Two questions about mollifiers
Let $f$ be a smooth function whose laplacian equals 1 everywhere in $mathbb{R}^{p}$ and $p>1$. Let $B(r)$ be the ball of radius $r>0$ and centered at the origin in $mathbb{R}^{p}$, and let $ chi_{r} $ designate the characteristic function of $B(r)$ (it equals 1 on the ball and 0 elsewhere).
Finally set $g=chi_{r}f$ and for $n=1,2,...$, let $psi_{n}$ be the $ C^{infty}-$regularization of $ g $, i.e.,
$$ psi_{n}(zeta)=int_{mathbb{R}^{p}}g(y)phi_{n}(zeta-y)dlambda_{p}(y) $$
with $ phi_{n}(t)=phi(1-n^{2}mid xmid^{2}) $, and
$phi(t)$ is equal to $C_{p}e^{-1/t^{}}$ for $|t|>0$ and equals $0$ for $|t|leq 0$, and $ C_{p} $ is a constant chosen so that
$$ sigma_{p}int_{0}^{1}t^{p-1}phi(1-t^{2} )dt=1.$$
My questions are:
1) If $R>r>0$ and for $n$ big enough, is each $psi_{n}$ a smooth function with compact support in $B(R)$; i.e., $psi_{n}in C_{c}^{infty}(B(R))$?
2) Is it true that $lim_{ntoinfty}Delta psi_{n}=chi_{r}?$($Delta$ is the laplacian).
My answers to both questions is yes. I am right?
real-analysis distribution-theory
asked Dec 3 '18 at 1:07
M. Rahmat
332212
|
show 3 more comments
Let $f$ be a smooth function whose laplacian equals 1 everywhere in $mathbb{R}^{p}$ and $p>1$. Let $B(r)$ be the ball of radius $r>0$ and centered at the origin in $mathbb{R}^{p}$, and let $ chi_{r} $ designate the characteristic function of $B(r)$ (it equals 1 on the ball and 0 elsewhere).
Finally set $g=chi_{r}f$ and for $n=1,2,...$, let $psi_{n}$ be the $ C^{infty}-$regularization of $ g $, i.e.,
$$ psi_{n}(zeta)=int_{mathbb{R}^{p}}g(y)phi_{n}(zeta-y)dlambda_{p}(y) $$
with $ phi_{n}(t)=phi(1-n^{2}mid xmid^{2}) $, and
$phi(t)$ is equal to $C_{p}e^{-1/t^{}}$ for $|t|>0$ and equals $0$ for $|t|leq 0$, and $ C_{p} $ is a constant chosen so that
$$ sigma_{p}int_{0}^{1}t^{p-1}phi(1-t^{2} )dt=1.$$
My questions are:
1) If $R>r>0$ and for $n$ big enough, is each $psi_{n}$ a smooth function with compact support in $B(R)$; i.e., $psi_{n}in C_{c}^{infty}(B(R))$?
2) Is it true that $lim_{ntoinfty}Delta psi_{n}=chi_{r}?$($Delta$ is the laplacian).
My answers to both questions is yes. I am right?
real-analysis distribution-theory
asked Dec 3 '18 at 1:07
M. Rahmat
332212
You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
1
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
1
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
1
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
1
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37
|
show 3 more comments
Let $f$ be a smooth function whose laplacian equals 1 everywhere in $mathbb{R}^{p}$ and $p>1$. Let $B(r)$ be the ball of radius $r>0$ and centered at the origin in $mathbb{R}^{p}$, and let $ chi_{r} $ designate the characteristic function of $B(r)$ (it equals 1 on the ball and 0 elsewhere).
Finally set $g=chi_{r}f$ and for $n=1,2,...$, let $psi_{n}$ be the $ C^{infty}-$regularization of $ g $, i.e.,
$$ psi_{n}(zeta)=int_{mathbb{R}^{p}}g(y)phi_{n}(zeta-y)dlambda_{p}(y) $$
with $ phi_{n}(t)=phi(1-n^{2}mid xmid^{2}) $, and
$phi(t)$ is equal to $C_{p}e^{-1/t^{}}$ for $|t|>0$ and equals $0$ for $|t|leq 0$, and $ C_{p} $ is a constant chosen so that
$$ sigma_{p}int_{0}^{1}t^{p-1}phi(1-t^{2} )dt=1.$$
My questions are:
1) If $R>r>0$ and for $n$ big enough, is each $psi_{n}$ a smooth function with compact support in $B(R)$; i.e., $psi_{n}in C_{c}^{infty}(B(R))$?
2) Is it true that $lim_{ntoinfty}Delta psi_{n}=chi_{r}?$($Delta$ is the laplacian).
My answers to both questions is yes. I am right?
real-analysis distribution-theory
asked Dec 3 '18 at 1:07
M. Rahmat
332212
Let $f$ be a smooth function whose laplacian equals 1 everywhere in $mathbb{R}^{p}$ and $p>1$. Let $B(r)$ be the ball of radius $r>0$ and centered at the origin in $mathbb{R}^{p}$, and let $ chi_{r} $ designate the characteristic function of $B(r)$ (it equals 1 on the ball and 0 elsewhere).
Finally set $g=chi_{r}f$ and for $n=1,2,...$, let $psi_{n}$ be the $ C^{infty}-$regularization of $ g $, i.e.,
$$ psi_{n}(zeta)=int_{mathbb{R}^{p}}g(y)phi_{n}(zeta-y)dlambda_{p}(y) $$
with $ phi_{n}(t)=phi(1-n^{2}mid xmid^{2}) $, and
$phi(t)$ is equal to $C_{p}e^{-1/t^{}}$ for $|t|>0$ and equals $0$ for $|t|leq 0$, and $ C_{p} $ is a constant chosen so that
$$ sigma_{p}int_{0}^{1}t^{p-1}phi(1-t^{2} )dt=1.$$
My questions are:
1) If $R>r>0$ and for $n$ big enough, is each $psi_{n}$ a smooth function with compact support in $B(R)$; i.e., $psi_{n}in C_{c}^{infty}(B(R))$?
2) Is it true that $lim_{ntoinfty}Delta psi_{n}=chi_{r}?$($Delta$ is the laplacian).
My answers to both questions is yes. I am right?
real-analysis distribution-theory
real-analysis distribution-theory
asked Dec 3 '18 at 1:07
M. Rahmat
332212
asked Dec 3 '18 at 1:07
M. Rahmat
332212
edited Dec 3 '18 at 1:57
asked Dec 3 '18 at 1:07
M. Rahmat
332212
asked Dec 3 '18 at 1:07
M. Rahmat
332212
asked Dec 3 '18 at 1:07
M. Rahmat
332212
332212
You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
1
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
1
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
1
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
1
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37
|
show 3 more comments
You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
1
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
1
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
1
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
1
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37
You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
1
1
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
1
1
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
1
1
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
1
1
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37
|
show 3 more comments
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You are right and thanks for letting me know. I corrected.
– M. Rahmat
Dec 3 '18 at 1:58
1
What is $lambda_p$ ?
– reuns
Dec 3 '18 at 2:17
1
If you meant $psi_n = f chi_r ast phi_n$ which is of course $C^infty_c$ then $Delta psi_n = chi_rDelta f ast phi_n+ f Delta chi_r ast phi_n + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) ast phi_n$. As $n to infty$ the $ ast phi_n$ disappear. For $|x| ne r$ the derivatives of $chi_r$ are $0$. Whence $lim_{n to infty}Delta psi_n = chi_rDelta f + T$ where $T = f Delta chi_r + sum_i (partial_{x_i} chi_r)(partial_{x_i} f) $ is a distribution supported on $|x|=r$.
– reuns
Dec 3 '18 at 2:26
1
$chi_r$ is constant on $|x|> r$ and $|x| < r$ so its derivatives vanish there. Hence the distributional derivatives of $chi_r$ are distributions supported on $|x|=r$.
– reuns
Dec 3 '18 at 4:11
1
The distributional derivative of $f(t)=frac{1+sign(t)}{2}$ is $delta(t)$. And the value of $f(0)$ doesn't play any role, we only care that $f$ is constant for $t >0$ and $t < 0$.
– reuns
Dec 3 '18 at 5:37