0
$\begingroup$

Let $f$ be a function such that $supp(f)=K$. Compute $supp(f_\varepsilon)$ where $f_\varepsilon$ is the regularization of $f$. Im not sure how to do this, since we have no information of the support of the convolution of two functions. We do know that $supp(\omega_\varepsilon)=B(0,\varepsilon)$ and $\omega_\varepsilon=e^{\frac{-1}{1-||x/\varepsilon||^2}} $ for all $\|x\|<\varepsilon$ and 0 otherwise.

$\endgroup$
0
$\begingroup$

The convolution is $$ (f\ast \omega_{\varepsilon})(x)=\int\ f(y)\omega_{\varepsilon}(x-y)\ dy $$ Then note that $f(y)\omega_{\varepsilon}(x-y)\neq 0$ implies $y\in{\rm supp}(f)$ and $x-y\in \bar{B}(0,\varepsilon)$ the closed ball around he origin of radius $\varepsilon$. So the support of the convolution is contained in ${\rm supp}(f)+\bar{B}(0,\varepsilon)$, i.e., the epsilon thickening of the support of $f$. I don't think one can in general compute the support of the convolution exactly because of possible sign cancellations. If $f$ and $\omega_{\varepsilon}$ are nonnegative then the above sum of sets is the support of the convolution.

$\endgroup$

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.