Posted by: hilbertthm90 | August 26, 2008

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.

Posted in analysis | Tags: , , , ,

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Responses

  1. 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.

    By: Todd Trimble on August 30, 2008
    at 1:11 am


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