# The Unapologetic Mathematician

## The Extremal Case of Hölder’s Inequality

We will soon need to know that Hölder’s inequality is in a sense the best we can do, at least for finite $p$. That is, not only do we know that for any $f\in L^p$ and $g\in L^q$ we have $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$, but for any $f\in L^p$ there is some $g\in L^q$ for which we actually have equality. We will actually prove that

$\displaystyle\lVert f\rVert_p=\max\left\{\left\lvert\int fg\,d\mu\right\rvert\Big\vert\lVert g\rVert_q\leq1\right\}$

That is, not only is the integral bounded above by $\lVert f\rVert_p\lVert g\rVert_q$ — and thus by $\lVert f\rVert_p$ — but there actually exists some $g$ in the unit ball which achieves this maximum.

Hölder’s inequality tells us that

$\displaystyle\left\lvert\int fg\,d\mu\right\rvert\leq\int\lvert fg\rvert\,d\mu\leq\lVert f\rVert_p\lVert g\rVert_q\leq\lVert f\rVert_p$

so $\lVert f\rVert_p$ must be at least as big as every element of the given set. If $\lVert f\rVert_p=0$, then it’s clear that the asserted equality holds, since $f=0$ a.e., and so $0$ is the only element of the set on the right. Thus from here we can assume $\lVert f\rVert_p>0$.

We now define a function $g$. At every point $x$ where $f(x)=0$ we set $g(x)=0$ as well. At all other $x$ we define

$\displaystyle g(x)=\lVert f\rVert_p^{1-p}\frac{\lvert f(x)\rvert^p}{f(x)}$

In the case where $p=1$ we will verify that $\lVert g\rVert_\infty=1$. That is, the essential supremum of $g$ is $1$. And, indeed, we find that $g(x)=1$ at points where $f(x)>0$, and $g(x)=-1$ at points where $f(x)<0$.

If $1<p<\infty$, then we check

$\displaystyle\begin{aligned}\lVert g\rVert_q&=\left(\int\lvert g\rvert^q\,d\mu\right)^\frac{1}{q}\\&=\left(\int\lVert f\rVert_p^{-p}\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\lVert f\rVert_p^p\right)^\frac{1}{q}\\&=1\end{aligned}$

In either case, it’s easy to see that

$\displaystyle\int fg\,d\mu=\lVert f\rVert_p$

as asserted.

Posted by John Armstrong | Analysis, Measure Theory | 2 Comments

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