3
$\begingroup$

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...

  • 0
    Can this also be tagged functional-analysis?2012-08-23

1 Answers 1

1

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.

  • 0
    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?)2012-08-23
  • 0
    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.2012-08-23
  • 0
    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 ?[...]2012-08-23
  • 0
    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 $$ \sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k$$.2012-08-23
  • 0
    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.2012-08-23
  • 0
    I'm sure that it's true. Check in the reference I gave about Hilbert spaces. I'll give details for the proof you need.2012-08-23
  • 0
    I proved that $(\alpha_k \langle x, f_k \rangle f_k)_{k \ge 0}$ is a summable family. This implies that the corresponding series is convergent, and that any rearrangement of this series is still convergent, with the same sum, as pointed out in what I said about general properties of summable families.2012-08-23
  • 0
    You said that $\sum_k \alpha_k \langle x, f_k \rangle f_k$ is summable iff $\sum_k \| \alpha_k \langle x, f_k \rangle f_k \|$ is summable. This is NOT true. As I said in my answer, summability is'nt equivalent to absolute summability for infinite dimensional Banach spaces. This is why I use instead the criterion for orthogonal system ...2012-08-23
  • 0
    Sorry that I still don't get it, but could you please be a little more specific how you proved that $( \alpha_k \langle x,f_k \rangle f_k)_{k\geq 0}$ is a summable sequence ? I my first comment I wrote it out, what you said in the first paragraph of youre answer. How can I combine that two pieces of information (Bessel inequality + equivalent condition for summability) to get that $( \alpha_k \langle x,f_k \rangle f_k)_{k\geq 0}$ is summable ?2012-08-23
  • 0
    I've already written details in my answer.2012-08-23
  • 0
    Ah, sorry, I didn't see it. One last question (and this was the problem I couldn't understand it earlier): Is it a theorem in analysis that $\sum_{k=1}^{\infty} a_k \cdot b_k$ is convergent if the sequence $(a_k)_k$ is bounded and $\sum_{k=1}^{\infty} b_k$ is convergent ?2012-08-23
  • 0
    For convergence, it's false ! (take $a_k = (-1)^k$ and $b_k = \frac{(-1)^k}{k}$) But it's true for absolute convergence : if $|a_k| \le M$ for all $k$, then $\sum_k |a_k b_k| \le M \sum_k |b_k|$, and so the LHS term is finite whenever the RHS term is.2012-08-23
  • 0
    Thanks A LOT! (will upvote it once I have enough reputation to do that)2012-08-23