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I have asked this question on MSE but did not receive an answer. I thought I could try it here.

Let $T$ be a self-adjoint trace-class operator on $L^2(\mathbb{R})$. Is is true that it can be represented as an integral operator.

I thought the kernel would be $$k_T(x,y) =\sum_{i=1}^\infty \lambda_i \phi_i(x) \bar\phi_i(y).$$

Here $\{\phi_i\}$ is an eigenbasis of $T$, i.e. $T=\sum_i \lambda_i |\phi_i\rangle\langle\phi_i|$. Then, we have $$\int k_T(\cdot,y) f(y) = \int\sum_i \lambda_i \phi_i(\cdot) \bar\phi_i(y) f(y) dy = \sum_i \lambda_i \phi_i \langle \phi_i, f\rangle=\sum_i \lambda_i |\phi_i\rangle\langle\phi_i|f\rangle = Tf.$$

Is this correct?

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    $\begingroup$ A comment about the close vote: I agree the question isn't research level, but if it goes unanswered on MSE for two days I think it's fair to answer it here. $\endgroup$ – Nik Weaver Feb 26 '16 at 1:36
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Yes, this is correct. Actually, "self-adjoint trace-class" is more than you need; any Hilbert-Schmidt operator can be represented as an integral operator. The Hilbert-Schmidt operators from $L^2(X)$ to $L^2(Y)$ are precisely the integral operators with kernel in $L^2(X\times Y)$ (at least for $\sigma$-finite $X$ and $Y$). This should be in a standard reference, probably Dunford-Schwartz, for example.

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    $\begingroup$ I was wondering about that because then, I can always calculate the trace of the operator with the formula $\int k_T(x,x) dx$. To be more precise, given an traceclass self-adjoint operator T, I can WLOG say that there exists an integral kernel $k_T(x,y)$ such that $tr T = \int k_T(x,x) dx$. However, given an integral kernel, I can only use the trace formula as long as this kernel is continuous. Are my thoughts correct? $\endgroup$ – Peter Feb 25 '16 at 16:22
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    $\begingroup$ Yeah, that's a little sketchy because the set $\{(x,x): x \in \mathbb{R}\}$ has measure zero. But yes, if the kernel is continuous then you're okay. $\endgroup$ – Nik Weaver Feb 25 '16 at 16:26
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    $\begingroup$ So my question was actually about the sketchy part. So given a trace-class, s.a. operator $T$. Then, one element of the equivalence class of the integral kernel is given by $k_T(x,y) =\sum_i \lambda_i \phi_i(x) \bar\phi_i(y)$. Then, $\int k_T(x,x) dx = \sum_i \lambda_i$ which is exactly the trace. This gives indeed the exact value of the trace, although $k_T(x,y)$ may not be continuous. I was wondering if this is correct. $\endgroup$ – Peter Feb 25 '16 at 16:35
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    $\begingroup$ Yes, I agree with you. As long as you're careful that the kernel is given in this way, that is correct. $\endgroup$ – Nik Weaver Feb 25 '16 at 16:37
  • $\begingroup$ Assume that $\text{tr}T = 1$. Can we rewrite $T$ under the form $T=\int |u\rangle\langle u| {\rm d}\mu(u)$ where ${\rm d}\mu$ is a Borel probability measure on the unit sphere?. $\endgroup$ – Muniain Jul 12 at 7:29

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