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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