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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?

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  • $\begingroup$ I am not sure even for compact operators, try the convolution with a good kernel. $\endgroup$ – checkmath May 23 '12 at 10:40
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    $\begingroup$ Consider the linear operator $\mathbb{R} \to C(\mathbb{R})$ sending $t$ to the constant function $f(x) = t$. Note that $\mathbb{R}$ can be considered as the space of continuous functons on a point. $\endgroup$ – t.b. May 23 '12 at 10:41
  • $\begingroup$ Considere a kernel $K(x,y)=e^{-x^2-y^2}$. $Tf(x)=\int_{R^2}K(x,y)f(y)dy$ $\endgroup$ – checkmath May 23 '12 at 11:01
  • $\begingroup$ @checkmath I am not able to figure out what you are saying . Can u help ? $\endgroup$ – Theorem May 23 '12 at 14:55
  • $\begingroup$ @Ananda can you prove that the above operator $T: L^{\infty}\to L^{\infty}$ is linear continuous? If yes prove that this operator doesn't satisfies your Thesis. $\endgroup$ – checkmath May 23 '12 at 22:50
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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.

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