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
- 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?
- Where do the $a_{n_k}$ come from? Are constructed analogously to the $x_{n_k}$?
- 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?
Best Answer
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.
For a fixed $k$, $a_{n_k}$ are these of $a_n$, for which $x_n$ belongs to a sequence $(x_{n_k})$.
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.