Equivalence of compactness in metric space

Lemma In a metric space {(X,d)}, if {\{C_k\}_{k=1}^\infty \subset X} is a sequence of compact sets such that {C_{k+1} \varsubsetneq C_k} for {k \in {\mathbb N}} and {\lim_{k \rightarrow \infty}} diam{C_k = 0}. Then {\cap_{k=1}^\infty C_k = \{c\}}, where {c} is a point in {X}.

Proof: First note that {\cap_{k=1}^\infty C_k} cannot contain more than 1 points. If {a,b \in \cap_{k=1}^\infty C_k}, then {a,b \in C_k \quad \forall k}. Since {\lim} diam{C_k}=0, it forces {a=b}.

It remains to show {\cap_{k=1}^\infty C_k} is nonempty. Assume it is empty, then {\cup_{k=1}^\infty (X \setminus C_k ) = X \setminus (\cap_{k=1}^\infty C_k) = X}. Since metric space is Hausdorff, {C_k} is closed for all {k} and {\{ X \setminus C_k \}} is an open cover for {C_1}. Since {C_1} is compact, there exists finite subcover such that {C_1 \subset \cup_{i=1}^m (X\setminus C_{k_i}) = X \setminus C_{k_m}}, which contradicts {C_{k_m} \varsubsetneq C_1}. \Box

Theorem In a metric space {(X,d)} (no matter it has countable basis or not), let {E} be a subset in {X}. The following statement are equivalent by assuming axiom of choice:

  1. {E} is compact;
  2. {E} is sequentially compact;
  3. {E} is complete and totally bounded.

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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}.

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Geometry Hahn Banach theorem for weak* topology

We have geometric Hahn Banach theorem in standard functional analysis course, saying that there exists a separating hyperplane separates two special sets in usual topology. However this theorem can be used for some special sets in weak* topology. This is the problem 9 in the functional analysis book by Prof. Brezis.

Theorem 1 Let {E} be a Banach space, {A,B \subset E^*} be two nonempty disjoint convex subsets. Assume {A} is open in weak* topology. Then there exist some {x \in E, x \neq 0} and a constant {\alpha} such that the hyperplane {\{ f \in E^* : \langle f, x \rangle_{E^* \times E} = \alpha \}} separates {A} and {B}.

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Composition of sobolev functions

The following materials are exercise 8.10 and 8.11 from the functional analysis book by Prof. Brezis. The questions arise from whether we can well define a composition of two W^{1,p} functions or not. From Corollary 8.11 we know if G \in C^1, u \in W^{1,p}, G(u) \in W^{1,p} is well defined. However bad things might happen if G \in W^{1,p}. For instance, G(x) = |x|, then G(u) might be bad on the set \{ u = 0\}. I thought we could still define things in the sense of distribution, but Prof. Brezis said in this special case we could have something more than that, because of the following propositions. The proof is not hard by following the hints, but the idea is really clever (IMO)!

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