analysis

Not Compact Unit Ball

Here is a beautiful little theorem. The unit ball in an infinite dimensional Hilbert space is not compact. The proof is quite simple. So the unit ball $B=\{v\in \mathcal{H} : \|v\|\leq 1\}$. Recall that since this is a Hilbert space, we have an inner product defining this norm $\langle v, v \rangle=\|v\|^2$.

Since our space is infinite dimensional, we can choose $\{v_1, v_2, \ldots \}$ to be linearly independent inductively. Basic application of Gram-Schmidt gets us this set to be orthonormal, in particular, each one has norm 1 and so is in the unit ball.

Now look at the distance between any two $\|v_i - v_j\|^2=\langle v_i - v_j, v_i - v_j \rangle$ $= \langle v_i, v_i \rangle -2\langle v_j, v_i \rangle +\langle v_j, v_j\rangle$ $=1+2(0)+1$ $=2$.

Thus all are $\sqrt{2}$ apart and hence there is no subsequence that converges. Since sequential compact and compact are the same here, we are done. (Note the inner product is not symmetric, but I knew the middle terms would be zero, so I went ahead and abused that instead of writing two zeros).

I know the result to be true in a Banach space as well, but I don’t see a quick fix without the inner product…

A lot more of these things will start popping up with my analysis prelim two weeks from today.

3 thoughts on “Not Compact Unit Ball”

1. Todd Trimble says:

A general result is that a normed vector space V is locally compact iff it is finite-dimensional. Sketch of proof: if V is locally compact, then the unit ball B can be covered by finitely many translates of the ball (1/2)B. Let H be the vector space generated by the translation vectors; then H is closed in V and we get an induced normed vector space structure on the quotient V/H, for which B = (1/2)B. V/H = 0 then follows.

2. Thomas M. says:

Thanks for the article! Needed to recall a few basic points of functional analysis for a seminar and found your post.
Interesting in Hilbert spaces, but I remember the following example for Banach spaces:

Take the Banach space (C0, sup||.||) and the sequence {sin(nx)}, n in IN, as points in it. This normed space is obviously an infinitely dimensional vector space.
These points, wrt the sup-norm, lie on the circumference of the unit sphere – since sup||sin(n.)||=1.
Furthermore, the “distance” between two arbitrary points is 2 – since sup||sin(n.)-sin(m.)||=2.
So the sequence doesn’t have a converging subsequence and is thus neither sequentially compact, or compact – since we’re dealing with metric spaces.

3. M. Thamban Nair says:

Here are simple proofs for:

(1) Closed unit ball in an infinite dimensional inner product space is not compact.

(2) Closed unit ball in an incomplete normed linear space is not compact.

Proof of (1): As already given earlier: Consider a sequence $(v_n)$ of orthonormal vectors. The entries belong to the closed unit sphere, but having no convergent subsequence, as $\|v_n-v_m\|\geq \sqrt{2}$ for every $n\neq m$.

Proof of (2): This has been observed recently by me and will appear elsewhere as well: Since the space is not complete there is a Cauchy sequence $(x_n)$ which does not converge. Since the sequence is Cauchy, it is bounded, say $\|x_n\|\leq M$ for all $n$ so that the sequence $(x_n/M)$ is a Cauchy sequence belonging to the closed unit ball which does not converge.

M. Thamban Nair