Showing that the set $D = \{\|x\|\ |\ x\in\ K\}$ of closed convex subset $K$ of a Hilbert space $H$ is itself closed.

Question

In an exercise I was asked to show that if $H$ is a vector space and $K\subset H$ is closed and convex, then $D = \{\|x\|\mid x\in K\}$ is closed in $\mathbb{R}$. My proof idea was to do something like this:

Take any sequence $\{d_n\}\subset D$, and define $\{x_n\}\subset K$ such that $\|x_n\|=d_n\forall n$ ($x_n$ is clearly not unique). By the "Bolzano-Weierstrass Theorem", there exists a convergent subsequence of $x_n$, call it $x_{n_k}\to x$, and hence $\lim_{n\to\infty}d_n=\lim_{k\to\infty}\|x_{n_k}\|=\|x\|$ for some $x\in K$, and hence $\lim_{n\to\infty}d\in D$.

Now clearly, the Bolzano-Weierstrass Theorem cannot be applied to an arbitrary Hilbert space (for example, this fails to hold in infinite dimensional spaces such as $\ell^2$). Is there any way I could still make this proof "work"? How could I modify the idea to show that $D$ is closed.

On that note, is $D$ even closed? The exercise doesn't ask to directly show this, but I wanted to use this as part of the proof of a larger question.

Answer 1

First, your space is a Hilbert space therefore metrisable. Then, compactness is equivalent to sequential compactness (which is what you call Weierstrass Bolzano).

Weierstrass Bolzano doesn't hold for the entire space, but holds for compact sets in metric spaces.

Faster solution $F: H \to \mathbb R$ defined by $F(x)=\| x\|$ is continuous, hence it takes compact sets to compact sets.

Answer 2

The answer by N.S. is missing something very important. In infinite dimensions it is not true that closed and bounded sets are compact. So even though it is true that $F$ maps compact sets to compact sets, that does not mean it maps closed and bounded sets to closed and bounded sets. In fact, it turns out that $F$ can map a closed and bounded set to a set that is not closed. Here is an example.

Take $\ell^2$ and consider the sequence $x_i:=(1-\frac1i)e_i$. Let $C$ denote the convex hull of the sequence and take $K$ as the closure of $C$. We can show that $D=[0,1)$, which is not closed.

First note that $\|x_i\|$ is always smaller than 1, but it gets arbitrarily close to it. From the triangle inequality it follows that even the convex hull can not admit vectors with norm 1 or bigger. That the closure can not admit vectors with norm 1 or bigger is less trivial, but the main argument is that any sequence in $C$ with norm converging to 1 can not be convergent, since it would need to keep moving on to higher dimensions.

Fun side note: If you choose $e_i$ carefully, this works for any infinite-dimensional normed vector space. If you want to try this, look into Riesz' lemma.