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Let H be the Hilbert space of square-summable sequences of reals.

A few years ago I thought I had proved that the subspace Z of real sequences with only finitely many nonzero terms, such that they sum to zero, is dense in H.

(Since then I've seen this confirmed in Rudin's text, Functional Analysis, 2nd ed., but only as a teensy subquestion in a terminally hairy exercise.)

Now I can't quite reproduce my proof, so it may have been wrong. Can someone please tell me how to prove that.

(What I tried was taking an arbitrary point c = (c1,...,ck,...) of H, and defining the point d(N) in H as follows for N $\ge$ 1:

First set AN = (c1+...+cN) / N. Then set d(N)k = ck - AN for k $\le$ N, and set d(N)k = 0 for k > N.

Clearly, the element d(N) of H lies in the subspace Z defined above. The squared Hilbert norm of its difference with c is of form N(AN)2 + T(N), where T(N) is just the squared norm of the tail of c, and so goes to 0 as N$\to\infty$.

I'm left with the expression N(AN)2, which so far I haven't been able to prove $\to$ 0 as N$\to\infty$. I suspect this is true, and it works on all the examples I've tried so far.)

NOTE: I'm only interested in a down-and-dirty proof that doesn't invoke anything but simple inequalities.

Thanks for any help you can offer.

user151644
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  • For a not LATEX-ed question, nice typesetting! – ajotatxe Oct 29 '14 at 17:41
  • Can you elaborate on how you arrive at "The squared Hilbert metric of its difference with c is of form $(A_N)^2 / N + T(N)$"? that is not what I see when I look at that. – Jason Knapp Oct 29 '14 at 17:49
  • Jason, I miswrote that (A_N)^2 / N. It should have been N(A_N)^2. (I had been working with the sum, but in this post I used the average.) Sorry about that. (Now fixed.) – user151644 Oct 29 '14 at 18:00

1 Answers1

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Given $x \in H$ and $\epsilon > 0$, there is $y \in H$ with $\|x - y\| < \epsilon/2$ such that only finitely many elements of $y$ are nonzero. Suppose the sum of those nonzero elements of $y$ is $s$. Consider $z$ obtained from $y$ by changing $n$ of its elements from $0$ to $-s/n$. Thus $z \in Z$, and $\|z - y\|^2 = n (s/n)^2 = s^2/n$. If $n$ is sufficiently large, this is less than $(\epsilon/2)^2$. So $\|x - z\| < \epsilon$. We conclude that $Z$ is dense in $H$.

Robert Israel
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  • I believe your proof, Robert, but I got down to essentially the same expression as your s^2 / n, and could not see a proof of why this gets arbitrarily small as n -> oo. – user151644 Oct 29 '14 at 17:55
  • OK, to be more specific: My expression N(A_N)^2 in the Question is the same as Robert Israel's expression s^2/n. – user151644 Nov 03 '14 at 23:30
  • I think you really need to do it in two stages, as I did. First get rid of all but finitely many nonzero elements, then shift a finite number of elements to make the sum $0$. – Robert Israel Nov 04 '14 at 00:10
  • To continue: In terms of the components of the original element c, s^2/n = (the sum of the squares of c_1 through c_n) + (twice the sum of all pairwise products c_^j c_^k with indices j,k such that 1 <= j < k <= n). Therefore, since 2ab <= a^2 + b^2, we can bound s^2 above with n SUM(n), where we define SUM(n) := ((c_1)^2 + ... + (c_n)^2). Hence we have s^2/n <= n SUM(n)/ n = SUM(n). Instead it approaches the squared norm of c. Maybe @RobertIsrael can elaborate on why his s^2/n approaches 0 as n -> oo. – – user151644 Nov 04 '14 at 01:24
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    Because my $s$ is constant. – Robert Israel Nov 04 '14 at 02:36