2
$\begingroup$

Let $f: S \to \mathbb{R}$ be a Lebesgue integrable function on a measure space $(S,\Sigma,\mu)$, then we may have \begin{align} \|f\|_{L^1} &:= \int_S |f| d\mu \\ \|f\|_{L^2} &:= \left(\int_S |f|^2 d\mu\right)^{\frac{1}{2}} \\ \|f\|_{L^\infty} &:= \inf\{C\ge0: |f|\le C \quad \text{a.e.-}\mu \} \end{align}

Is there any inequality like \begin{align} \|f\|_{L^1} \le \|f\|_{L^2} \le \|f\|_{L^\infty} \end{align} And why?

I am asking because I saw an inequality in some handout that \begin{align} |(uv,w)_{L^2}| \lesssim \|u\|_{L^\infty}\|v\|_{L^2}\|w\|_{L^2} \end{align} I know the first step the Cauchy-Schwarz inequality is applied so \begin{align} |(uv,w)_{L^2}| \lesssim \|u v\|_{L^2}\|w\|_{L^2} \end{align} but I don't understand why there holds \begin{align} \|u v\|_{L^2}\|w\|_{L^2} \lesssim \|u\|_{L^\infty}\|v\|_{L^2}\|w\|_{L^2} \end{align}

In particular, why does there hold \begin{align} \|u v\|_{L^2} \lesssim \|u\|_{L^\infty}\|v\|_{L^2} \end{align}

It looks like Holder's inequality applied but the $p,q$ do not satisfy conjugates. Can you show me how to use Holder's inequality to derive this inequality?

$\endgroup$
2
  • $\begingroup$ If $u\leq M$ a.s. then $\sqrt{ \int u^2v^2 } \leq \sqrt{ \int M^2v^2 } =M \sqrt{ \int v^2 } $ $\endgroup$
    – clark
    May 21, 2018 at 21:47
  • $\begingroup$ Thank you for your simple and clear answer! But I still want to know if we can, perhaps, derive this from Holder's inequality. $\endgroup$ May 21, 2018 at 21:55

2 Answers 2

4
$\begingroup$

It's not Hölder. You have $$ \|uv\|_2^2=\int |uv|^2=\int |u|^2\,|v|^2\leq\|u\|_\infty^2\,\int |v|^2=\|u\|_\infty\,\|v\|_2^2. $$ It's just that $|u|\leq\|u\|_\infty$ a.e.

$\endgroup$
2
$\begingroup$

This is the Holder's inequality for $q=1$ and $p=\infty$ i.e.

\begin{align*} \|v u\|_2 &=\sqrt{\|v^2 u^2 \|_1}\\ &\leq \sqrt{\| v^2\|_\infty}\sqrt{ \| u^2\|_1}\\ &=\|v\|_\infty \|u\|_2 \end{align*} And for the proof someone can see Martin's answer.

In certain cases the previous version of Holder's can be, easily, derived from the more popular one, i.e. when $1<p,q<\infty$. Indeeed, $$ \int v^2 u^2 dx \leq \sqrt[p]{\int v^{2p}} \sqrt[q]{\int u^{2q}} $$

For $\frac{1}{q}+\frac{1}{p}=1$. When $ q\longrightarrow 1$ then $p\longrightarrow \infty$.

Someone can argue, in certain cases, that $\sqrt[p]{\int v^{2p}}\longrightarrow \| u\|^2 _\infty$.

Again, under certain conditions, it can be shown that $\int u^{2q} \rightarrow \int u^2 $.

$\endgroup$
2
  • $\begingroup$ Any written criticism, would be welcome:), as I am not sure what is wrong with the post. $\endgroup$
    – clark
    May 22, 2018 at 0:56
  • $\begingroup$ I am not really sure, but I believe that for Hölder's inequality to hold, the requirement between exponents is that $\frac{1}{p}+\frac{1}{q} = 1$, and in you case $p=2, q=\infty$ it don´t fullfill it. Also about the inequalities for individual norms, I make the same question recently here $\endgroup$
    – Joako
    Oct 18, 2021 at 13:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .