Limit of a convolution 2
$begingroup$
Let $phi$ be a non-negative test function with total integral =1 and with support in the ball of radius 1 and centered at the origin in $mathbb{R}^{n}$. Set $phi_{t}(x) =t^{-n}phi(x/t)$. Is the following limit
$$limsup_{trightarrow0}int_{mathbb{R}^{n}}|Deltaphi_{t}(y)|dy$$
bounded? ($Delta$ is the laplacian)
real-analysis distribution-theory
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
$endgroup$
add a comment |
$begingroup$
Let $phi$ be a non-negative test function with total integral =1 and with support in the ball of radius 1 and centered at the origin in $mathbb{R}^{n}$. Set $phi_{t}(x) =t^{-n}phi(x/t)$. Is the following limit
$$limsup_{trightarrow0}int_{mathbb{R}^{n}}|Deltaphi_{t}(y)|dy$$
bounded? ($Delta$ is the laplacian)
real-analysis distribution-theory
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
$endgroup$
$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17
add a comment |
$begingroup$
Let $phi$ be a non-negative test function with total integral =1 and with support in the ball of radius 1 and centered at the origin in $mathbb{R}^{n}$. Set $phi_{t}(x) =t^{-n}phi(x/t)$. Is the following limit
$$limsup_{trightarrow0}int_{mathbb{R}^{n}}|Deltaphi_{t}(y)|dy$$
bounded? ($Delta$ is the laplacian)
real-analysis distribution-theory
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
$endgroup$
Let $phi$ be a non-negative test function with total integral =1 and with support in the ball of radius 1 and centered at the origin in $mathbb{R}^{n}$. Set $phi_{t}(x) =t^{-n}phi(x/t)$. Is the following limit
$$limsup_{trightarrow0}int_{mathbb{R}^{n}}|Deltaphi_{t}(y)|dy$$
bounded? ($Delta$ is the laplacian)
real-analysis distribution-theory
real-analysis distribution-theory
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
edited Dec 17 '18 at 5:33
M. Rahmat
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
asked Sep 30 '17 at 1:58
M. RahmatM. Rahmat
332212
332212
$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17
add a comment |
$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17
$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17
$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
For $t>0$ we can write by definition that
$$int_{Bbb R^n} |Delta phi_t(y)|dy = int_{Bbb R^n} t^{-n}| (partial_{y_1}^2 +ldots+partial_{y_n}^2) phi(y/t)|dy = int_{Bbb R^n} t^{-n-2}| (Delta phi)(y/t)|dy $$
$$= int_{Bbb R^n} t^{-2}| (Delta phi)(u)|du. $$
Therefore, unless $int_{Bbb R^n}| (Delta phi)(u)|du=0$, you limit diverges to $+infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $phi$ you have $int_{Bbb R^n}| (Delta phi)(u)|du>0$.
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $t>0$ we can write by definition that
$$int_{Bbb R^n} |Delta phi_t(y)|dy = int_{Bbb R^n} t^{-n}| (partial_{y_1}^2 +ldots+partial_{y_n}^2) phi(y/t)|dy = int_{Bbb R^n} t^{-n-2}| (Delta phi)(y/t)|dy $$
$$= int_{Bbb R^n} t^{-2}| (Delta phi)(u)|du. $$
Therefore, unless $int_{Bbb R^n}| (Delta phi)(u)|du=0$, you limit diverges to $+infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $phi$ you have $int_{Bbb R^n}| (Delta phi)(u)|du>0$.
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
add a comment |
$begingroup$
For $t>0$ we can write by definition that
$$int_{Bbb R^n} |Delta phi_t(y)|dy = int_{Bbb R^n} t^{-n}| (partial_{y_1}^2 +ldots+partial_{y_n}^2) phi(y/t)|dy = int_{Bbb R^n} t^{-n-2}| (Delta phi)(y/t)|dy $$
$$= int_{Bbb R^n} t^{-2}| (Delta phi)(u)|du. $$
Therefore, unless $int_{Bbb R^n}| (Delta phi)(u)|du=0$, you limit diverges to $+infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $phi$ you have $int_{Bbb R^n}| (Delta phi)(u)|du>0$.
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
add a comment |
$begingroup$
For $t>0$ we can write by definition that
$$int_{Bbb R^n} |Delta phi_t(y)|dy = int_{Bbb R^n} t^{-n}| (partial_{y_1}^2 +ldots+partial_{y_n}^2) phi(y/t)|dy = int_{Bbb R^n} t^{-n-2}| (Delta phi)(y/t)|dy $$
$$= int_{Bbb R^n} t^{-2}| (Delta phi)(u)|du. $$
Therefore, unless $int_{Bbb R^n}| (Delta phi)(u)|du=0$, you limit diverges to $+infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $phi$ you have $int_{Bbb R^n}| (Delta phi)(u)|du>0$.
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
For $t>0$ we can write by definition that
$$int_{Bbb R^n} |Delta phi_t(y)|dy = int_{Bbb R^n} t^{-n}| (partial_{y_1}^2 +ldots+partial_{y_n}^2) phi(y/t)|dy = int_{Bbb R^n} t^{-n-2}| (Delta phi)(y/t)|dy $$
$$= int_{Bbb R^n} t^{-2}| (Delta phi)(u)|du. $$
Therefore, unless $int_{Bbb R^n}| (Delta phi)(u)|du=0$, you limit diverges to $+infty$. On the other hand, compactly supported harmonic functions are identically zero, therefore for your function $phi$ you have $int_{Bbb R^n}| (Delta phi)(u)|du>0$.
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
answered Oct 4 '17 at 8:42
TZakrevskiyTZakrevskiy
20.1k12354
20.1k12354
add a comment |
add a comment |
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$begingroup$
Where is the convolution here?
$endgroup$
– TZakrevskiy
Oct 4 '17 at 8:17