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 . Recall that since this is a Hilbert space, we have an inner product defining this norm .

Since our space is infinite dimensional, we can choose 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

.

Thus all are 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.

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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.

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.

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 of orthonormal vectors. The entries belong to the closed unit sphere, but having no convergent subsequence, as for every .

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 which does not converge. Since the sequence is Cauchy, it is bounded, say for all $n$ so that the sequence is a Cauchy sequence belonging to the closed unit ball which does not converge.

M. Thamban Nair