# Bounded Sequence in $L^\infty$ and Interpolation in $L^p$

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

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

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