Proof of Gelfand-Phillips Theorem

\textbf{Theorem.} Let X be a Banach space and A\subseteq X. Prove that A is precompact (in the norm) if and only if for every w^*-convergent to \textbf{0} sequence \{x_n^*\}_{n\ge 1}\subseteq X^* it holds that \{x_n^*\}_{n\ge 1} converges uniformly to \textbf{0} on  A.

 

\textbf{Proof.} If A is precompact the result follows from Arzela-Ascoli theorem (combined with the fact that w^*-convergent sequences are norm bounded).

Now we prove the reverse direction. 

First we proof the following characterization of compactness in normed spaces: A is compact if and only if A is bounded and for any \varepsilon>0 there exists a finite-dimensional space F such that A\subseteq F+\varepsilon\mathbf{B}.

\textit{Proof}. 

If A is precompact, then for any \varepsilon>0 there exists \{a_i\}_{i=1}^n\subseteq A such that A\subseteq\bigcup_{i=1}^n\mathbf{B}_{\varepsilon}(a_i). In particular A is bounded and
A\subseteq \text{span}(\{a_i\}_{i=1}^n)+\varepsilon\textbf{B}.
For  the reverse direction, let \varepsilon>0 and F be finite-dimensional space such that A\subseteq F+\frac{\varepsilon}{3}\mathbf{B}. For each a\in A choose f_a\in F with \|a-f_a\|\le\varepsilon/3. Since A is bounded and F is finite-dimensional, the set \{f_a\}_{ a\in A} is precompact.
Thus there exists \{f_{a_i}\}_{i=1}^n\subseteq F such that \{f_a\}_{a\in A}\subseteq \bigcup_{i=1}^n\textbf{B}_{\varepsilon/3}(f_{a_i}).
Now let a\in A be arbitrary. Then there exists i\in \{1,2,\ldots,n\} such that \|f_a-f_{a_i}\|<\varepsilon/3. Consequently
\|a-a_i\|\le\|a-f_a\|+\|f_a-f_{a_i}\|+\|f_{a_i}-a_i\|<\varepsilon,
hence \{a_i\}_{i=1}^n\subseteq A is a finite \varepsilon-net for A. \square

Now let A be a set which is not precompact. If it is unbounded, the results follows easily. Assume otherwise. Thus there exists \varepsilon>0 such that for any finite-dimensional space F, there exists a\in A with \text{d}(a,F)\ge\varepsilon. Choose a sequence \{x_i\}_{i\ge 1}\subseteq X which is dense in X. Let F_n:=\text{span}(\{x_i\}_{i=1}^n). Choose a_n\in A with \text{d}(a_n,F_n)\ge\varepsilon. Using Hahn-Banach construct x_n^*\in X^* such that \langle x_n^*,a_n\rangle\ge\varepsilon, \|x_n^*\|=1 and F_n\subseteq\text{ker}(x_n^*).
Thus for every n \sup_{a\in A}|\langle x_n^*,a\rangle|\ge\varepsilon,
hence \{x_n^*\}_{n\ge 1} does not converge uniformly to \mathbf{0} on A. On the other hand \bigcup_{n\ge 1}F_n is dense in X. Fix x\in X and let \delta>0 be arbitrary. Then there exists m such that for some z_m\in F_m it holds that \|z_m-x\|<\delta. For n\ge m it holds that \langle x_n^*,z_m\rangle=0, hence
|\langle x_n^*,x\rangle|= |\langle x_n^*,z_m-x\rangle|\le \|x_n^*\|\|z_m-x\|<\delta.
Consequently \{x_n^*\}_{n\ge 1} is weak^* convergent to \mathbf{0}.

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