# Kolmogorov-Riesz theorem

I will show an ${L^1}$ version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.

Definition In a metric space, a set is totally bounded, if for any fixed ${\varepsilon>0}$, the set can be covered by finitely many open balls of radius ${\varepsilon}$.

Lemma In a metric space, a set ${E}$ is compact if and only if ${E}$ is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain)) Let ${\mathcal{F}}$ be a bounded set of ${L^1({\mathbb R}^n)}$ functions, Assume that ${\forall \varepsilon >0}$, there exists a ${\delta >0}$ such that

$\displaystyle \| f(\cdot + y) - f \|_{L^1({\mathbb R}^n)} \le \varepsilon$

whenever ${|y| < \delta}$, for all ${f \in \mathcal{F}}$. Then ${\mathcal{F}_{| \Omega}}$ has a compact closure in ${L^1(\Omega)}$ for any finite measure set ${\Omega \subset {\mathbb R}^n}$.

Proof: Since ${L^1 (\Omega)}$ is complete, by the above lemma we need to show ${\mathcal{F}_{| \Omega}}$ is totally bounded. The idea is first find a bounded domain ${E \subset \Omega}$ with almost same measure, use Arzela-Ascoli theorem to get compactness for ${\mathcal{F}_{| E}}$, then we have finite open balls covering ${\mathcal{F}_{| E}}$. Enlarge those balls a bit to get a finite cover for ${\mathcal{F}_{| \Omega}}$.

Let ${\{ \rho_n \}}$ be a sequence of ${C_c^\infty}$ mollifiers with supp${\rho_n \subset B(0, \frac{1}{n})}$. Fix ${\varepsilon > 0}$, also fix ${n}$ such that ${\frac{1}{n} < \delta}$, we have

$\displaystyle \| \rho_n * f - f \|_{L^1({\mathbb R}^n)} \le \int_{{\mathbb R}^n} \int_{B(0, \frac{1}{n})} |\rho_n(y)| |f(x-y) - f(x)| \,dydx < \varepsilon \quad (1)$

, for any ${f \in \mathcal{F}}$. Direct verification shows for any ${f \in \mathcal{F}}$,

$\displaystyle \| \rho_n * f \|_{L^\infty} \le \| \rho_n \|_{L^\infty} \| f\|_{L^1} \le C(n) \quad (2)$

$\displaystyle \| \nabla(\rho_n * f)\|_{L^\infty} = \| \nabla\rho_n * f \|_{L^\infty} \le \| \nabla\rho_n \|_{L^\infty} \| f\|_{L^1} \le C(n)$

which implies

$\displaystyle |\rho_n * f(x_2) - \rho_n * f(x_1)| \le C(n) |x_2-x_1| \quad (3)$

By (2) and the inner regularity of measurable sets, we can find a bounded measurable set ${E}$ such that ${\| \rho_n * f \|_{L^1(\Omega \setminus E)} < \varepsilon}$. Therefore

$\displaystyle \| f \|_{L^1(\Omega \setminus E)} \le \| \rho_n * f - f \|_{L^1({\mathbb R}^n)} + \| \rho_n * f \|_{L^1(\Omega \setminus E)} < 2 \varepsilon \quad (4)$

By (2) and (3) we know the functions in the set ${\rho_n* \mathcal{F}}$ are bounded and equicontinuous, so we can apply Arzela-Ascoli theorem and hence ${(\rho_n* \mathcal{F})_{|E}}$ has a compact closure in ${C(\bar{E)}}$. This means for every sequence in ${(\rho_n* \mathcal{F})_{|E}}$, there exists a uniformly convergent subsequence. Since the subsequence also converges in ${L^1(E)}$, we know that ${(\rho_n* \mathcal{F})_{|E}}$ has a compact closure in ${L^1(E)}$. Therefore, ${(\rho_n* \mathcal{F})_{|E}}$ can be covered by finite ball of radius ${\varepsilon}$ in ${L^1(E)}$, say ${B(g_i , \varepsilon)}$. i.e.

$\displaystyle (\rho_n* \mathcal{F})_{|E} \subset \cup_{i=1}^m B(g_i , \varepsilon)$

, where ${g_i \in L^1(E)}$. Now define

$\displaystyle \tilde{g}_i(x) \in L^1(\Omega) = \begin{cases} g_i(x), \quad &\text{if } x\in E\\ 0, &\text{if } x\in \Omega \setminus E \end{cases}$

Finally I claim that ${\mathcal{F}_{| \Omega} \subset \cup_{i=1}^m B(\tilde{g}_i , 4\varepsilon)}$. For any ${f \in \mathcal{F}_{| \Omega}}$,

$\displaystyle \| f - \tilde{g}_i \|_{L^1(\Omega)} \le \|f\|_{L^1(\Omega \setminus E)} + \| f - g_i\|_{L^1(E)}$

$\displaystyle \le \|f\|_{L^1(\Omega \setminus E)} + \| \rho_n * f - f \|_{L^1({\mathbb R}^n)} +\| \rho_n * f - g_i \|_{L^1(E)}$

Choose ${g_i}$ such that ${\rho_n * f \in B(g_i , \varepsilon)}$, along with the estimate (1) and (4), we have

$\displaystyle \| f - \tilde{g}_i \|_{L^1(\Omega)} < 4 \varepsilon.$

This proves that ${\mathcal{F}_{| \Omega}}$ is totally bounded, hence ${\mathcal{F}_{| \Omega}}$ has a compact closure in ${L^1(\Omega)}$. $\Box$

Theorem 2 (Kolmogorov Riesz Theorem (infinite measure domain)) Let ${\mathcal{F}}$ be a bounded set of ${L^1({\mathbb R}^n)}$ functions, Assume that ${\forall \varepsilon >0}$, there exists a ${\delta >0}$ such that

$\displaystyle \| f(\cdot + y) - f \|_{L^1({\mathbb R}^n)} \le \varepsilon$

whenever ${|y| < \delta}$, for all ${f \in \mathcal{F}}$. Furthermore, ${\forall \varepsilon >0}$, there exists an ${\Omega \subset {\mathbb R}^n}$ with finite measure such that

$\displaystyle \|f\|_{L^1({\mathbb R}^n \setminus \Omega)} < \varepsilon, \quad \forall f \in \mathcal{F}. \quad (5)$

Then ${\mathcal{F}}$ has a compact closure in ${L^1({\mathbb R}^n)}$.

Proof: By the previous theorem, fix ${\varepsilon > 0}$, ${\mathcal{F}_{| \Omega} \subset \cup_{i=1}^m B(g_i , \varepsilon)}$. Define

$\displaystyle \tilde{g}_i(x) \in L^1({\mathbb R}^n) = \begin{cases} g_i(x), \quad &\text{if } x\in \Omega\\ 0, &\text{if } x\in {\mathbb R}^n \setminus \Omega \end{cases}$

Then by condition (5) and similar calculation in the previous theorem, ${\mathcal{F} \subset \cup_{i=1}^m B(\tilde{g}_i , 2\varepsilon)}$. $\Box$

Remark The condition (5) for infinite measure domain is essential. A counterexample without (5) would be as following.Let ${f}$ be any ${C_c^\infty({\mathbb R})}$ function, with supp${f \subset (0,1)}$. Let ${f_n(x) := f(x+n)}$, and ${\mathcal{F} = \{ f_n \}_{n = 1}^\infty}$. So ${\mathcal{F}}$ is a bounded set,

$\displaystyle \| f_n(\cdot + y) - f_n \|_{L^1({\mathbb R})} = \| f(\cdot + y) - f \|_{L^1({\mathbb R})} <\varepsilon$

whenever ${|y| < \delta(\varepsilon)}$. However,

$\displaystyle \| f_n - f_m\|_{L^1({\mathbb R})} = 2\| f \|_{L^1({\mathbb R})}$

if ${n \neq m}$. This implies ${\{ f_n\}}$ does not have any convergent subsequence. Hence ${\mathcal{F}}$ does not have compact closure in ${L^1({\mathbb R})}$.