Density of a closed and convex subset of a Banach space.

Question

There's this homework problem that I've been battling on-and-off for about a week now. I've managed to reduce it to the following lemma.

Lemma. Let $X$ denote a Banach space, and consider closed and convex $C \subseteq X$ such that for all $x \in X$, we can find a positive integer $n$ such that $x \in nC$. Suppose there is a sequence $0_*$ in $X$ that converges to $0$, such that for all natural $i$ we have that $0_i \notin C$. Then for all $c \in C$, there is a sequence $c_*$ in $X$ that converges to $c$, such that for all natural $i$ we have that $c_i \notin C$.

Any ideas how to prove this? I just want some help getting started.

Answer 1

The Lemma you want to prove is not true (even worse, the negation of your statement is true).

By assumption,

$$ X = \bigcup_n nC. $$

As $C$ is closed, so is each $nC$. By Baire's category theorem, we see $(nC)^\circ \neq \emptyset$ for some $n \in \Bbb{N}$ and hence $C^\circ \neq \emptyset$. ($M^\circ$ denotes the interior of $M$).

So let $c \in C^\circ$ be arbitrary. For each sequence $(x_n)_n$ with $x_n \to c$, we then have $x_n \in C$ for $n$ large enough, because $C$ is a neighborhood of $c$.

But this shows that what you want to show is false (take $(x_n)_n = c_\ast$ in your notation).

EDIT: Ok, it could still be that your Lemma is true. But this can only be the case if there is no set $C$ at all that satisfies all requirements of your Lemma.

EDIT 2: Ok, your statement is true.

Hints for proving it:

1. first show that $C':=C\cap (-C)$ also satisfies $X = \bigcup_n n\cdot C'$.

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2. as above, show $(C')^\circ \neq \emptyset$.

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3. conclude $0 \in (C')^\circ$.

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4. Use this to derive a contradiction.

The following is the complete proof, do not read on if you want to solve the exercise yourself.

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First observe that $0 \in nC$ for some $n \in \Bbb{N}$, which implies $0 \in C$.

Now, if $x/n \in C$ for some $n$ and $m \geq n$, this implies $$\frac{x}{m} = \frac{n}{m}\cdot \frac{x}{n} + \left(1-\frac{n}{m}\right) \cdot 0 \in C,$$ because $C$ is convex. Hence, if $x \in nC$, then $x \in mC$ for all $m \geq n$.

Now, let $x \in X$ be arbitrary. Then $x \in n_1 C$ for some $n_1$ and $-x \in n_2 C$ for some $n_2$. Together, we see

$$ x \in N\cdot C \text{ and } -x \in N \cdot C \text{ for } N := \max\{n_1, n_2\}, $$

which implies $x \in N \cdot C'$ for $C' := C \cap (-C)$.

All in all, we have shown

$$ X = \bigcup_n n\cdot C'. $$

Using the same argument as above, $(C')^\circ \neq 0$. So, let $x \in (C')^\circ$ be arbitrary. Because of $-C' = C'$, we see $-x \in (C')^\circ$. It is an easy exercise to show that the interior of a convex set is convex. Hence,

$$ 0 = \frac{1}{2} x + \frac{1}{2} (-x) \in (C')^\circ, $$

In particular, $0 \in C^\circ$, because of $C' \subset C$. But this easily implies that the sequence $0_\ast$ in your requirements can not exist.