Support of a regularized functionRelationship of the support of a test function with the support of...
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Support of a regularized function
Relationship of the support of a test function with the support of distribution.Support of a generalized functionWhen does the regularization of a function converges to the function?what does support of convolution of functions says geometrically?Support of convolutionSupport of derived functionProperty singular support of the convolution of distributionsConvolution support and propertyConcept of supportTest functions-support function
$begingroup$
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.
real-analysis distribution-theory regularity-theory-of-pdes
asked Mar 12 at 3:15
user593295user593295
628
$endgroup$
add a comment |
$begingroup$
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.
real-analysis distribution-theory regularity-theory-of-pdes
asked Mar 12 at 3:15
user593295user593295
628
$endgroup$
add a comment |
$begingroup$
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.
real-analysis distribution-theory regularity-theory-of-pdes
asked Mar 12 at 3:15
user593295user593295
628
$endgroup$
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.
real-analysis distribution-theory regularity-theory-of-pdes
real-analysis distribution-theory regularity-theory-of-pdes
asked Mar 12 at 3:15
user593295user593295
628
asked Mar 12 at 3:15
user593295user593295
628
asked Mar 12 at 3:15
user593295user593295
628
asked Mar 12 at 3:15
user593295user593295
628
asked Mar 12 at 3:15
user593295user593295
628
628
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The convolution is
$$
(fast omega_{varepsilon})(x)=int f(y)omega_{varepsilon}(x-y) dy
$$
Then note that $f(y)omega_{varepsilon}(x-y)neq 0$ implies $yin{rm supp}(f)$ and $x-yin bar{B}(0,varepsilon)$ 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.
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
$endgroup$
add a comment |
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$begingroup$
The convolution is
$$
(fast omega_{varepsilon})(x)=int f(y)omega_{varepsilon}(x-y) dy
$$
Then note that $f(y)omega_{varepsilon}(x-y)neq 0$ implies $yin{rm supp}(f)$ and $x-yin bar{B}(0,varepsilon)$ 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.
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
$endgroup$
add a comment |
$begingroup$
The convolution is
$$
(fast omega_{varepsilon})(x)=int f(y)omega_{varepsilon}(x-y) dy
$$
Then note that $f(y)omega_{varepsilon}(x-y)neq 0$ implies $yin{rm supp}(f)$ and $x-yin bar{B}(0,varepsilon)$ 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.
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
$endgroup$
add a comment |
$begingroup$
The convolution is
$$
(fast omega_{varepsilon})(x)=int f(y)omega_{varepsilon}(x-y) dy
$$
Then note that $f(y)omega_{varepsilon}(x-y)neq 0$ implies $yin{rm supp}(f)$ and $x-yin bar{B}(0,varepsilon)$ 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.
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
$endgroup$
The convolution is
$$
(fast omega_{varepsilon})(x)=int f(y)omega_{varepsilon}(x-y) dy
$$
Then note that $f(y)omega_{varepsilon}(x-y)neq 0$ implies $yin{rm supp}(f)$ and $x-yin bar{B}(0,varepsilon)$ 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.
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
answered Mar 12 at 9:56
Abdelmalek AbdesselamAbdelmalek Abdesselam
756311
756311
add a comment |
add a comment |
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