# On Mazur's original proof of his theorem: in Banach spaces convex hulls of precompact sets are precompact.

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?

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 $$(x_{n_k})$$.