Approximate Identity
$begingroup$
Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space.
Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
begin{equation}
lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
end{equation}
convolution
asked Jan 8 at 19:45
ManiMani
92
$endgroup$
add a comment |
$begingroup$
Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space.
Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
begin{equation}
lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
end{equation}
convolution
asked Jan 8 at 19:45
ManiMani
92
$endgroup$
$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17
add a comment |
$begingroup$
Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space.
Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
begin{equation}
lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
end{equation}
convolution
asked Jan 8 at 19:45
ManiMani
92
$endgroup$
Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space.
Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
begin{equation}
lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
end{equation}
convolution
convolution
asked Jan 8 at 19:45
ManiMani
92
asked Jan 8 at 19:45
ManiMani
92
edited Jan 21 at 23:03
Mani
asked Jan 8 at 19:45
ManiMani
92
asked Jan 8 at 19:45
ManiMani
92
asked Jan 8 at 19:45
ManiMani
92
92
$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17
add a comment |
$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17
$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
$begingroup$
It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
$begingroup$
It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
edited Jan 16 at 0:33
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Jan 16 at 0:22
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
17.8k12247
add a comment |
add a comment |
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$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30
$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06
$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17