# If $\phi \in C^1_c(\mathbb R)$ then $\lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$.

Let $\phi \in C^1_c(\mathbb R)$. Prove that $$\lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0).$$

Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. $$\vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}.$$ Now, I would use the well-known fact that $$\int_\mathbb R \frac{\sin x}{x} \, dx = \pi.$$ On the other hand, by substitution rule, we have also $$\int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi.$$ Indeed, I would like to estimate the quantity $$\begin{split} & \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ & = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ & \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert \phi(x)-\phi(0) \right\vert dx \end{split}$$ but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.

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## 4 Answers

Note that $$\small \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\pi\phi(0)= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\phi(0)\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}dx= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx$$ Denote $$\small I_{m}(n):=\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\qquad I(n):=\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx$$ We claim that $I_m(n)$ converges to $I(n)$ uniformly by $n\in\mathbb{N}$ when $m\to\infty$. Indeed, since $\phi$ is compactly supported $$\small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|I_m(n)-I(n)\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\right|=$$ $$\small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}(-\phi(0))dx\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\phi(0)\int\limits_{\mathbb{R}\setminus[-\pi mn,\pi mn]}\frac{\sin(y)}{y}dy\right|=$$ $$\small |\phi(0)|\lim\limits_{m\to\infty}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(y)}{y}dy\right|=0$$ Since convergence is uniform by $n\in\mathbb{N}$ we can say $$\small \lim\limits_{n\to\infty} I(n)= \lim\limits_{n\to\infty} \lim\limits_{m\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx$$ Since $\varphi\in C_c^1(\mathbb{R})$, then the function $x^{-1}(\varphi(x)-\varphi(0))$ is in $L^1([-\pi m,\pi m])$ for all $m\in\mathbb{N}$. Then by Riemann–Lebesgue lemma $$\small \lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx=0$$ so $\small\lim\limits_{n\to\infty}I(n)=0$. This is exactly what we wanted to prove.

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What a simple and excellent proof! Thank you very much for your kindness. – Romeo Jul 24 '12 at 21:19
@Romeo, not at all – Norbert Jul 24 '12 at 21:20
I saw the edit, but not how to get the bound. – Davide Giraudo Jul 24 '12 at 21:40
Davide, it is more possible for me to mistake then you not to see the bound – Norbert Jul 24 '12 at 21:41
Note that $\varphi$ is compcatly supported, then does its derivative. So we can integrate over support of $\phi$. The measure  of support is finite (since support is bounded) and derivative is bounded since it is continuous – Norbert Jul 24 '12 at 21:44
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A quick proof of this can be given using the Riemann-Lebesgue lemma, which is covered in Rudin and a number of other texts. Write your limit as $$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}\, dx + \lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(0)\chi_{[-1,1]}(x)}{x}\, dx$$ Here $\chi_{[-1,1]}(x)$ denotes the characteristic function of $[-1,1]$. Since $\phi(x) \in C_c^1({\mathbb R})$, the function ${\displaystyle \frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}}$ is a bounded function with compact support; the only place you have to worry about is $x = 0$ and you can use the mean value theorem for example to show it's bounded near $x = 0$. Since the function is bounded function with compact support it is in $L^1$, which is enough to apply the Riemann-Lebesgue lemma and say the first term goes to zero. As for the second term, after changing variables to $y = nx$ we may rewrite it as $$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(y)\frac{\phi(0)\chi_{[-n,n]}(y)}{y}\, dy$$ $$= \phi(0)\lim_{n \rightarrow \infty} \int_{-n}^n \frac{\sin(y)}{y}\, dy$$ $$= \phi(0)\int_\mathbb R \frac{\sin(y)}{y}\, dy$$ $$= \pi \phi(0)$$ So this will be the overall limit.

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Nice, and doesn't use further derivatives than first one. – Davide Giraudo Jul 26 '12 at 15:56
Only need that $\phi(x)$ is a compactly supported Lipschitz function... – Zarrax Jul 26 '12 at 17:49
If you assume it, then the argument for the convergence on $0$ may not follow by mean value theorem, but just using boundedness of the ratio $\frac{\phi(x)-\phi(0)}x$. – Davide Giraudo Jul 26 '12 at 18:33

We can assume WLOG that $\phi\in C^3_c(\Bbb R)$, since this subset is dense for the supremum norm and $\frac{\sin x}x$ is integrable over compact subsets. We have $$\phi(x)-\phi(0)=x\phi'(0)+x\int_0^1(1-s)\phi''(sx),$$ hence, if the support of $\phi$ is contained in $[-R,R]$ \begin{align} \small \int_{-\infty}^{+\infty}\frac{\sin(nx)}x\phi(x)dx&=\small\phi(0)\int_{-R}^R\frac{\sin(nx)}xdx+\phi'(0)\int_{-R}^R\sin(nx)dx+\int_{—R}^R\sin(nx)\int_0^1(1-s)\phi''(sx)dsdx\\ &=\small\phi(0)\int_{-nR}^{nR}\frac{\sin t}tdt-\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R+\int_{—R}^R\frac{\cos(nx)}n\int_0^1 s(1-s)\phi'''(sx)dsdx \end{align} We have $$\lim_{n\to+\infty}\int_{-nR}^{nR}\frac{\sin t}tdt=\int_{-\infty}^{+\infty}\frac{\sin t}tdt;$$ $$\left|\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R\right|\leq \frac 2n\sup_{t\in \Bbb R}|\phi''(t)|,$$ and $$\left|\int_{—R}^R\frac{\cos(nx)}n\int_0^1 s(1-s)\phi'''(sx)dsdx\right|\leq \frac{2R}n\sup_{t\in \Bbb R}|\phi'''(t)|$$

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Assume that $\phi(x)$ is supported in $|x|< L$. Since $\phi$ is differentiable, $\frac{\phi(x)-\phi(0)}{x}$ is bounded and therefore integrable on $|x|<L$. $$\begin{align} \int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x &=\pi\,\phi(0)+\int_{-\infty}^\infty\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x\\ &=\pi\,\phi(0)+\color{#C00000}{\int_{-L}^L\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x}\\ &-2\,\phi(0)\color{#00A000}{\int_{nL}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x}\\ \end{align}$$ As $n\to\infty$, the red integral vanishes by the Riemann-Lebesgue Lemma and the green integral vanishes because The Dirichlet Integral converges. This leaves us with $$\lim_{n\to\infty}\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x=\pi\,\phi(0)$$

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