2
$\begingroup$

I have a question in my mind. I would like to know whether a linear operator actually preserves compact support? Or is there any relation? Does it make sense to ask this question, at all?

  • 0
    @NateEldredge I didn't bother to specify a space, but can you explain me how it would make a difference ?2012-05-24

1 Answers 1

2

Consider the Kernel $K(x,y)=e^{-x^2-y^2}$ integrating by polar coordinates we get $\pi=\int_{\mathbb{R}^2 }K(x,y)dxdy$

Now considere the operator $T:L^{\infty}\to L^\infty$ given by

$Tf(x)=\int_{\mathbb{R} }K(x,y)f(y)dy$ This operator is linear and bounded because

$|Tf(x)|\leq\int_{\mathbb{R} }K(x,y)|f(y)|dy\leq \int_{\mathbb{R} }K(x,y)||f||_{\infty}dy \leq e^{-x^2}\int_{\mathbb{R}}e^{-y^2}||f||_{\infty}dy\leq e^{-x^2}\sqrt{\pi}||f||_{\infty}\leq \sqrt{\pi}||f||_\infty$

Then $||Tf||_{\infty}\leq\sqrt{\pi}||f||_\infty$.

But considere $\phi\geq 0$ with support in $(-2,2)$ such $\phi=1$ in $(-1,1)$

Then $T\phi(x)=Tf(x)=\int_{\mathbb{R} }K(x,y)\phi(y)dy\geq \int_{-1}^{1}e^{-x^2-y^2}1 dy>0$

For any $x \in \mathbb{R}$. That is with no compact support.