536 CHAPTER 19. HILBERT SPACES

Therefore, the open sets, B(x, 1

2

)for x ∈ S are disjoint and cover S. Since H is assumed

to be separable, there exists a point from a countable dense set in each of these disjointballs showing there can only be countably many of the balls and that consequently, S iscountable as claimed.

It remains to verify 19.5.26 and that span(S) is dense. If span(S) is not dense, thenspan(S) is a closed proper subspace of H and letting y /∈ span(S),

z≡ y−Py||y−Py||

∈ span(S)⊥ .

But then S∪{z} would be a larger orthonormal set of vectors contradicting the maximalityof S.

It remains to verify 19.5.26. Let S = {xi}∞

i=1 and consider the problem of choosing theconstants, ck in such a way as to minimize the expression∣∣∣∣∣

∣∣∣∣∣x− n

∑k=1

ckxk

∣∣∣∣∣∣∣∣∣∣2

=

||x||2 +n

∑k=1|ck|2−

n

∑k=1

ck (x,xk)−n

∑k=1

ck(x,xk).

This equals

||x||2 +n

∑k=1|ck− (x,xk)|2−

n

∑k=1|(x,xk)|2

and therefore, this minimum is achieved when ck = (x,xk) and equals

||x||2−n

∑k=1|(x,xk)|2

Now since span(S) is dense, there exists n large enough that for some choice of constants,ck, ∣∣∣∣∣

∣∣∣∣∣x− n

∑k=1

ckxk

∣∣∣∣∣∣∣∣∣∣2

< ε.

However, from what was just shown,∣∣∣∣∣∣∣∣∣∣x− n

∑i=1

(x,xi)xi

∣∣∣∣∣∣∣∣∣∣2

≤

∣∣∣∣∣∣∣∣∣∣x− n

∑k=1

ckxk

∣∣∣∣∣∣∣∣∣∣2

< ε

showing that limn→∞ ∑ni=1 (x,xi)xi = x as claimed. This proves the theorem.

The proof of this theorem contains the following corollary.

Corollary 19.5.3 Let S be any orthonormal set of vectors and let

{x1, · · · ,xn} ⊆ S.