\mathbb{D}}\left((1-|z|)^2\int_{\mathbb{D}\setminus D(0,R)}
	\left|\left(\sum_{k=0}^{\infty}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)(k+1)}{n+k+1}z^{n+k+1}\right)\overline{w}^k\right)
    \right.\right.\\
	&\qquad\quad\cdot\left.\left.
    \left(\sum_{j=0}^{\infty}(j+1)z^i\overline{w}^j\right)\right|^2
    (1-|w|)^2
    \,dA(w)\right)\\
	&\quad=\sup_{z\in\mathbb{D}}(1-|z|)^2
    \int_{\mathbb{D}\setminus D(0,R)}
	\left| \sum_{m=0}^{\infty} \left(\sum_{k=0}^{m}\left(\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)(k+1)}{n+k+1}z^{n+k+1}\right)(m-k+1)z^{m-k}\right)
    \overline{w}^m\right|^2\\
	&\qquad\quad\cdot(1-|w|)^2\,dA(w)

    \\
	&\quad\asymp\sup_{z\in\mathbb{D}}(1-|z|)^2\Bigg(\sum_{m=0}^{\infty}\left|\sum_{k=0}^m\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)(k+1)(m-k+1)}{n+k+1}z^{n+m+1}	\right|^2

    \int_R^1s^{2m+1}(1-s)^2\,ds\Bigg)\\
	&\quad\gtrsim\sup_{0<r<1}\Bigg((1-r)^2\Bigg(\sum_{m=0}^{\infty}r^{2m}\left(\int_R^1s^{2m+1}(1-s)^2\,ds\right)\\
	&\qquad\quad\cdot\int_0^{2\pi}\left|\sum_{n=0}^{\infty}(n+1)\widehat{g}(n+1)\left(\sum_{k=0}^{m}\frac{(k+1)(m-k+1)}{n+k+1}\right)r^{n+1}e^{i\theta(n+1)}\right|^2
    \Bigg)\,d\theta\\
	&\quad\gtrsim\sup_{0<r<1}|\widehat{g}(N+1)|^2(N+1)^2(1-r)^2r^{2N+2}\sum_{m=0}^{\infty}r^{2m}\left(\int_R^1s^{2m+1}(1-s)^2\,ds\right)\left(\sum_{k=0}^{m}(m-k+1)\right)^2\\
	&\quad\gtrsim R^{2N+1}(1-R)^2\int_R^1(1-s)^2\left(\sum_{m=0}^{\infty}(m+1)^4(Rs)^{2m+1}\right)\,ds\\
	&\quad\asymp R^{2N+1}(1-R)^2\int_R^1\frac{(1-s)^2}{(1-Rs)^5}ds\\
	&\quad\asymp R^{2N+1},\quad 0<R<1.
	\end{split}
	\end{equation*}
By letting $R\rightarrow1^-$ we obtain a contradiction with \eqref{eq:compactdirichlet}. Therefore $g$ must be a constant if $T_g: D^1_0\rightarrow H^\infty$ is compact. This finishes the proof of the theorem.
\hfill$\Box$

\bigskip

\noindent{\emph {Proof of}} Theorem~\ref{Theorem:p=1'}. By using the proof of Theorem~\ref{Theorem:p=1}(i) and standard arguments, we deduce
	\begin{equation*}
	\begin{split}
	\|T_g\|_{D^1_0\to H^\infty}
	&\lesssim\|T_g\|_{H^1\to H^\infty}
	\lesssim\|T_g\|_{\mathord{\rm HL}_1\to H^\infty}
	\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{\mathord{\rm HL}_{\infty}}\\
	&\asymp
	\sup_{k\in\mathbb{N}\cup\{0\}}\left((k+1)\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{n+k+1}\right)
	\end{split}
	\end{equation*}
because $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$ by the hypothesis. Thus $T_g:X_1\to H^\infty$ is bounded if \eqref{Eq:p=1} is satisfied.

Conversely, if $T_g:X_1\to H^\infty$ is bounded, then $T_g:D^1_0\to H^\infty$ is bounded by \eqref{Hardy1}. Therefore Lemma~\ref{lemma:dual-Dirichlet-2}, \cite[Theorem~2.2]{CPPR} and \eqref{Eq-alpha-bloch} yield
	\begin{equation*}
	\begin{split}
	\|T_g\|_{D^1_0\to H^\infty}
	&\asymp\sup_{z\in\mathbb{D}}\left(\left(\sup_{w\in\mathbb{D}}\left|\overline{\int_0^z\frac{g'(\zeta)\zeta}{(1-\overline{w}\zeta)^{3}}d\zeta}\right|(1-|w|)^2\right)+|g(z)-g(0)|\right)\\
	&\asymp\sup_{z\in\mathbb{D}}\left(\sup_{w\in\mathbb{D}}\left|\overline{\int_0^z\frac{g'(\zeta)}{(1-\overline{w}\zeta)^{2}}d\zeta}\right|(1-|w|)\right).
	\end{split}
	\end{equation*}
Now that $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$ by the hypothesis, standard arguments yield
	\begin{equation}\label{mierda}
	\|T_g\|_{D^1_0\to H^\infty}
	\asymp\sup_{0\le s<1}\left(\left(\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\frac{(n+1)(k+1)\widehat{g}(n+1)}{(n+k+1)}s^{k}\right)(1-s)\right).
	\end{equation}
Since the coefficients
	$$
	\sum_{n=0}^{\infty}\frac{(n+1)(k+1)\widehat{g}(n+1)}{(n+k+1)}
	$$
are positive for all $k$, and increasing in $k$, we deduce
	\begin{equation*}
	\begin{split}
	\|T_g\|_{D^1_0\to H^\infty}
	&\gtrsim\sup_{K\in\mathbb{N}\cup\{0\}}\left(\left(\sum_{n=0}^{\infty}\frac{(n+1)(K+1)\widehat{g}(n+1)}{(n+K+1)}\right)\limsup_{s\to1^-}\sum_{k=K}^{\infty}s^{k}(1-s)\right)\\
	&=\sup_{K\in\mathbb{N}\cup\{0\}}\left((K+1)\sum_{n=0}^{\infty}\frac{(n+1)\widehat{g}(n+1)}{(n+K+1)}\right),
	\end{split}
	\end{equation*}
and thus \eqref{Eq:p=1} is satisfied. The norm estimate \eqref{eq:j11'} is an immediate consequence of the proof just established.
This finishes the proof.
\hfill$\Box$

\section{$\mathord{\rm BMOA}$, Bloch space and $H^\infty_{\log}$}\label{Sec:Bloch}

This section is devoted to the proof of Theorem~\ref{Bounded}. Unlike the other main results, its proof does not require much tools, but can be carried out with relatively straightforward arguments.

\medskip

\noindent{\emph {Proof of}} Theorem~\ref{Bounded}.
The chain of inequalities \eqref{Eq:embedding} shows that (i)$\Rightarrow$(ii)$\Rightarrow$(iii) and
	\begin{equation}\label{Eq:1}
	\|T_g\|_{\mathord{\rm BMOA}\rightarrow H^\infty}\lesssim\|T_g\|_{X\rightarrow H^\infty}\lesssim \|T_g\|_{H_{\log}^\infty\rightarrow H^\infty}.
	\end{equation}
If (iii) is satisfied, then, by \cite[Theorem 2.5 (v)]{CPPR}, we have
	\begin{equation*
	\begin{split}
	\|T_g\|_{\mathord{\rm BMOA}\rightarrow H^\infty}
	&\asymp\sup_{z\in\mathbb{D}}\|G^{H^2}_{g,z}\|_{H^1}
	=\sup_{z\in\mathbb{D}}\int_{0}^{2\pi}\left|\overline{\int_0^z\frac{g'(\z)}{1-\z e^{-i\theta}}\,d\zeta}\right|d\theta\\
	&=\sup_{z\in\mathbb{D}}\int_0^{2\pi}\left|\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{(n+1)\overline{\widehat{g}(n+1)}\overline{z}^{n+k+1}}{n+k+1}e^{ik
\theta}\right|d\theta.
	\end{split}
	\end{equation*}
Since $\widehat{g}(n)\geq0$ for all $n\in\mathbb{N}\cup\{0\}$ by the hypothesis, Hardy's inequality \cite[p.~48]{D} and Fatou's lemma yield
	\begin{equation*}
	\begin{split}
	\|T_g\|_{\mathord{\rm BMOA}\rightarrow H^\infty}
	&\gtrsim\sup_{z\in\mathbb{D}}\sum_{k=0}^{\infty}\left|\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)\overline{z}^{n+k+1}}{(n+k+1)(k+1)}\right|\\
	&\ge\sup_{0\le r<1}\sum_{k=0}^{\infty}\sum_{n=0}^\infty\frac{(n+1)\widehat{g}(n+1)r^{n+k+1}}{(n+k+1)(k+1)}\\
	&=\sum_{n=0}^\infty(n+1)\widehat{g}(n+1)\sum_{k=0}^{\infty}\frac{1}{(k+1)(n+k+1)}\\
	&\gtrsim\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2).
	\end{split}
	\end{equation*}
Therefore, (iii) implies (iv), and
	\begin{equation}\label{Eq:2}
	\|T_g\|_{\mathord{\rm BMOA}\rightarrow H^\infty}\gtrsim\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2).
	\end{equation}

If (iv) is satisfied, then
	\begin{equation*}
	\begin{split}
	\int_0^1M_\infty(t,g')\log\frac{e}{1-t}\,dt
	&=\int_0^1 \left(\sum_{n=0}^\infty (n+1)\widehat{g}(n+1)t^n \right)\log\frac{e}{1-t}\,dt\\
	&=\sum_{n=0}^\infty (n+1)\widehat{g}(n+1) \int_0^1  t^{n}\log\frac{e}{1-t}\,dt\\
	&\asymp\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2).
	\end{split}
	\end{equation*}
Thus (v) holds and
	\begin{equation}\label{Eq:3}
	\sum_{n=0}^\infty\widehat{g}(n+1)\log(n+2)
	\asymp\int_0^1M_\infty(t,g')\log\frac{e}{1-t}dt.
	\end{equation}

Finally, assume (v). Then
	\begin{equation*}
	\begin{split}
	\|T_g(f)\|_{H^\infty}
	=\sup_{z\in\mathbb{D}}\left|\int_0^zf(\z)g'(\z)d\z\right|
	&\le\sup_{0\le r<1}\int_0^rM_\infty(s,f)M_\infty(s,g')\,ds\\
	&\le\|f\|_{H_{\log}^\infty}\int_0^1M_{\infty}(s,g')\log\frac{e}{1-s}\,ds,\quad f\in\H(\mathbb{D}),
	\end{split}
	\end{equation*}
and therefore
	\begin{equation}\label{Eq:4}
	\|T_g\|_{H_{\log}^\infty\rightarrow H^\infty}\lesssim\int_0^1M_\infty(r,g')\log\frac{e}{1-r}dr.
	\end{equation}
To complete the proof of the theorem it is now enough to show that $T_g: H^\infty_{\log}\to H^\infty$ is compact, that is
 $\lim_{n\to \infty}\|T_g(f_n)\|_{H^\infty}=0$ for each sequence $\{ f_n\}$ of analytic functions in $\mathbb{D}$ such that $\sup_n\|f_n\|_{H^\infty_{\log}}<\infty$
and $f_n\to0$ uniformly on compact subsets of $\mathbb{D}$.
Let $\varepsilon>0$. Fix $R=R(\varepsilon)\in(0,1)$ such that
	$$
	\int_R^1M_{\infty}(r,g')\log\frac{e}{1-r}dr<\varepsilon,
	$$
and then pick up $N=N(\varepsilon,R)\in\mathbb{N}$ such that $|f_n(z)|\le\varepsilon$ for all $ z \in\overline{D(0,R)}$ and $n\ge N$. Then, if $n\ge N$, we have
	\begin{equation*}
	\begin{split}
	\|T_g(f_n)\|_{H^\infty}
	&=\limsup_{|z|\to1^-}\left|\int_0^zf_n(\z)g'(\z)d\z\right|
	\le\limsup_{r\to1^-}\int_0^rM_\infty(s,f_n)M_\infty(s,g')\,ds\\
	&=\int_0^RM_\infty(s,f_n)M_\infty(s,g')\,ds
	+\int_R^1M_\infty(s,f_n)M_\infty(s,g')\,ds\\
	&\le\varepsilon\int_0^RM_\infty(s,g')\,ds
	+\|f_n\|_{H_{\log}^\infty}\int_R^1M_{\infty}(s,g')\log\frac{e}{1-s}\,ds
	\lesssim\varepsilon.
	\end{split}
	\end{equation*}
Thus $T_g:~H^\infty_{\log}\mapsto H^\infty$ is compact. The last thing to do is to observe that \eqref{Eq:1}--\eqref{Eq:4} imply the norm estimates \eqref{Eq:norm-estimates}. This finishes the proof.
\hfill$\Box$

\medskip

We finish the section and the paper by discussing briefly other natural choices for the space $X$ in Theorem~\ref{Bounded}. For a  nonnegative function  $\om\in L^1([0,1))$, the extension to $\mathbb{D}$, defined by $\om(z)=\om(|z|)$ for all $z\in\mathbb{D}$, is called a radial weight. For $0<p<\infty$ and such an $\omega$, the Lebesgue space $L^p_\om$ consists of complex-valued measurable functions $f$ on $\mathbb{D}$ such that
    $$
    \|f\|_{L^p_\omega}^p=\int_\mathbb{D}|f(z)|^p\omega(z)\,dA(z)<\infty.
    $$
The corresponding weighted Bergman space is $A^p_\om=L^p_\omega\cap\H(\mathbb{D})$. For a radial weight $\om$, its associated weight $\om^\star$ is defined by $\om^\star(z)=\int_{|z|}^1 s \log\frac{s}{|z|}\om(s)\,ds$ for all $z\in\mathbb{D}\setminus\{0\}$. It arises naturally when the Hardy-Stein-Spencer formula is applied to the dilatation $f_r$ in order to establish the identity
	$$
	\|f\|_{A^p_\om}^p=\|\Delta|f|^p\|_{L^1_{\om^\star}}+\om(\mathbb{D})|f(0)|^p,\quad f\in\H(\mathbb{D}),
	$$
see~\cite[Theorem~4.2]{PR} for details. Because the Laplacian of $|f|^p$ contains the factor $|f'|^2$, which can be interpreted as the Jacobian of the non-univalent change of variable $w=f(z)$, this equivalent norm is useful, for example, in the study of composition operators~\cite{PelRatSc}. But the associated weight comes to the picture also in some other instances which are more closely related to the topic of the present paper. To explain this, we say that a radial weight $\om$ belongs to the class~$\widehat{\mathcal{D}}$ if there exists a constant $C=C(\om)\ge1$ such that $\widehat{\om}(r)\le C\widehat{\om}(\frac{1+r}{2})$ for all $0\le r<1$. Moreover, if there exist $K=K(\om)>1$ and $C=C(\om)>1$ such that $\widehat{\om}(r)\ge C\widehat{\om}\left(1-\frac{1-r}{K}\right)$ for all $0\le r<1$, then we write $\om\in\widecheck{\mathcal{D}}$. The intersection $\widehat{\mathcal{D}}\cap\widecheck{\mathcal{D}}$ is denoted by $\mathcal{D}$. For $\om\in\widehat{\mathcal{D}}$, the space $C^1(\om^\star)$ consists of $f\in\H(\mathbb{D})$ such that the measure $|f'|^2\om^\star\,dA$ is a $1$-Carleson measure for $A^p_\om$~\cite[Theorem 6.1]{PelSum14}. As usual, we say that a positive Borel measure $\mu$ on $\mathbb{D}$ is a $p$-Carleson measure for $X$ if $X$ is continuously embedded into~$L^p_\mu$. The space $C^1(\om^\star)$ arises in the study of integration operators acting on weighted Bergman spaces. Indeed, it is known that, for each $0<p<\infty$, the operator $T_g$ is bounded from $A^p_\om$ into itself if and only if $g \in C^1(\om^\star)$~\cite[Theorem~6.4]{PelSum14}. For $\om\in\mathcal{D}$, the space $C^1(\om^\star)$ is nothing else but the Bloch space by the proof of \cite[Theorem~6.1(C)]{PelSum14}, but it may be a proper subspace of $\mathcal{B}$ by \cite[Theorem 6.1(D)]{PelSum14}, yet it always contains $\mathord{\rm BMOA}$. Therefore we may choose $X=C^1(\om^\star)$ in Theorem~\ref{Bounded}. It is worth observing that while $\mathord{\rm BMOA}$ and $\mathcal{B}$ are conformally invariant, there exists $\om\in\widehat{\mathcal{D}}\setminus\mathcal{D}$ such that $C^1(\om^\star)$ is not \cite[Proposition~5.6]{PR}. Recall that a Banach space $X\subset\H(\mathbb{D})$, equipped with a semi-norm $\rho_X$, is conformally invariant if there exists a constant $C=C(X)>0$ such that $\rho_X(f\circ\varphi) \le C \rho (f)_X$ for all $f\in X$ and for all automorphisms $\varphi$ of $\mathbb{D}$.

redpajama_set_name: RedPajamaArXiv


Section 22 of the Ethics in Public Office Act 1995 (the 1995 Act) allows certain categories of person to make complaints to the Standards in Public Office Commission (the Standards Commission) about alleged contraventions of the 1995 Act by different classes of person who have obligations under that Act.
Section 4 of the Standards in Public Office Act 2001 (the 2001 Act) allows any person to make a complaint to the Standards Commission about an alleged "specified act" by a "specified person".
These provisions are set out in more detail below.
While it is open to any person to make a complaint to the Standards Commission in line with the provisions of the Ethics Acts the Commission requests that all local avenues should be exhausted before such a complaint is made.
Accordingly, if a person considers that an employee or a member of a local authority may have contravened a provision of Part 15 of the Local Government Act 2001, the Commission is of the view that the matter should be raised with the local authority ethics registrar in the first instance. If a person is not satisfied with the outcome of the consideration of the matter following referral to the ethics registrar, he or she may make a complaint to the Standards Commission.
set out in detail the basis for alleging that contravention.
Where documentary material is relevant, copies of that material should be furnished with the complaint where possible.
Section 8 of the 2001 Act precludes the Standards Commission from investigating a complaint made to it unless the identity of the person making the complaint is disclosed. However, the Standards Commission may, if it considers it appropriate to do so, restrict the disclosure of such identity for the purposes of the investigation of the complaint or otherwise in the interests of justice.
3. contravened a provision of the Electoral Act 1997, as amended (the Electoral Acts).
the matter is one of significant public importance.
The procedures which the Standards Commission follow in any case where it considers that an investigation under the Ethics Acts into the matter complained of is appropriate, are set out in its 'Statement of Intended Procedures'.
In addition to the investigation of complaints under the Ethics Acts, the Standards Commission has a separate supervisory function to ensure compliance with the provisions of the Electoral Act 1997 (as amended). Where a person considers that the Electoral Acts may have been contravened, whether by a "specified person" or otherwise, he /she may contact the Standards Commission, which, if it considers it appropriate to do so, will investigate the matter.
Section 4(4) of the Electoral Acts provides that the Standards Commission may make such enquiries as it considers appropriate and may require any person to furnish any information or document which may be required for the purposes of its duties under the Electoral Acts.


Gaithersburg Mayor, Jud Ashman, gives MCM an update on the status of Lakeforest Mall. Officials say the JC Penney and Starbucks located there will be closing. Ashman says the decline at Lakeforest Mall is due to the evolution of malls and the increase in online shopping. Please watch the video for more details.


text:
Company:Minera Alumbrera, Goldcorp, Xstrata, Yamana Gold Operational status:In production Materials extracted:copper, gold Type of mine:open pit Main issues:water, public health, biodiversity, pollution
In 1992, Yacimientos Mineros de Agua Dionisio, a state owned mining company, awarded the Alumbrera concession to Canadian owned company International Musto Exploration[1]. Six years later, the Canadian mining company GoldCorp began mining operations Bajo de Alumbrera. Today, the mine is among the ten largest copper enterprises in the world and among the fifteen largest gold enterprises in the world. Every year, the mine generates $680 million in revenue.[2] The mine is located in the northeastern province of Catamarca in Argentina's major mining region. Despite the immense wealth that has been generated by the mine, the Alumbrera mine is a source of great conflict and tension in the region.
The mine has been a source of conflict for the inhabitants of the nearby communities since its inception. Community members have been mainly concerned over the potential negative environmental impacts as well as the potentially harmful effects on local social and economic development. The environmental threats to surrounding areas include soil contamination, water contamination, air pollution, and water depletion. The community members' concerns over water are predicated on the fact that the mine uses around 50 million liters of water daily, in order to process the 360 tons of rocks moved every day[3]. This causes a significant decrease in the availability of uncontaminated and unprocessed water, which spawned great discontent for the people living near the mine.
The conflict between the mining companies and the local inhabitants began shortly after operations commenced at the mine in 1998. In April 1999, Juan González, a biologist at the National University of Tucumán as well as the secretary of the Environment for the province of Tucumán, accused the authorities of Minera Alumbrera of contaminating the DP2 Canal with toxic fluids[4]. Though the attorney of Tucumán, Antonio Gustavo Gómez, rejected his accusation, it brought heightened attention to issues surrounding the mine. Concerns over the environmental threats resurfaced in 2002 when Oscar Beltrán, the defense lawer of the nearby village of Santiago, inspected the Salí Dulce river, close to the Alumbrera mine, and confirmed that it was contaminated because of the industrial waste from the Alumbrera plant[5]. In light of Beltrán's findings, Gomez launched an investigation into the Minera Alumbrera company.
In 2006, local community members began to take a more central role in the conflict when they demanded that the president of Minera La Alumbrera, Julián Rooney, be prosecuted by Gomez in the Federal Court of Appeals on charges of water contamination and other crimes against the environment[6]. They further demonstrated their opposition to the mine on May 19, 2008 when environmental activists from Tinogasta and other areas in Catamarca staged a successful roadblock of the regional route to Chile's ports, used by many in the mining industry to get from Bajo de Alumbrera to transport the processed minerals for export. Activists stated that the roadblock and protest was in defense of their livelihood and the self-determination of their villages, as well as against the government for their lack of accountability to the people in these areas.[7] Their tactics proved somewhat successful: on June 8, 2008, the Federal Chambers of Tucamán brought criminal charges of contamination against Julián Rooney, president of Minera La Alumbrera, because of the complaint filed by Juan González in 1998.[8] This case was the first time environmental charges have been pressed against a mining company in all of Latin America.
However, production continued at Bajo de Alumbrera. In response, activists and community members continued to put pressure on the government and GoldCorp through public demonstrations and roadblocks which often ended in severe police repression. One roadblock in February 2012 engendered a brutal police response with led to the hospitalization of at least thirteen protestors due to injuries from rubber bullets and batons.[9] GoldCorp has recently reported that there has been a decline in productivity at the Alumbrera mine since 2011, so they will soon begin to phase out the Alumbrera project and focus their efforts towards the Agua Rica deposit.[10]
[1] http://www.mining-technology.com/projects/bajo/
[2] http://www.goldcorp.com/Unrivalled-Assets/Mines-and-Projects/Central-and-South-America/Operations/Alumbrera/Overview-and-Operating-Highlights/default.aspx
[3] http://www.noalamina.org/mineria-argentina/mineria-salta/profesionales-metan-advirtieron-peligros-explotacion-minera-zona
[4] http://www.noalamina.org/mineria-argentina/mineria-tucuman/presion-vecinos-procesamiento-presidente-alumbrera
[5] http://www.noalamina.org/mineria-preguntas-frecuentes/mineria-mineras
[7] http://www.noalamina.org/mineria-argentina/mineria-catamarca/catorce-camiones-alumbrera-retenidos-tinogasta
[8] http://www.noalamina.org/english-sites/general/mining-corporation-charged-crimes-against-environment-argentina
[9] http://www.lacapital.com.ar/contenidos/2012/02/11/noticia_0011.html
[10] http://www.goldcorp.com/English/Unrivalled-Assets/Mines-and-Projects/Central-and-South-America/Operations/Alumbrera/Overview-and-Operating-Highlights/default.aspx
Timeline of Key Events

redpajama_set_name: RedPajamaCommonCrawl


text:
if you can't see it, then you are useless in the fight to stop it.
We are builders in a constant struggle with destroyers. let's win this.
Republicans are radicals, not conservatives.
When your entire life is steeped in white supremacy, equality feels like discrimination.
JFC, are there no editors left at that goddamn rag?
Peak wingnut was a lie.
Come on, man.
They traffic in fear. it is their only currency. if we are fearful, they are winning.
Historically it was a little unusual for the president to be an incoherent babbling moron.
Let us savor the impending downfall of lawless scoundrels who richly deserve the trouble barreling their way.
Eh, that's media spin. biden's health is fine and he's doing a good job.
I've spoken to my cat about this, but it doesn't seem to do any good.
That's my take and I am available for criticism at this time.
You are here: Home / Open Threads / Excellent Links / Excellent Read: January 6th – 'Many Things Transpired That Day'
Excellent Read: January 6th – 'Many Things Transpired That Day'
by Anne Laurie | May 18, 2021 10:43 am | 56 Comments
This post is in: Excellent Links, Republican Venality, Violent Insurrection at the Capitol
In Exclusive Jailhouse Letter, Capitol Riot Defendant Explains Motives, Remains Boastful https://t.co/eLtBxTe1cb
— Jennifer Mercieca (@jenmercieca) May 17, 2021
Jailhouse bards are almost as common as jailhouse lawyers, but this is a somewhat terrifying report on what happens when the GOP is treasonous and the Qrazy is contagious:
In a letter sent from behind bars, a key defendant in the Jan. 6 riot at the U.S. Capitol said he and fellow inmates have bonded in jail, and boasted that those attacking the building could have overthrown the government if they had wanted.
The letter is signed "the 1/6ers" and expresses no remorse for the assault on the Capitol, in which five people died. While no names appeared on it, ProPublica was able to determine, through interviews with his family and a review of his correspondence from jail, that it was penned by Guy Reffitt, a member of the Three Percenter right-wing militant group accused of participating in the riot. The letter said the inmates arrested for their role in the attack regularly recite the Pledge of Allegiance inside the Washington, D.C. jail and sing the national anthem "all in unison, loud and proud most everyday."
"January 6th was nothing short of a satirical way to overthrow a government," said the letter, written by hand on yellow lined paper. "If overthrow was the quest, it would have no doubt been overthrown."…
Reffitt, 48, worked most of his adult life on oil rigs, an occupation that took him and sometimes his family around the world, including three years in Malaysia. But when the coronavirus hit in 2020, work dried up and he intensified his political activity, focusing on the Black Lives Matter movement, which he viewed as destructive.
Reffitt saw his actions on Jan. 6 as a critical step in protecting his wife and kids from what he viewed as a decades-long American slide toward "tyranny," according to his text messages…
In their most recent filing, prosecutors added new evidence to their case against Guy Reffitt. They obtained a recording of a Jan. 10 Zoom meeting involving Reffitt and two other Three Percenters. In it, Reffitt allegedly said he helped lead the charge on the Capitol with a .40-caliber pistol at his side, at one point telling a U.S. Capitol Police officer who was firing nonlethal rounds at him, "Sorry, darling. You better get a bigger damn gun."
Reffitt went on to describe how the group might be able to disable a social media company's servers by using a sniper rifle to disable the generators at a nearby Texas facility. According to court records, he said attacking the servers would "make them feel it back" in Washington, D.C. He added: "Then they won't know we're coming next time."
In court filings, his lawyer said that prosecutors have "relied on bragging" and that none of the government's video or photographs from the Capitol show Reffitt to be armed. Reffitt has not been charged with a gun crime…
After then-President Donald Trump lost his bid for reelection, Guy Reffitt began to sequester himself in the front room of his suburban brick home, glued to Newsmax as it reported theories of how the vote was rigged.
On Dec. 19, Reffitt found a new obsession, his family said, when Trump tweeted: "Big protest in D.C. on January 6th. Be there, will be wild!"
From then on, Reffitt's texts bounced between plans for shopping and cooking prime rib for Christmas and talk of going to D.C. to "shock the world."
"It's the government that is going to be destroyed in this fight," Reffitt texted his family on Dec. 21. "Congress has made fatal mistakes this time."…
If his family is to be believed, Reffitt was a major drama queen. That doesn't make him, or his similarly inclined fellows, any less dangerous. Consider John Wilkes Booth and his co-conspirators, who were laughably inept and yet still changed the course of our history — much for the worse.
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If the former guy is ever taken into custody, people like Reffitt (the ones not already locked up) will go apeshit.
"If overthrow was the quest, it would have no doubt been overthrown."…
I'm not even gonna try to understand this.
Just punt all these shits into the Sun already.
Dorothy A. Winsor
What this clown describes sound like a gang to me.
Then his son turned him in and left home.
Reading the son's account I think Reffit would have met the requirements for a "red flag" law action.
They are state laws that authorize courts to issue a special type of protection order, allowing the police to temporarily confiscate firearms from people who are deemed by a judge to be a danger to themselves or to others.
Often, the request for the order will come from relatives or friends concerned about a loved one who owns one or more guns and has expressed suicidal thoughts or discussed shooting people. The authorities may also request an order.
For me, if my heavily armed Right wing nut relative started ranting about doing "something big" I think it's completely reasonable to think a mass murder is possible. I mean, we can read his manifesto later or take the guns- your choice.
They must be punished. Lack of consequence will embolden them. Because the BJP was not punished and instead was rewarded after they demolished the Babri Masjid 1992, India is paying the price now.
Goku (aka Amerikan Baka)
@Kay:
I had a professor in nursing school, a psych nurse no less (& black to boot; I think she was from Appalachia originally) actually take the time during a lecture to talk about how wrong "red flag" laws are, I shit you not
That guy really can't process that not only did his FG lose and fail (etc), he and his fellow 'satirists' similarly lost and failed, big time, on the big screen, in front of all.
dr. bloor
Quite a letter. At least we know where John Blutarsky is now.
@mrmoshpotato: I think they're trying to say that they aren't fuck-ups who fucked up and believed TFG, then got theirselves busted.
BruceFromOhio
But when the coronavirus hit in 2020, work dried up and he intensified his political activity, focusing on the Black Lives Matter movement, which he viewed as destructive.
Yes, you did forget "God Bless America," and this version of "the American spirit" seems better suited to a Ku Klux Klan robe. Rot in jail, please.
What irritates the ever living fuck out of me is that these two-bit ratfuck soulless racists will bring the end of what they claim to hold dear, to the ruin of all. Not the Chinese army, not the Russian nuclear weapons, not the COVID virus, but stupid knuckle-dragging racists.
@Goku (aka Amerikan Baka):
Well, every state has some variety and combination of "pink slipping" or "temporary hold" so it's really a choice for these people- they can come in under the narrow red flag laws or they can come in under the much more restrictive holds, but no one has to put up with imminent, credible threats of mass murder.
He thought his father was going to kill a lot of people, and soon. Given what his father said and how he behaved this was a rational fear. It's definitely a balancing of rights, a mental health hold suspends almost all rights pending hearing, no one ever pretends it doesn't. It's treated like a profound infringement on rights, because that's what it is. That doesn't mean it can't be done. It is done, and often.
opiejeanne
Satirical way to overthrow the government, he says. Five people died, satirically?
I think this guy needs a dictionary with a concise definition of satire, plus a clueX4 to the head. There is no way the actions of 1/6 could have overthrown the government, not with so few co-conspirators, unless TFG had managed to suddenly become competent.
@germy:
I welcome this. Gotta know where the dirt is to get it cleaned up. We're seeing it now in Kay's school board, GA, FL, TX, AZ legislatures, that ridiculous AZ "election audit" : can't fight what you can't see.
@Wapiti: Whatever. Fuck them all. Destination Sun but also (too) destination F#%Ked.
different-church-lady
…he and fellow inmates have bonded in jail…
Well, ain't that how prison gangs work, after all?
Fuck 'em
Shoot them into the sun
Rescue them
Fuck 'em again
Shoot them back into the sun
Rand Careaga
@different-church-lady: Chain gangs too.
Hkedi [Kang T. Q.]
@opiejeanne:
I call Popehat's rule of goats your honor.
You can say to her "yes, this is definitely an infringement on rights", because it is. These people seem to come to a screeching halt at the most obvious things that everyone else already considered and then stop thinking at all.
Her objection doesn't put it away. She's not done yet. Who are the sets of people with rights and how can we balance all of them. This idea these people have that yelling "my rights!" is some kind of final word just isn't true. The prospective shooters rights give way to the public's right to stay alive. We're definitely picking and choosing here, no one pretends we aren't. All right out in the open.
@Hkedi [Kang T. Q.]: Had to look that up because I had forgotten that gem. Appropriate.
Tony Jay
"Nuh-uh. Nuthin' treasonous or criminal about us. We totally could have overthrown the Government if we wanted to, 'cos we're so awesome and American and whatnot, but as it's still there, we obviously didn't. D'uh! You can say "Thank You" any time you like, Snowflake."
"That's satire, Libtard. Look it up!"
Either lock this prick away for the good of the gene pool or make him a three-term State Congressman for somewhere Red and Poor. There is no middle ground.
When are the organizers and inciters getting charged?
Having brownshirt thugs like this assclown locked up is good for the country, but the string-pullers are going to try again if they're not taken down hard.
Oh you don't have to convince me. I couldn't believe what I was hearing at the time
@different-church-lady: I like your plan.
Amir Khalid
He spent three years in Malaysia? Hmm. I don't think I ever met the guy.
@opiejeanne: I didn't know about Popehat's Rule of goats. Now I do.
Enhanced Voting Techniques
What kind of adult writes like that? And the rest of it, that's how upset teens talk. These guys really are frozen at age 14.
Consider John Wilkes Booth and his co-conspirators, who were laughably inept and yet still changed the course of our history — much for the worse.
Good Lord, yes, Booth even stopped to preen on a stage after shooting Lincoln.
If there's a silver-lining to this, it looks like Reffitt's son is going to be okay.
I didn't know about Popehat's Rule of goats.
Neither did I. Might as well put it in the comments for everyone else:
The Rule of Goats: even if you say you're only fucking goats ironically, you're still a goatfucker
If he's convicted, that letter will be Exhibit A to the government's sentencing memorandum, and probably guarantees that he will do the max. And the Bureau of Prisons will be fully justified in assigning him to a Level 5 (maximum security) facility.
I imagine his lawyers are having a f***ing stroke.
J R in WV
For a successful coup, you need all the brass at all the cop shops, along with all the front line storm-troopers, as well as all the National Guard command and their front line troopers, plus the USAF, the Army, and the Marines. No way TFG ever gets that level of support then, even more impossible today.
Benw
They like their army cosplay so much, I think he meant sartorial way to overthrow a govt.
Clearly taking a cue from TFG: cannot keep his mouth shut even when it is in his best interest to do so.
I feel like some people are reachable if you can get them to the next step. They see conspiracies and plots everywhere but it's really much more boring than that :)
You file an ex parte petition in a probate court in Ohio. Nothing mysterious or deep state about it. 90% of the time you don't have to file and you can do something less restrictive. The goal is to get consent, not to drag them off in chains.
@burnspbesq:
That thought crossed my mind as I was reading his…whatever it is.
Things weren't going well for this man and his whole family knew it. He was traveling the country using false license plates- several! This was going no where good. And how DID it go for him? He's in jail. No bigger loss of liberty than that.
His son is the only sensible person in that family and arguably the only one who was actually looking out for the father. It could have been worse- much worse.
cmorenc
Unfortunately, Lincoln's security detail at Ford's Theater that night were even more inept than John Wilkes Booth & company. He had but a single security guard who went to a nearby tavern during intermission and may not have been back when Booth approached the Presidential box. Even so, it's astonishing that in the immediate aftermath of Lee's surrender at Appomattox, when the possibility that a bitter southern sympathizer might attempt to assassinate Lincoln, that he wasn't constantly being guarded by an entire platoon of Union troops, and not so easily approached, especially without at least a thorough check for weapons.
@Dorothy A. Winsor:
"When you're a Jet, you're a Jet for life."
Needs more Natalie Wood.
Urban Suburbanite
Good on the kid for getting out of there. I suspect this guy had a history of explosive behavior way before he was unemployed and hooked on ragebait.
There's a pattern with this guy that's definitely not "satirical". And I suspect the wife is working real hard to downplay his behavior because she knows that it's going to be coming up in court. He was stalking activists, "standing guard" with a gun (I will be shocked if his social media accounts aren't loaded with murder fantasies), and was involved with the Threepers way beyond buying a hoodie or putting a sticker on his truck.
@Urban Suburbanite:
I have some sympathy for the families. It is really, really hard on them. The exchange where he snapped at her and told her to go to work and mind her own business makes me think she's afraid of him.
Another Scott
I saw this on the ProPublica page a few days ago. Didn't bother to click it.
I'm kinda taken with the penmanship. 1) It's printed. Apparently cursive writing is no longer a thing these days, so I assume the writer is substantially younger than me. 2) The "h"s have a backward tilt (while the "t"s don't) which makes me think that the writer is left-handed.
Sherlock Holmes would probably figure out that they guy grew up in a particular neighborhood in Indiana, but I'm not that observant…
Scott.
@Another Scott: Guy Reffitt is 48. In pictures, he's holding his water bottle in his left hand.
The Pale Scot
God Bless America is not a war cry.
Sounds like the "sea shanty" scene in Taxi, which I cannot find
the Bureau of Prisons will be fully justified in assigning him to a Level 5 (maximum security) facility.
No, save the money and drop him an Alabama farm prison. Ten years of getting up at 5am to pick those butternuts.
EmbraceYourInnerCrone
@Danielx: That's one of the things that always astounds me about these fools. It's right there in the Miranda rights speech "You have the right to remain silent, if you give up that right anything you say may be used against you…" Most of these idiots lawyers first advice would be do not talk about the incident. First rule of fight club, don't talk about fight club. But they HAVE to brag and posture…. The only thing that gives me hope is that his son had the courage, at only 18, to turn him in, because it was the right thing to do.
This isn't the first piece of reporting on this guy. His kids testified in court about his threats to them. And yeah, I wouldn't be surprised if the wife is afraid of him while downplaying his behavior at the same time (it's not abnormal, unfortunately)
https://www.nbcdfw.com/news/judge-refuses-to-free-north-texas-capitol-riot-suspect-linked-to-militia/2632042/?amp
My Alma Mater, Marshall University, in Huntington WV, won the NCAA National Championship in Division 1 Soccer — 1-nil — versus Indiana U playing in Cary NC last night.
Whoot~!!~ They also won the NCAA Football championship in Div II some years back, of which I have the sweatshirt. I'll need another few shirts asap.
I'm not a big soccer fan, really, but when your old school does that well out of nowhere, ya gotta cheer a little bit.
Also — WVU frequently wins the NCAA rifle target shooting championship, having won more than any of the military academies. Look it up. Whoot! That would be the alma mater of every other member of my immediate family, parents, bro RWNJ, wife. We grew up leaving for Morgantown at 5 am at least once a year to see the Old Gold and Blue play Pitt or Penn State in the Very long ago.
Soprano2
One thing that stuck out to me is that he thinks our founding documents are "resting" in the Capitol building. They aren't, they "rest" in the National Archives. I actually saw them myself! It's amazing to me how these people think they know simple things, but are wrong about them
ETA – the "God Bless America" thing. It's one of the things I hate Bud Selig for, codifying the idea of singing that song in the 7th inning. It's not that I don't like the song, it's that it doesn't need to be sung at a ballgame! They always make you stand for it. As far as I know, there's no requirement to stand up when "God Bless America" is sung.
narya
I want all of these jamokes in prison–in SEPARATE prisons, preferably ones that don't have fellow travelers in the population, if there's a way to determine that.
Calouste
@Enhanced Voting Techniques: Yep, it's my theory that these people's emotional development comes to a screeching halt one time when their mom asks them to clean up their room or eat their broccoli or something. Around age 13-14.
@Soprano2: Right there with you on the God Bless America thing..I also hate the military flyovers and unfurling giant flags on the field for sporting events. It conflates sporting events with military might etc and irritates the hell out of me. There are way too many "macho" behaviors expected and encouraged in sports already.
rikyrah
@schrodingers_cat:
They must be punished. Lack of consequence will embolden them.
ICAM.
These muthaphuckas would have KILLED Madam Speaker if they had gotten their hands on her.
You know it. I know it.
Pence's azz WAS DEAD TOO.
They had the phucking gallows prepared for him.
@rikyrah:
Yes, but it would have been a peaceful lynching, right? NOT!!
Actually, I don't think the "gallows" those monsters built was capable of actually hanging anyone, but it sure made their point — We're not a peaceful demonstration, we're a lynch mob!!~
Steeplejack on TGIFriday Morning Open Thread: Busy, Busy, Busy (Jan 27, 2023 @ 12:46pm)
Geminid on TGIFriday Morning Open Thread: Busy, Busy, Busy (Jan 27, 2023 @ 12:44pm)
Ken on TGIFriday Morning Open Thread: Busy, Busy, Busy (Jan 27, 2023 @ 12:43pm)
mrmoshpotato on TGIFriday Morning Open Thread: Busy, Busy, Busy (Jan 27, 2023 @ 12:43pm)


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