Let $(X,d)$ be a nondegenerate compact connected metric space, and $$2^X=\{A\subset X|A\textrm{ is compact and not empty}\}.$$ Given $A,B\in 2^X$ and $\epsilon>0$, define $$N(\epsilon,A)=\{x\in X|d(x,a)<\epsilon\textrm{ for some }a\in A\}$$ and $$H(A,B)=\inf\{\epsilon>0|A\subset N(\epsilon,B)\textrm{ and }B\subset N(\epsilon,A)\}.$$ It can be proven that $(2^M,H)$ is a metric space.

Now, if $(M_1,d_1)$ and $(M_2,d_2)$ are metric spaces, a function $f:M_1\to M_2$ is called nonexpansive if for every $x_1,x_2\in M_1$ we have $$d_2(f(x_1),f(x_2))\leq d_1(x_1,x_2).$$ Notice that every isometry is nonexpansive.

Let $(L,||\cdot||)$ be a Banach space and for any $A\subset L$ let $conv(A)$ denote the convex hull of $A$. Let $X$ be a nondegenerate compact convex subset of $L$ with the metric obtained from the norm on $L$. It is well-known (see Dunford and Schwartz's Linear Operators: Part I: General Theory, page 416) that if $A$ is a compact subset of $L$, then so is $conv(A)$. So, we can regard $conv$ as a function from $(2^X,H)$ into $(2^X,H)$.

$conv$ is nonexpansive:

Let $A,B\in 2^X$. Assume that $\epsilon>H(A,B)$, so that $A\subset N(\epsilon,B)$ and $B\subset N(\epsilon,A)$. Let $x\in conv(A)$. Then, we have two cases:

  1. $x=\sum_{k=1}^N\lambda_kx_k$ with $x_1,\dots,x_N\in A$ and $\sum_{k=1}^N\lambda_k=1$: In this case, we take any set $\{y_1,\dots,y_N\}\subset B$ such that $d(x_k,y_k)<\epsilon$(*) for $k=1,\dots,N$, so that $y=\sum_{k=1}^N\lambda_ky_k\in conv(B)$. So, $$d(x,y)=||x-y||=||\sum_{k=1}^N\lambda_kx_k-\sum_{k=1}^N\lambda_ky_k||=||\sum_{k=1}^N\lambda_k(x_k-y_k)||\leq\sum_{k=1}^N\lambda_k||x_k-y_k||$$ $$<\sum_{k=1}^N\lambda_k\epsilon=\epsilon,$$ so that $x\in N(\epsilon,conv(B))$. Analogously, if $y$ is a convex combination of points of $B$, then $y\in N(\epsilon,conv(A))$.
  2. $x=\lim_{n\to\infty}s_n$ where each $s_n$ is a convex combination of points of $A$, as in the previous case. In this case, define $s'_n$ as $y$ in the previous case: using the same coefficients for points in $B$ which satisfy (*). The sequence $s'_n$ may not be convergent, but since $conv(B)$ is compact, it has a convergent subsequence $s'_{n_k}$ to a point $y\in conv(B)$. Since $s_n\to x$, then $s_{n_k}\to x$, so $$d(x,y)=||x-y||=||\lim_{k\to\infty}s_{n_k}-\lim_{k\to\infty}s'_{n_k}||=||\lim_{k\to\infty}(s_{n_k}-s'_{n_k})||=\lim_{k\to\infty}||s_{n_k}-s'_{n_k}||\leq\epsilon,$$ since $s_{n_k}$ and $s'_{n-k}$ have the same form as $x$ and $y$ in the previous case. So, $x\in N(\delta,conv(B))\forall\delta>\epsilon$. Analogously, $y\in N(\delta,conv(A))\forall\delta>\epsilon$.

In conclusion, $\forall\delta>\epsilon$ we have that $conv(A)\subset N(\delta,conv(B))$ and $conv(B)\subset N(\delta,conv(A))$, so that $$H(conv(A),conv(B))=\inf\{\delta>0|conv(A)\subset N(\delta,conv(B))\textrm{ and }conv(B)\subset N(\delta,conv(A))\}\leq\epsilon$$ $\forall\epsilon>H(A,B)$, so $H(conv(A),conv(B))\leq H(A,B)$.

Can you give an example of a Banach space $(L,||\cdot||)$ and a convex compact subset $X$ of $L$ such that $conv:2^X\to2^X$ is not an isometry?


Taking convex hull is not even an injective operation. Consider the subsets $$A=\{-1,1\} \quad \text{and}\quad B=\{-1,0,1\}$$ of the real line (so, $L=\mathbb R$). These two sets have the same convex hull, namely $[-1,1]$.


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