Integral operator defined on $two$ distinct dense subspaces

Question

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?

Answer 1

Unfortunately, "its definition is identical" is not a precise mathematical notion, it does not exclude the possibility that the operator is defined using cases.

You can take, as suggested by Christian Remling, the two dense subspaces with trivial intersection given by polynomial functions and by $C_0^\infty=\{C^\infty$ functions that vanish in a nbh of the $\partial$ $\}$ inside $L^2(0,1)$, and take the definition of your operator $A$ to be $$ A(f)=\begin{cases} \int_0^1 f(x)dx&\text{if $f$ is a polynomial} \\ f(\frac12) &\text{if } f\in C_0^\infty \end{cases} $$