$lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$.
$begingroup$
How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?
The definition is,
$lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.
I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.
analysis fourier-analysis fourier-transform harmonic-analysis
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
$endgroup$
add a comment |
$begingroup$
How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?
The definition is,
$lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.
I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.
analysis fourier-analysis fourier-transform harmonic-analysis
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
$endgroup$
add a comment |
$begingroup$
How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?
The definition is,
$lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.
I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.
analysis fourier-analysis fourier-transform harmonic-analysis
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
$endgroup$
How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?
The definition is,
$lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.
I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.
analysis fourier-analysis fourier-transform harmonic-analysis
analysis fourier-analysis fourier-transform harmonic-analysis
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
asked Dec 16 '18 at 3:22
IdkwhatIdkwhat
236
236
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Idea: If you take $f(x) = delta(x)$, then
you see that $hat f(xi) = text{const}$, say $1$. Hence we have
begin{align}
|f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
end{align}
However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.
Spoiler:
Let $d=1$. Take the sequence
$f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
$endgroup$
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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active
oldest
votes
$begingroup$
Idea: If you take $f(x) = delta(x)$, then
you see that $hat f(xi) = text{const}$, say $1$. Hence we have
begin{align}
|f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
end{align}
However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.
Spoiler:
Let $d=1$. Take the sequence
$f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
$endgroup$
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
add a comment |
$begingroup$
Idea: If you take $f(x) = delta(x)$, then
you see that $hat f(xi) = text{const}$, say $1$. Hence we have
begin{align}
|f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
end{align}
However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.
Spoiler:
Let $d=1$. Take the sequence
$f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
$endgroup$
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
add a comment |
$begingroup$
Idea: If you take $f(x) = delta(x)$, then
you see that $hat f(xi) = text{const}$, say $1$. Hence we have
begin{align}
|f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
end{align}
However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.
Spoiler:
Let $d=1$. Take the sequence
$f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
$endgroup$
Idea: If you take $f(x) = delta(x)$, then
you see that $hat f(xi) = text{const}$, say $1$. Hence we have
begin{align}
|f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
end{align}
However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.
Spoiler:
Let $d=1$. Take the sequence
$f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
answered Dec 16 '18 at 5:24
Jacky ChongJacky Chong
18.8k21129
18.8k21129
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
add a comment |
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
$endgroup$
– Idkwhat
Dec 16 '18 at 9:13
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Delta function is not a function.
$endgroup$
– Jacky Chong
Dec 16 '18 at 18:10
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Ooh, thank you.
$endgroup$
– Idkwhat
Dec 17 '18 at 5:49
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
$endgroup$
– Idkwhat
Dec 17 '18 at 8:37
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
$begingroup$
Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
$endgroup$
– Jacky Chong
Dec 17 '18 at 13:51
add a comment |
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