Kolmogorov - M. Riesz - Fréchet Theorem

Question

I need help to understand, some steps of the proof of this theorem.

(Kolmogorov-M. Riesz-Fréchet) Let $\mathcal{F}$ be a bounded set in $L^p(\mathbb{R}^N)$ with $1\leq p < \infty$. Assume that

\begin{equation} \lim\limits_{|h|\longrightarrow 0 }\|\tau_hf-f\|_p=0 \ uniformly \; in \, f \in \mathcal{F}, \end{equation}

i.e $\forall \varepsilon >0 \; \exists \delta >0$ such that $\|\tau_hf-f\|_p<\varepsilon \; \forall f \in \mathcal{F}, \; \forall h \in \mathbb{R}^N$ with $|h|<\delta$. Then the closure of $\mathcal{F}_{|\Omega}$ is compact for any measurable set in $\Omega \subset \mathbb{R}^N$ with finite measure.

Well, you can find it in Haim Brezis, Functional Analysis, Sobolev Spaces and PDE. page 111

In step 1: We claim that \begin{equation} \|(\rho_n*f)-f\|_{L^p(\mathbb{R}^N)}\leq \varepsilon \ \forall f \in \mathcal{F}, \ \forall n > 1/n. \label{4.26_kolmogorov_23} \end{equation}

And we have, \begin{equation*} \begin{split} |(\rho_n*f)(x)-f(x)| &\leq \int |(f(x-y)-f(x))\rho_n(y)dy|\\ &\leq \left[ \int |(f(x-y)-f(x))|^p\rho_n(y)dy \right]^{1/p} \end{split} \end{equation*} By Hölder´s inequality

So, i can´t understand, how use the Hölder´s inequality in the last inequality, please help me!!

Answer 1

If $\rho_n$ is a probability density function and $\mu_n(B)=\int_B\rho_n(x)dx$, then \begin{align}\int |f(x-y)-f(x)|\rho_n(y)dy&=\int|f(x-y)-f(x)|d\mu_n\\ &\le \left(\int|f(x-y)-f(x)|^{p}d\mu_n\right)^{1/p}\left(\int |1|^{q}d\mu_n\right)^{1/q}\\ &=\left(\int|f(x-y)-f(x)|^{p}\rho_n(y)dy\right)^{1/p} \end{align}

I know that you didn't mention that $\rho_n$ was a density but given that $\rho_n$ is not written as to the power of p after Hölder and the q seems to be missing this is my best guess of what happened.

Answer 2

$$\int|f(x-y)-f(x)|\rho_n(y)dy=\int|f(x-y)-f(x)|{(\rho_n(y))}^{1/p}{(\rho_n(y))^{1-1/p}}dy$$ Then use the Holder's inequality and $\int \rho_n=1$, get the result.