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It is with some sort of reverential fear that I've come here to write. I've been reading you for a long time, but writing is another story... In any case, I suppose it is too late now to back out!

Then, I am looking for (as many as possible) references to known "different" proofs of the classical spectral theorem for compact (linear) operators (on complex Banach spaces) with a special focus on the point where we are given to show that all non zero elements in the spectrum are, in fact, eigenvalues. I am well aware of the "usual one" (as basically drafted in this Wikipedia entry - just look at the ideas since at present the proof is flawed in some parts, as outlined by Prof. Johnson below in the comments) and I have tidings of a proof based on the Fredholm alternative (though I don't know any explicit reference in this case). Indeed, I'm wondering if there are some others around. Thanks so much for any clues.

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Thank you, this settles for the premise. Now then, what about "different" proofs (and references thereof)? – Salvo Tringali Jul 18 '11 at 20:02
The proof in Wikipedia implicitly assumes that the unit ball of the Banach space is weakly sequentially compact (i.e., that the space is reflexive) and should be modified. – Bill Johnson Jul 19 '11 at 14:01
I must confess that I've read it with little attention. Thank you for pointing this out, BJ. I'm editing the original question accordingly. – Salvo Tringali Jul 19 '11 at 18:22
up vote 1 down vote accepted

Let $T$ be a compact operator on the Banach space $X$ and $\lambda$ a non zero point in the spectrum $\sigma(T)$ of $T$. Then $\lambda$ is in the boundary of $\sigma(T)$ since $T$ is compact and hence is an approximate eigenvalue of $T$. Take a net $x_a$ of norm one vectors in $X$ s.t. $Tx_a$ converges and $\lambda x_a - T x_a$ converges to zero. Since $\lambda \not= 0$, this forces $x_a$ to converge to an eigenvector of $T$ with eigenvalue $\lambda$.

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Voilà! This is (almost) exactly the proof that I had in mind myself. Any reference? – Salvo Tringali Jul 19 '11 at 18:06
No; I took your question as a "challenge" and came up with this proof. It is not quite as elementary as the usual proof. – Bill Johnson Jul 19 '11 at 18:23
Ah! I had completely missed that... The only difference between your proof and mine is that I'm stating it in the language of $\varepsilon$-pseudospectra, but it makes no real difference, indeed. – Salvo Tringali Jul 19 '11 at 18:43

There is a proof of the spectral theorem for arbitrary bounded normal operators based on the theory of commutative Banach algebras. Then the general theorem is easily specified for the case of compact operators. For the details see, for example,

K. Maurin, Methods of Hilbert spaces. Warsaw, 1967.

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If I understand it correctly, the question is on general compact operators. – András Bátkai Jul 18 '11 at 20:14
Yes indeed, I am basically interest in the case of arbitrary compact (linear) operators on a (complex) Banach space. No need for them to be normal. But nonetheless, thank you for the contribution, AK. – Salvo Tringali Jul 18 '11 at 21:44

I don't know how "different" you will consider this proof, but if you want a quick and efficient exposition I would suggest to have a look to Klaus Deimling's Nonlinear Functional Analysis.

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