a) Let $1\leq p_1\leq p\leq p_2\leq \infty$ and for $\alpha \in [0,1]$

$\frac {1}{p}=\frac {\alpha}{p_1}+\frac {1- \alpha}{p_2}$

Prove that if $f\in L^{p_1}\cap L^{p_2}$, then $f\in L^p$ and we have the following inequality:

$||f||_{L^p}\leq ||f||^\alpha_{L_{p_1}}||f||^{1-\alpha}_{L_{p_2}}$

b) Let $f_n$ be a bounded sequence in $L^\infty([0,1])$ and $f_n\to f$ in $L^2([0,1])$. Prove that $f$ is in $L^\infty([0,1])$ and that $f_n\to f$ in $L^p$ for any $2\leq p\leq+\infty$


a) From: $\frac {1}{p}=\frac {\alpha}{p_1}+\frac {1- \alpha}{p_2}$ we have that:

$1=\frac {\alpha p}{p_1}+\frac {(1- \alpha)p}{p_2}$

So we use the H$\ddot{o}$lder inequality with $p=\frac{p_1}{\alpha p},q=\frac{p_2}{(1-\alpha)p}$

$||f||_{L^p}^p=\int_0^1|f|^pdx=\int_0^1|f|^{\alpha p}|f|^{(1-\alpha) p}dx\leq(\int_0^1(|f|^{\alpha p })^\frac{p_1}{\alpha p }dx)^\frac{\alpha p}{p_1}(\int_0^1(|f|^{(1-\alpha) p })^\frac{p_2}{(1-\alpha )p }dx)^\frac{(1-\alpha) p}{p_2}=(||f||^\alpha_{L_{p_1}}||f||^{1-\alpha}_{L_{p_2}})^p$

b) From the previous Interpolation inequality we can take $p_1=2$ and $p_2=\infty$ and have

$\frac{1}{p}=\frac{\alpha}{2}+0$ so $\alpha = \frac{2}{p} \in [0,1]$ since $p\geq2$

We have:

$||f-f_n||_{L^p}\leq ||f-f_n||_{L^2}^{\frac{2}{p}}||f-f_n||_{L^\infty}^{1-\frac{2}{p}} $

By assumption we have the $L^2$ convergence $f_n\to f$, so it is left to show that $L^\infty$ norm of $f_n-f$ is bounded, this is where I have a small confusion.

There is a sequential form of Banach-Alaoglu which gives us the existence of a weak-$\star$ converging subsequence with $||f||_{L^\infty}\leq M$ but this result is quite strong and somethings tells me that there must be a more straightforward way to show it. My attempt is the following:

Since $f_n\to f$ in $L^2$, there exists a subsequence $f_{n_k}$, which converges pointwise to $f$ almost everywhere. We have:

$||f_n-f||_{L^\infty}=\operatorname{esssup}_{x\in[0,1]}|f_n(x)-f(x)|\leq \operatorname{esssup}_{x\in[0,1]}|f_n(x)|+\operatorname{esssup}_{x\in[0,1]}|f(x)|=||f_n||_{L^\infty}+\operatorname{esssup}\sup_{x\in[0,1]}\lim_{k\to \infty}|f_{{n}_k}(x)|\leq ||f_n||_{L^\infty}+\liminf_{k\to \infty}\operatorname{esssup}_{x\in[0,1]}|f_{{n}_k}(x)|=||f_n||_{L^\infty}+||f_{n_k}||_{L^\infty}\leq 2M$.

Could You please take some time and go through my proof? I would really appreciate it.

Sorry for a long post

  • 2
    $\begingroup$ You could leave out $f_n$ in the last big equation just to simplify it. The post looks fine. $\endgroup$ – daw Feb 13 at 11:00
  • $\begingroup$ For part b) you could alternatively show that $f\in (L^1)^\ast=L^\infty$. $\endgroup$ – MaoWao Feb 13 at 11:02

What you did is essentially correct. Here is a remark.

Before we write things like $\left\lVert f-f_n\right\rVert_\infty$, we have to be sure that $f-f_n$ is in $\mathbb L^\infty$. Since $f_n\in\mathbb L^\infty$, we have to prove that $f\in\mathbb L^\infty$. We can follow the ideas you use: let $\left(f_{n_k}\right)_{k\geqslant 1}$ be an almost everywhere convergent subsequence. Then for almost every $x$, $$\left\lvert f(x)\right\rvert\leqslant \lim_{k\to +\infty}\left\lvert f_{n_k}(x)\right\rvert\leqslant \lim_{k\to +\infty}\left\lVert f_{n_k} \right\rVert_\infty\leqslant \sup_{n\geqslant 1}\left\lVert f_{n} \right\rVert_\infty,$$ which shows not only that $f\in\mathbb L^\infty$ but also that $\left\lVert f \right\rVert_\infty\leqslant \sup_{n\geqslant 1}\left\lVert f_{n} \right\rVert_\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.