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I'm coursing an introduction to Hilbert spaces and I recently saw in the book "A Course in Modern Mathematical Physics" by P. Szekeres a proof that I don't completely understand.

The aim is to prove that $l^2$ (the set of all complex sequences where $u_i \in \mathbb{C}$ such that

$\sum_{i=1}^{\infty} |u_i|^2 < \infty$

) is a Hilbert space. It begins showing that this space is linear and has a well-defined inner product. But then this comes:

Quote:

The norm defined by this inner product is

$||u|| = \sqrt{\sum_{i=1}^{\infty} |u_i|^2} \leq \infty$.

For any integer $M$ and $n,m > N$

$\sum_{i=1}^M |u_i^{(m)} - u_i^{(n)}|^2 \leq \sum_{i=1}^{\infty} |u_i^{(m)} - u_i^{(n)}|^2 = ||u^{(m)} - u^{(n)}||^2 < \varepsilon^2$,

and taking the limit $n \to \infty$ we have

$\sum_{i=1}^M |u_i^{(m)} - u_i|^2 \leq \varepsilon^2$.

In the limit $M \to \infty$

$\sum_{i=1}^{\infty} |u_i^{(m)} - u_i|^2 \leq \varepsilon^2$

so that $u^{(m)} - u \in l^2$. Hence $u = u^{(m)} - (u^{(m)} - u)$ belongs to $l^2$ since it is the difference of two vectors from $l^2$ and it is the limit of the sequence $u^{(m)}$ since $||u^{(m)} - u|| < \varepsilon$ for all $m > N$. It turns out, as we shall see, that $l^2$ is isomorphic to most Hilbert spaces of interest - the so-called $\textit{separable Hilbert spaces}$.

Unquote

The problem that I've found is why he can assure that the sequence has limit $u_i$: is it an assumption or has it to be proved? I don't really understand how the last paragraph concludes it.

Anne Bauval
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ErikLAndre
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1 Answers1

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The part of the proof you are showing is that one that proves that $u$ is in $\ell^2$. Before that, in a part you are not quoting, the author likely deduced that $$ |u^{(m)}_i-u^{(n)}_i|\leq\|u^{(m)}-u^{(n)}\|_2. $$ This shows that for each $i$ the sequence of numbers $\{u_i^{(m)}\}$ is Cauchy. Because $\mathbb C$ (or $\mathbb R$, depending on what the author's choice is) is complete, there is a limit $u_i=\lim_m u^{(m)}_i$. Then comes the part of the proof you quoted, that shows that $u$ is in $\ell^2$ and it is the limit of $\{u^{(m)}\}$ in the $\|\cdot\|_2$-norm.

Martin Argerami
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  • I testify the quote was faithful and complete. Sadly, the author deduces nowhere that $

    • |u^{(m)}_i-u^{(n)}_i|\leq|u^{(m)}-u^{(n)}|_2. $ He even doesn't explicitly begin his proof with something like "let $(u^{(n)})$ be a Cauchy sequence in $\ell^2$. For all $\varepsilon>0,$" 2) The quote mentions the choice of $\mathbb C$ (not $\mathbb R$). 3) The previous example of Hilbert space in this book was $\mathbb C^n$ but I have no access to that page. In view of the quote, it may also be that his (sketch of) proof of completeness for $\ell^2$ rather relies on that for $\mathbb C^M$.

    – Anne Bauval Oct 08 '22 at 20:24
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    Well, I know nothing else about the book, but if that's the case then it is very poorly written. – Martin Argerami Oct 08 '22 at 20:43
  • I don't know how to prove that the linearity holds, so how can i prove that ⟨(zn)+(vn),(wn)⟩=⟨(zn),(wn)⟩+⟨(vn,wn)⟩ ? More specifically, I don't understand why we can split the infinite sum while proving this. How did you do this @MartinArgerami? – user33 Apr 05 '25 at 08:28
  • In $\mathbb C$ the limit of the sum is the sum of the limits. \begin{align} \langle z+v,w\rangle &=\sum_n(z+v)n\overline{w_n} =\sum_n(z_n+v_n)\overline{w_n}=\sum_n z_n\overline{w_n}+v_n\overline{w_n}\ &=\lim{N\to\infty}\sum_{n=1}^N z_n\overline{w_n}+v_n\overline{w_n} =\lim_{N\to\infty}\sum_{n=1}^N z_n\overline{w_n}+\sum_{n=1}^Nv_n\overline{w_n}\ &=\lim_{N\to\infty}\sum_{n=1}^N z_n\overline{w_n}+\lim_{N\to\infty}\sum_{n=1}^Nv_n\overline{w_n} =\sum_{n=1}^\infty z_n\overline{w_n}+\sum_{n=1}^\infty v_n\overline{w_n} \ &=\langle z,w\rangle+\langle v,w\rangle. \end{align} – Martin Argerami Apr 05 '25 at 10:06