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?

  • 3
    $\begingroup$ 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). $\endgroup$ – Anthony Quas May 17 '14 at 3:54
  • $\begingroup$ Alright. Guess I'll delete it then. $\endgroup$ – Cameron Williams May 17 '14 at 3:55
  • $\begingroup$ 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). $\endgroup$ – Christian Remling May 17 '14 at 3:56
  • 4
    $\begingroup$ 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. $\endgroup$ – fedja May 18 '14 at 1:42
  • 2
    $\begingroup$ @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. $\endgroup$ – Cameron Williams May 18 '14 at 3:39

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.