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})}.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s