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Definition ((closed) Convex Hull) Let $M \subset X$. Then $$ \text{co}(M) := \left\{ \sum_{i = 1}^{N} \lambda_i x_i: N \in \mathbb{N}, \lambda_i \in [0,1], \sum_{i = 1}^{N} \lambda_i = 1, x_i \in M \ \forall i \in \{1, \ldots, N\} \right\} $$ Furthermore, the closed convex hull $\overline{\text{co}}(M)$ is the closure of $\text{co}(M)$ with respect to the norm on $X$.

Here (attention: this is in German and apparently "kompakt" actually means "relativ kompakt") Mazur proves the following

Theorem Let $Z$ be a precompact subset of a Banach space $X$. Then the smallest convex superset of $Z$, $W$ precompact as well.

I have some questions regarding the proof, which are detailed below, so I will try to lay out the proof here for you. I am a German native speaker so there shouldn't be anything lost in translation.

Since the statement is trivial for finite $Z$, we assume that $Z$ is infinite. Since $Z$ is precompact there exists a countable dense subset $A$. Let $A = (x_k)_{k\in \mathbb{N}} \subset Z$ be an enumerate. We define $$ V := \left\{ \sum_{n \in \mathbb{N}} a_n x_n: a_k \ge 0 \ \forall k \in \mathbb{N}, \sum_{n \in \mathbb{N}} a_n = 1 \right\}. $$ We know show that $V$ is precompact.

Let $\varepsilon > 0$. Since $A \subset Z$ is totally bounded, there exists a finite subset $(x_{n_j})_{j \in \{1, \ldots, p\}} \subset (x_n)_{n \in \mathbb{N}}$, such that for every $x_n$ there exists a $k \in \{1, \ldots, p\}$ such that \begin{align} \tag{$\star$} \| x_n - x_{n_k} \| \le \frac{\varepsilon}{2} \end{align} holds. Therefore, one can partition $(x_n)_{n \in \mathbb{N}}$ into $p$ subsequences $(x_{n_k, 1}, x_{n_k,2})_{k = 1}^{p}$, such that for all $k \in \{1, \ldots, p\}$ and all $\ell$ \begin{equation*} \| x_{n_{k, \ell}} - x_{n_k} \| \le \frac{\varepsilon}{2} \end{equation*} holds. Define \begin{equation*} S := \left\{ \sum_{k = 1}^{p} b_k x_{n_k}: b_k \ge 0 \ \forall k \in \{1, \ldots, p\}, \sum_{k = 1}^{p} b_k = 1 \right\}, \end{equation*} which is totally bounded. Therefore there exist $s_1, \ldots s_q \in S$ such that $$ \forall y \in S \ \exists k \in \{1, \ldots, q\}: \| y - y_k \| \le \frac{\varepsilon}{2}. $$ Lastly we show that $$ \forall x \in V \ \exists k\in \{1, \ldots, q\}: \| x - y_k \| < \varepsilon $$ holds. If $x \in V$, it is of the form $x = \sum_{n = 1}^{\infty} a_n x_n$. We define $$ b_k := \sum_{j = 1}^{p} a_{n_{k,j}}, \quad y := \sum_{k = 1}^{p} b_k x_{n_k} \in S. $$ Therefore there exists a $k \in \{1, \ldots, q \}$, such that $\| y - y_k \| \le \frac{\varepsilon}{2}$. Since we have \begin{align*} x - y = & a_{n_{1,1}}(x_{n_{1,1}} - x_{n_1}) + a_{n_{1,2}}(x_{n_{1,2}} - x_{n_1}) + \ldots + \\ & + a_{n_{2,1}}(x_{n_{2,1}} - x_{n_2}) + a_{n_{2,2}}(x_{n_{2,2}} - x_{n_2}) + \ldots + \\ & + \ldots \\ & + a_{n_{p,1}}(x_{n_{p,1}} - x_{n_2}) + a_{n_{p,2}}(x_{n_{p,2}} - x_{n_p}) + \ldots \end{align*} and ($\star$) implies $\| x - y \| \le \frac{\varepsilon}{2}$ we conclude \begin{equation*} \| x - y_k \| \le \| x - y \| + \| y - y_k \| \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{equation*}

Since $V$ is precompact and convex, so is $\overline{V}$. Since we have $Z = \overline{A}$ and therefore $Z \subset \overline{A}$ and also $A \subset V$ we have $Z \subset \overline{V}$.

Since $W$ is the smallest convex superset of $Z$, we have $W \subset \overline{V}$ which is precompact.

My Questions

  1. Where does the partitioning statement come from? I get that since we have one finite subsequence, at least one of the partitions is finite, but why does the inequality hold?
  2. Where do the $a_{n_k}$ come from? Are constructed analogously to the $x_{n_k}$?
  3. I am aware of this answer proving the statement in question. I suspect that his answer just omits all the technical detail from this proof but takes exactly the same approach. Is this correct?
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  1. Since $A=(x_k)$ is totally bounded, it admits a finite $\frac {\varepsilon}{2}$-net $(x_{n_j})$ consisting of its points. It provides the partition of $A$ into $p$ subsequences. Their (in)finiteness is irrelevant to the proof.

  2. For a fixed $k$, $a_{n_k}$ are these of $a_n$, for which $x_n$ belongs to a sequence $(x_{n_k})$.

  3. I think copper.hat’s proof is almost the same, but it is a bit more simple and straightforward. The omitted technical details are clear to a professional.

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