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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 $2\leq p\leq+\infty$

Proof:

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

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  • 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
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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 $\left\lVert f \right\rVert_\infty\leqslant \sup_{n\geqslant 1}\left\lVert f_{n} \right\rVert_\infty$.

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