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This is a deeper question of this question I asked: If all of the sequences in $U$ are not convergent, is $U$ closed?.

I know that $U=\left\{(1+1 / n) e_{n}\right\}$ is closed, where $\left\{e_{n}\right\}$ is an orthonormal sequence in an infinite dimensional Hilbert space $X$.

Now the question is: Is the convex hull of $U$ closed?

In fact, I am trying to construct a conterexample of this statement:

Let $X$ be a Banach space, $U \subset X$ a closed, convex set, and $x_{0} \in X \backslash U .$ Then there exists a point $y \in U$ such that $\left\|y-x_{0}\right\|=d\left(U, x_{0}\right)$.

This example works when we do not require the convexity of $U$: just let $x_0$ be the origin. When it comes to convexity, I also think about using Hahn-Banach to find a hyperplane to separate $\{x_0\}$ and $U$, but it seems useless. Or maybe the statement is true and how do we prove it?

bernard
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1 Answers1

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The last question has been answered in the link provided by David Mitra.

The answer to the first question is NO. Let $x=\sum \frac 1 {2^{n}} (1+\frac 1 n) e_n$. Then $x$ belongs to the closure of the convex hull of $((1+\frac 1 n) e_n)$ but it does not even belong to linear span of these vectors.

To see that $x$ belongs to closure of the convex hull consider the vectors $x_N=\frac {\sum\limits_{k=1}^{N} 2^{-k} (1+\frac 1 K)e_k} {\sum\limits_{k=1}^{N} 2^{-k}}$. Show that $X_N \to x$.

Kavi Rama Murthy
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