# Rearranging the spectral theorem

The spectral theorem for selfadjoint compact operators $L$ with infinite range says that $$Lx=\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k,$$ where the $f_k$'s form an orthonormal system and the $\alpha_k$'s are real nonzero eigenvalues, such that they tend to zero. Now my question is, if we rearrange this sum arbitrary, will it still converge ?

My hunch would be "yes", since the proof of this theorem seems to work for any arrangement, but I'm feeling uneasy accepting that...

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Can this also be tagged functional-analysis? –  Matt N. Aug 23 '12 at 12:38

It's a standard fact from hilbertian analysis that if $(x_i)_{i\in I }$ is an orthogonal (not necessarily orthonormal) system, then $(x_i)_{i\in I}$ is a summable family iff $(||x_i||^2)_{i \in I}$ is summable.

An application of Bessel's inequality and the previous fact shows that $(a_k \langle x, f_k \rangle f_k)_k$ is summable, and hence you can rearrange arbitrary this series and still get a convergent series with the same sum.

Indeed, since $(\alpha_k \langle x, f_k \rangle f_k)_{k \ge 0}$ is an orthogonal system, it's summable if and only if $\sum_{k\ge 0} \| \alpha_k \langle x, f_k \rangle f_k \|^2$ is finite, that is $\sum_{k\ge 0} | \alpha_k|^2 | \langle x, f_k \rangle |^2 \| f_k \|^2$ is finite.

But $(\alpha_k)_k$ is a convergent, and hence bounded, sequence, and $\|f_k \| = 1$, so we are reduced to prove that $\sum_{k \ge 0} | \langle x, f_k \rangle |^2$ is finite; but it's exactly the content of Bessel's inequality.

EDIT : A few facts about summable families

The precise definition of a summable family is as follows : a family $(x_i)_{i \in I}$ (where the set $I$ can be uncountable) of vectors in a Banach space $E$ is summable, with sum $x$, if for every $\epsilon > 0$, there exists a finite set $J \subset I$ such that for every finite set $K \subset I$ with $J \subset K$ one has $\| \sum_{i \in K} x_i - x \| \le \epsilon$. This can be equivalently characterised by the convergence of an appropriated net. If a family is summable, its sum is unique. We donote it by $x = \sum_{i \in I} x_i$.

This notion is invariant under rearrangement of indices : if $\sigma : J \to I$ is a bijection, then $(x_i)_{i \in I}$ is summable iff $(x_{\sigma(j)})_{j \in J}$ is summable, and sums are equal.

If the family of real numbers $(\| x_i \|)_{i \in I}$ is summable, then $(x_i)_{i \in I}$ is itself summable (provided $E$ is a Banach space, completeness is the key point here). The converse is not true in general, except if $E$ is finite dimensional.

In the special case where $I = \mathbb{N}$, summability implies the convergence of the series associated, but the converse is not true, even if $E$ is finite dimensionnal. Indeed, if $E$ is finite dimensional (which covers the case of a family of real numbers), by what was said previously, a family $(x_n)_{n \in \mathbb{N}}$ is summable iff the series $\sum_{n = 0}^{+ \infty} x_n$ is absolutely convergent, that is $\sum_{n = 0}^{+ \infty} ||x_n|| < \infty$.

You can find further details and proofs in "General Topology" by N. Bourbaki. They cover the wider case of summable families in topological groups, but you can easily specialize to Banach space if you are not interested in such a generalization.

Now, I'll turn to summable families in Hilbert space.

Bessel's inequality asserts that if $(e_i)_{i \in I}$ is an orthonormal system of an Hilbert space $H$, then for every $x \in H$, the family $( |\langle x, e_i \rangle|^2)_{i \in I}$ of real numbers is summable, and its sum is less or equal to $\|x \|^2$.

Parseval identity says that if $(e_i)_{i \in I}$ is a complete orthonormal system in $H$, then for every $x \in H$, the family $( \langle x, e_i \rangle e_i)_{i \in I}$ is summable, and its sum is exactly $x$. Moreover, Bessel's inequality turns out to be actually an equality.

For all of this, and for the fact recalled at the beginning of my answer, you can take a look at "The Elements of Operator Theory" by C.S. Kubrusly.

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Before I accept your answer, could you please give me a reference for the fact stated in the first paragraph ? (do you mean with "summable" that $\sum_{i\in I} ||x_i||^2$ converges?) –  user38525 Aug 23 '12 at 10:26
I provided some definitions and results about summable family. I also improve the very first statement in order to cover correctly the case you're considering. –  Ahriman Aug 23 '12 at 12:29
Wow, that was very detailed. But I'm not sure how you arrived at proving that the sequence of the RHS is summable. The Bessel inequality tells me that $$\sum_k | \langle x,f_k \rangle |^2 \leq ||x||^2.$$ On the other hand I know that $$\sum_k \alpha_k \langle x,f_k \rangle f_k$$ is summable iff $$\sum_k ||\alpha_k \langle x,f_k \rangle f_k||=\sum_k |\alpha_k| | \langle x,f_k \rangle |$$ is summable. But how do I combine these two to get what I want ?[...] –  user38525 Aug 23 '12 at 15:14
Somehow the notion of convergence needs to come into play since I need to deduce the summability of $$\alpha_k \langle x,f_k \rangle f_k$$ from the convergence of $$\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k$$. I have denoted a convergent sum by $$\sum_{k=1}^{\infty}$$ and an unconditional convergent sum (since I think that that is what I get from a summable sequence) by $$\sum_k$$. Note that originally I only have $$\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k$$. –  user38525 Aug 23 '12 at 15:17
Are you also sure, that "$(x_i)_{i\in I}$ is a summable family iff $(||x_i||^2)_{i \in I}$ is summable" is true ? I know of a theorem that says, that for an ONS $(x_k)_k$ we have $\sum_{k=1}^{\infty} \beta_k x_k$ is convergent sum iff $\sum_{k=1}^{\infty} |\beta_k|^2 x$ is convergent". This is related, but not the same. –  user38525 Aug 23 '12 at 15:21