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Recently I was told that there is a version of spectral theorem for compact operator $T:H_1\to H_2$ where $H_1,H_2$ are possibly different separable Hilbert spaces.

I do not remember details, but I think there was some additional assumption on $TT^*$ and the thesis of spectral theorem was something like: there is a sequence $\mu_n\to 0$ and Hilbert basis $(e_n)$ and $(f_n)$ such that $$Tx=\sum_{n=1}^{\infty}\mu_n(x,e_n)f_n$$ for all $x\in H_1.$

I want to know exact statement of such version of spectral thoerm. Do you know the book which covers it? (there is one for sure)

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    Not sure, but would start with Brezis and Kutateladze (two different books). – 2017-01-23
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    If you know the spectral theorem for selfadjoint operators $H\to H$, then the proof of this theorem is not very long. If you want, I can write it down. – 2017-01-23
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    @LeBtz The point of this question is to provide version of spectral theorem where $H_1\neq H_2.$ I wrote "spectral theorem" with " " in the title to defend myself from such misunderstanding. – 2017-01-23
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    @FallenApart That was clear. I meant that the proof of your version is simple, knowing the version where $H_1 = H_2$. But if you are not interested, thats no problem. – 2017-01-23
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    @LeBtz But mayby you are right and the spectral theorem was applied to $TT^*$ or $\sqrt(TT*),$ but still I would like to see some book which considers two different spaces, because I was told that there is one. – 2017-01-23
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    @FallenApart I think we don't understand eachother. Yes, there is a theorem where two different spaces are considered. The proof still works by looking at $T^\ast T$. Why is that a problem if it proves the general statement? – 2017-01-23
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    @LeBtz I wrote last comment poorly. I just have blurry image in my head with what I have been told once. I know that there is a book which covers two different spaces, but I am not sure whether thesis applies to $T$ or $TT*.$ If the latter one, the theorem is immidiete. But I just counting on that someone would give me the refference to the exact book which covers such approach. – 2017-01-23
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    @LeBtz So is the statement from the question corret? It would be great if you can provide me with the proof. – 2017-01-23

1 Answers 1

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Theorem: Let $H,K$ be Hilbert spaces, not necessarily separable, and $T:H\to K$ compact (for simplicity, we assume it not to have finite rank, otherwise we need to modify the proof slightly). Then there are orthonormal systems $(e_j)_{j\in\mathbb N}$ in $H$, $(f_j)_{j\in\mathbb N}$ in $K$ and a positive sequence $(s_n)_{n\in\mathbb N}$ with $\lim\limits_{n\to\infty}s_n = 0$ such that $$Tx = \sum_{j=0}^\infty s_j\langle x,e_j\rangle f_j$$ for each $x\in H$. Moreover, the ${s_j}^2$ are exactly the nonzero eigenvalues of $T^\ast T$.

Proof: The operator $T^\ast T:H\to H$ is compact and self-adjoint, hence, by the spectral theorem, there is an orthonormal system $(e_j)_{j\in\mathbb N}$ such that $$T^\ast Tx = \sum_{j=0}^\infty\lambda_j \langle x,e_j\rangle e_j,$$ for each $x\in H$, where the $\lambda_j$ are exactly the nonzero eigenvalues of $T^\ast T$, hence $\lim\limits_{j\to\infty}\lambda_j = 0$ and $\lambda_j>0$ for each $j\in\mathbb N$ since $T^\ast T$ is self-adjoint and positive. Set $s_j := \sqrt{\lambda_j}$ and $f_j := \frac{Te_j}{\|Te_j\|}$. This is well defined because $T$ can't vanish on $e_j$, otherwise $T^\ast T$ would also vanish on $e_j$ but $e_j$ is an eigenvector for a nonzero eigenvalue of $T^\ast T$.

$(f_j)_{j\in\mathbb N}$ is an orthonormal system because $\langle Te_j, Te_k\rangle = \langle T^\ast Te_j, e_k\rangle = \lambda_j\langle e_j,e_k\rangle = 0$ for $j\neq k$ and $\|f_j\| = 1$ is obvious.

Define $S:H\to K, x\mapsto \sum_{j=0}^\infty s_j\langle x,e_j\rangle f_j$. We claim that $T=S$. Since $S$ and $T$ are both continuous, it suffices to check the equality on an orthonormal basis of $H$.

Therefore, let $(e_j)_{j\in J}$ be an orthonormal basis of $H$ that contains $(e_j)_{j\in\mathbb N}$, i.e. $\mathbb N\subset J$. If $j\in J\setminus \mathbb N$, then $T^\ast Te_j = 0$. It follows $\|Te_j\|^2 = \langle Te_j, Te_j\rangle = \langle T^\ast Te_j, e_j\rangle = 0$, hence $Te_j = 0$ and also $Se_j=0$ because $\langle e_j, e_k\rangle = 0$ for each $k\in\mathbb N$.

If $j\in \mathbb N$, then $Se_j = s_j f_j = s_j\frac{Te_j}{\|Te_j\|}$, so it suffices to show ${s_j}^2 = \|Te_j\|^2$. This is correct because $\|Te_j\|^2 = \langle T^\ast Te_j, e_j\rangle = \lambda_j\langle e_j, e_j\rangle = {s_j}^2$.

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    I misunderstood your first comment, to my question. This is precisely what I wanted. Thank you very much. – 2017-01-23
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    You are welcome. – 2017-01-23