I have decided to edit my post a bit heavily for clarity. I was trying to be fairly general but it's hard to see what I'm asking so I've decided to limit myself to a specific example which will hopefully make it clearer.

Suppose I have an integral operator $T$ that can be defined on two distinct (zero intersection) subspaces $S_1$ and $S_2$ of $L^2(\mathbb{R})$ which is not *a priori* bounded. That is to say that for $f\in S_1$ or $f\in S_2$, $Tf$ exists pointwise almost everywhere and is given by

$$Tf(y) = \int_{-\infty}^{\infty} k(x,y)f(x) dx$$

for some kernel $k$.

Suppose then that on $S_1$ I can actually show that $Tf\in L^2(\mathbb{R})$ and $\|Tf\|\le C\|f\|$. Since $S_1$ is dense in $L^2(\mathbb{R})$, $T$ can be extended to a bounded operator on all of $L^2(\mathbb{R})$. Call the extension $\widetilde{T}$.

However this extension need not (and in many cases will not) be an integral operator on all of $L^2(\mathbb{R})$. Clearly, $\widetilde{T}|_{S_2}$ is a bounded operator but is it the case that $\widetilde{T}|_{S_2}$ is actually an integral operator? If so, does it agree identically with the definition above?

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