Let $X$ be a compact metric space and $\mu$ be strictly positive Borel measure on $X$. Let $T:L^2(X,\mu)\rightarrow L^2(X,\mu)$ be a self-adjoint, compact, and positive operator on the Hilbert space $L^2(X,\mu)$ with continuous kernel $K:X\times X\rightarrow\mathbb{R}$. In other words, $$Tf(x)=\int_X K(x,y)f(y) d\mu(y)$$ for any $x\in X$ and $f\in L^2(X,\mu)$. My question is the following one.

Q1: It is known that there exist the unique square root operator $T^{1/2}$ of $T$. Then, this $T^{1/2}$ is also an integral operator with some continuous kernel $K':X\times X\rightarrow\mathbb{R}$?

Since I am not an expert of operator theory, maybe this is not research level question. But at least I could not find any literature answer this.

My guess is, probably the answer is false. But I could not come up with any counterexamples.

Q2: If the answer is false, then what is counterexample?

Q3: By adding some mild conditions, are we able to make the statement true?

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    $\begingroup$ As you suspected, this need not hold, and the third answer to this question (by BigM) provides a counterexample: mathoverflow.net/questions/70969/… (To see that the square root is not an integral operator with a continuous kernel, recall that such an operator would be Hilbert-Schmidt.) $\endgroup$ – Christian Remling Nov 20 '19 at 22:33
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    $\begingroup$ And we don't really need any very fancy stuff here, just a continuous function whose Fourier coefficients are not in $\ell^1$. $\endgroup$ – Christian Remling Nov 20 '19 at 22:38
  • $\begingroup$ Thank you! as I expected, the answer for Q1 is false. Then, suppose that the summation of all the eigenvalues of $T$ are finite. This implies that $T^{1/2}$ is actually Hilbert-Schimidt operator. Therefore, there must be $K'\in L^2(\mu\otimes\mu)$ generating the operator $T^{1/2}$ (I saw this theorem from "Methods of Modern Mathematical Physics"). $\endgroup$ – S.Lim Nov 21 '19 at 4:28
  • $\begingroup$ Then, next question is, how can we guarantee the "continuity" of $K'$? Are there some references about this question? Under what condition can we guarantee continuity of kernel for Hilbert-Schmidt operator? $\endgroup$ – S.Lim Nov 21 '19 at 4:29
  • $\begingroup$ This preprint seems to be related to the question (although it does, probably, not answer the question itself). $\endgroup$ – Jochen Glueck Nov 22 '19 at 10:06

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