On compactness in a Hilbert space $H$

Question

If the set $\{h \in H : \|h\|\le1\}$ is compact, then show that dim $H$ is finite. How do I proceed ?

Answer 1

We can prove something more namely, for a normed space $X$ suppose the unit ball $S:=\{x\in X: ||X||≤1\}$ is compact then $ X$ is finite dimensional. You can use Riesz lemma which says that for a proper closed subspace $X_0$ of $X$ and for every $\theta $ with $0<\theta<1$, there exists $x_{\theta}\in X$ with $||x_{\theta}||=1$ and $||x-x_{\theta}||≥\theta$ for $x\in X_0$.

Now you argue by contradiction i.e assume $X$ is not finite dimensional and consider a subspace spanned by some $x_1\not =0$, then find $x_2\in S$ with $||x_2-x_1||≥1/2$. Next since finite dimensional subspaces are closed, $span (\{x_1,x_2\})$ is closed in $X$ . So find $x_3\in S$ with $||x_3-x||≥1/2$ for each $x\in span (\{x_1,x_2\})$ and so on. What you find a sequence $\{x_j\}\subseteq S$ such that $||x_k-x_j||≥1/2$ for $j\not =k$ i.e. $S$is not compact.

Answer 2

If dim $H$ were infinite, then there we could find a set of orthonormal vectors $\left \{ e_i \right \}^{\infty}_{i=1}\subseteq \{h \in H : \|h\|\le1\}=H_1.$ But $\left \{ e_i \right \}^{\infty}_{i=1}$ constitute a sequence in $H_1$ with no convergent subsequence, (because $\|e_n-e_m\|=\sqrt 2$), so $H_1$ cannot be compact