In a Hilbert space $H$, if the closed unit ball $\{x\in H\colon \|x\|\leqslant 1\}$ is compact, then how can it be proved that $H$ is finite-dimensional?


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    $\begingroup$ If $H$ is infinite dimensional, you can construct a sequence $(e_n)$ of norm one vectors that has no convergent subsequence. You can even make them orthonormal. $\endgroup$ – David Mitra Aug 24 '14 at 11:43

Since the closed unit ball $B[0,1]$ is compact, there exists a finite subset $F \subseteq B[0,1]$ such that

$$B(0,1) \subseteq \bigcup_{x \in F} B\left(x, \frac{1}{2} \right) = F+B \left(0,\frac{1}{2} \right). \tag{1}$$

Obviously, $L := \text{span} F$ defines a linear finite-dimensional subspace of $H$. In particular, $L$ is complete; hence closed in $H$. Iterating $(1)$, we get

$$B(0,1) \subseteq L + B \left(0,\frac{1}{4} \right) \subseteq \ldots \subseteq L + B \left(0, \frac{1}{2^n} \right).$$

Consequently, we can find for any $x \in B(0,1)$ a point $x_n \in L$ such that $|x_n-x| <\frac{1}{2^n}$. Since $L$ is closed, we conclude $x \in L$. As $L$ is a linear subspace, this proves $L=H$.

  • $\begingroup$ How do we iterate (1)? I think we want $B(0,\frac{1}{2})\subset L+B(0,\frac{1}{4})$, etc, so that $B(0,1)\subset L+(L+B(0,\frac{1}{4}))=L+B(0,\frac{1}{4})$, but how does this first $\subset$ appear? $\endgroup$ – Ryan Unger Dec 7 '16 at 5:17
  • $\begingroup$ @0celo7 Clearly, $(1)$ gives $$B(0,1) \subseteq L + B \left(0, \frac{1}{2} \right).$$ Since $L$ is a subspace, this actually implies $$B(0,r) \subseteq rL + B \left(0, \frac{r}{2} \right) = L + B \left(0, \frac{r}{2} \right)$$ for any $r>0$. $\endgroup$ – saz Dec 7 '16 at 7:30
  • $\begingroup$ Right, it's not clear to me why shrinking everything works like that. $\endgroup$ – Ryan Unger Dec 7 '16 at 13:45
  • $\begingroup$ @0celo7 Take $x \in B(0,r)$, then $r^{-1}x \in B(0,1)$. By the 2nd line of my previous comment, there exist $y \in L$ and $z \in B(0,1/2)$ such that $r^{-1} x = y+z$. Hence, $$x = r (r^{-1} x) = \underbrace{r y}_{\in L} + \underbrace{r z}_{\in B(0,r/2)}.$$ $\endgroup$ – saz Dec 7 '16 at 13:48
  • $\begingroup$ I just worked that out in the shower and was about to tell you "never mind." Thanks anyhow. Interesting proof! $\endgroup$ – Ryan Unger Dec 7 '16 at 14:06

By contraposition: show that if $H$ is infinite-dimensional then the unit ball of $H$ is not compact. Use the Riesz lemma for this.


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