Consider the Hilbert space $L\_2=L\_2[0,1]$. Is it true that for each nuclear (trace-class) operator on $L\_2$ there exists a function $K\in L\_1(L\_2)$ such that


$$Tf = \int\limits\_0^1 K(s) f(s) \,\mu(\mbox{d}s)?$$


The above is an integral in the sense of Bochner.


It seems to be a part of the folkolre but I was unable to trace a reference. Is it true for nuclear operator on $L\_p$ where $p\in (1,\infty)$?



Nuclear vs Integral operators on Hilbert spaces
Each trace class operator on $L\_2(0,1)$ is a Hilbert-Schmidt operator, but not conversely.


Each $L\_2(0,1)$-function is in $L\_1(0,1)$, but not conversely.


Each Hilbert-Schmidt operator on $L\_2$ is uniquely determined by an $L\_2(L\_2)$-function - indeed, it is an integral operator with $L\_2(L\_2)$ kernel.


So, the answer to your first question is yes, but I guess the answer to the second one is no.


The relation between tensor products and $L\_p$-spaces is horrible, cf. the final page of this note (<http://arxiv.org/abs/0903.2038>), so I doubt that your last question cannot have an affirmative answer.


