# Integral operator defined on $two$ distinct dense subspaces

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?

• I think cross-posting is quite strongly discouraged (as it causes wasted effort). If in doubt, a reasonable strategy is probably to post on MSE; wait a reasonable amount of time (at least a couple of days); then move to MO at that time, indicating that is what you've done on your original post. (Others may be able to give better etiquette advice here). – Anthony Quas May 17 '14 at 3:54
• Alright. Guess I'll delete it then. – Cameron Williams May 17 '14 at 3:55
• I'm assuming that by "identical on both subspaces" you mean that the operators agree on $H_1\cap H_2$. Then what you're suggesting will not work. For example, $H_2=H_1+ L(x)$ with $x\notin H_1$, and now you can define $Tx$ arbitrarily (since you didn't insist on $T_2$ bounded). – Christian Remling May 17 '14 at 3:56
• It is easy to show that it does not need to have the same kernel: Take $S_1$ be the space of functions that are in $L^1\cap L^2$ and have $0$ integral (it is dense in $L^2$!) and let $S_2$ be the set of all functions with compact support. Now just put $k(x,y)=1$. The extension from $S_1$ is then the zero operator. It is still integral, though, so the first of your questions requires a bit more thought. – fedja May 18 '14 at 1:42
• @fedja well that is disheartening. Seems like it can't really be done unfortunately. If you want to make this an answer (counter example), I'll gladly accept it. – Cameron Williams May 18 '14 at 3:39

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 $$A(f)=\begin{cases} \int_0^1 f(x)dx&\text{if f is a polynomial} \\ f(\frac12) &\text{if } f\in C_0^\infty \end{cases}$$