I want to prove the following theorem:
Theorem (Kolmogorov-Riesz)
Let $1\leq p<\infty$ and $\mathcal F\subseteq L^p := L^p(\mathbb R^d,\mathcal B,\mu)$. Then it holds that $\mathcal F$ is relatively compact iff
- $\sup_{f\in\mathcal F}\Vert f\Vert_{L^p} < \infty$
- for each $\epsilon>0$ there exists an $R>0$ such that $\Vert 1_{\mathbb R^d\setminus B_R(0)} f\Vert_{L^p} < \epsilon$ for each $f\in\mathcal F$, and
- for each $\epsilon>0$ there exists an $\rho>0$ such that $\Vert S_yf - f\Vert_{L^p} < \epsilon$ for each $h\in\mathbb R^d$ with $\Vert y\Vert < \rho$ and each $f\in\mathcal F$.
Here $S_y:L^p \rightarrow L^p$ is the shift operator given by $S_y(f)(x) = f(x + y)$.
I try to follow the proof given in this ressource: https://www.math.ntnu.no/conservation/2009/037.pdf (Theorem 5). However, there are some steps which I don't understand. I quoted the proof and added my comments/questions. Since the proof is not short, I am grateful to anyone who goes through these lines.
Proof
"$\Rightarrow$": Let $Q\subseteq\mathbb R^d$ be an open cube at the origin so that $\Vert y\Vert < \frac 12\rho$ whenever $y\in Q$.
Why does $Q$ have to be a cube? Why not take $Q = B_{\frac 12\rho}(0)$. Then for each $y\in Q$ it holds that $\Vert y - 0\Vert = \Vert y \Vert < \frac 12\rho$. I further don't see where $\Vert y\Vert < \frac 12\rho$ is used in the remainder of the proof.
Let $Q_1, Q_2, \dots, Q_N$ be mutually non-overlapping translates of $Q$ so that the closure of $\bigcup_{i=1}^N Q_i$ contains $B_R(0)$.
What's the point in defining theses $Q_i$'s? I think I know why they have to be disjoint, but why "translates"? The translate (by $b$) of a set $A$ is defined as $A+b$, isn't? I guess the idea is to cover $B_R(0)$, so why not chose a partition of $B_R(0)$? I also don't see why it's not enough that the $Q_i's$ cover $B_R(0)$, but why we need that the closure of $\bigcup_{i=1}^N Q_i$ contains $B_R(0)$.
Let $P:L^p\rightarrow L^p$ be the projection map of $L^p$ onto the linear span of $\{\mathsf 1_{Q_1}, \mathsf 1_{Q_2}, \dots, \mathsf 1_{Q_N}\}$ given by $$Pf(x) = \sum_{i=1}^N\mathsf 1_{Q_i}(x)\frac 1{\mu(Q_i)}\int_{Q_i} f(z)\ \mathrm dz.$$
Why do I want to project onto the linear span of $\{\mathsf 1_{Q_1}, \mathsf 1_{Q_2}, \dots, \mathsf 1_{Q_N}\}$?
From 2. and the definition of $Pf$ it follows for $f\in\mathcal F$ that \begin{align*}\Vert f - Pf\Vert_{L^p}^p &< \epsilon^p + \sum_{i=1}^N \int_{Q_i}\vert f(x) - Pf(x)\vert^p\ \mathrm dx \newline &= \epsilon^p + \sum_{i=1}^N \int_{Q_i}\left\vert \frac 1{\mu(Q_i)}\int_{Q_i}(f(x) - f(z)) \mathrm dz\right\vert^p\ \mathrm dx.\end{align*}
I think I know what's going on here: $$\Vert f - Pf\Vert_{L^p}^p = \int_{\mathbb R^d\setminus B_R(0)} \vert f(x) - Pf(x)\vert^p\ \mathrm dx + \int_{B_R(0)} \vert f(x) - Pf(x)\vert^p\ \mathrm dx$$ The $Pf(x)$ in the first term is zero for all $x\in\mathbb R^d\setminus B_R(0)$. So only $$\int_{\mathbb R^d\setminus B_R(0)}\vert f(x)\vert^p\ \mathrm dx$$ remains, which can be bounded by $\epsilon^p$ according to 2. Since $B_R(0)$ is contained in the closure of $\bigcup_{i=1}^NQ_i$, we have that
$$\int_{B_R(0)} \vert f(x) - Pf(x)\vert^p\ \mathrm dx \leq \int_{\bigcup_{i=1}^NQ_i} \vert f(x) - Pf(x)\vert^p\ \mathrm dx = \sum_{i=1}^N\int_{Q_i} \vert f(x) - Pf(x)\vert^p\ \mathrm dx,$$
where the equality follows from the fact that the $Q_i$'s are disjoint. Are my thoughts correct here? If so, why am I integrating over $\bigcup_{i=1}^NQ_i$ and not $\operatorname{cl}\left(\bigcup_{i=1}^NQ_i\right)$?
The equality in the last line is clear; it's just simple rearranging.
Next we use Jensen's inequality and change a variable of integration, where we note that $x-z\in 2Q$ when $x,z\in Q$: \begin{align*}\Vert f-Pf\Vert_{L^p}^p &< \epsilon^p + \sum_{i=1}^N\int_{Q_i}\frac{1}{\mu(Q_i)}\int_{Q_i}\vert f(x) - f(z)\vert^p\ \mathrm dz\ \mathrm dx \newline &\leq \epsilon^p + \sum_{i=1}^N\int_{Q_i}\frac{1}{\mu(Q_i)}\int_{2Q}\vert f(x) - f(x+y)\vert^p\ \mathrm dy\ \mathrm dx\end{align*}
The change of variables is given by $z := x + y$. But where does the $2Q$ in the integration domain come from?
\begin{align*} &\leq \frac 1{\mu(Q)}\int_{2Q}\int_{\mathbb R^d}\left\vert f(x) - f(x+y)\right\vert^p\ \mathrm dx\ \mathrm dy \newline &< \epsilon^p + \frac 1{\mu(Q)}\int_{2Q}\epsilon^p\ \mathrm dy = (2^d + 1)\epsilon^p \end{align*} by 3.
Fubini's Theorem allows interchanging the order of integration. Since $\sum_{i=1}^N\int_{Q_i} = \int_{\bigcup_{i=1}^NQ_i}\leq\int_{\mathbb R^d}$ it is clear where the $\mathbb R^d$ in the integration domain comes from. However, I don't know why $\frac1{\mu(Q_i)}$ becomes $\frac1{\mu(Q)}$. Is it because the $Q_i$ are translates of $Q$ and the Lebesgue measure $\mu$ is invariant unter translations?
The inequality in the second line is clear; it is an immediate consequence of 3.
I think it is also clear that $\frac 1{\mu(Q)}\int_{2Q} 1\ \mathrm dy = \frac{\mu(2Q)}{\mu(Q)} = 2^d$ as $\mu(T(Q)) = \det(T)\mu(Q)$ for any linear transformation $T:\mathbb R^d\rightarrow\mathbb R^d$. In particular, the translation $T$ given by $q\mapsto 2q$ has $\det(T) = 2^d$.
Thus $\Vert f- Pf\Vert_{L^p} < (2^d+1)^{\frac 1p}\epsilon$, and $\Vert f\Vert_{L^p} < (2^d + 1)^{\frac 1p}\epsilon + \Vert Pf\Vert_{L^p}$.
I suppose the second inequality follows from the fact that $P$ is a projection and hence $\vert\vert\vert P\vert\vert\vert = 1$. Therefore $\Vert Pf\Vert_{L^p} \leq \Vert f\Vert_{L^p}$, and the inverse Triangle inequality yields $$0\leq \vert\Vert f\Vert_{L^p} - \Vert Pf\Vert_{L^p}\vert \leq \Vert f - Pf\Vert_{L^p} < (2^d + 1)^{\frac 1p}\epsilon.$$ The absolute value is thus obsolete and adding $\Vert Pf\Vert_{L^p}$ yields the second inequality. Correct?
By the linearity of $P$, if $f,g\in\mathcal F$ and $\Vert Pf - Pg\Vert_{L^p}<\epsilon$ then $\Vert f-g\Vert_{L^p} < \big((2^d + 1)^{\frac 1p} + 1\big)\epsilon$. Moreover, since $P$ is bounded (in fact $\vert\vert\vert P\vert\vert\vert = 1$) and $\mathcal F$ is bounded by 1., the image $P[\mathcal F]$ is bounded. Since the image of $P$ is finite-dimensional, $P[\mathcal F]$ is totally bounded. Thus $\mathcal F$ is totally bounded by Lemma 1.
The image $P[\mathcal F]$ is fine-dimensional since $P$ projects onto a finte-dimensional space (the linear span of $\{\mathsf 1_{Q_1}, \mathsf 1_{Q_2}, \dots, \mathsf 1_{Q_N}\}$), right?
The remainder is clear.
Thank you very much!