What Soboblev embedding i could use in this case?Sobolev spaces doubtSobelev inequalityHölder's inequality and log convexity of $L^p$ normWhat is the norm of the pre-multiplication by a fixed matrix operator?Sobolev embedding theorem, inequalitiesRelation between Besov and Sobolev spaces (Littlewood-Paley-theory)Equivalence of norms on $W^1, p(I)$If $uin W^1,2(mathbb R^n)$, then $L^2^*-$norm of $u$ controlled by $L^2-$norm of $nabla u$?Comparison of Sobolev embedding for $mathbbR^n$ and a compact manifold $X^n$.An inequality in Sobolev spaces
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What Soboblev embedding i could use in this case?
Sobolev spaces doubtSobelev inequalityHölder's inequality and log convexity of $L^p$ normWhat is the norm of the pre-multiplication by a fixed matrix operator?Sobolev embedding theorem, inequalitiesRelation between Besov and Sobolev spaces (Littlewood-Paley-theory)Equivalence of norms on $W^1, p(I)$If $uin W^1,2(mathbb R^n)$, then $L^2^*-$norm of $u$ controlled by $L^2-$norm of $nabla u$?Comparison of Sobolev embedding for $mathbbR^n$ and a compact manifold $X^n$.An inequality in Sobolev spaces
$begingroup$
I'm trying to understand why this inequality is true
$$ Vert fVert_L^2^pcdot hleft(Vert fVert_inftyright) leq c Vert fVert_k+1^p cdot hleft(Vert fVert_k+1right), $$
where $ Vert cdotVert_k+1 $ denotes the usual norm of the Sobolev Space $ H^k+1(R^n) $, $ h(y) $ denotes a nondecreasing nonnegative and continuous function, c is a costant, $k geq 0 $ and $ p> 1$.
Maybe i can use an embedding theorem? which one? Can anyone please help me?
real-analysis inequality sobolev-spaces
asked Mar 28 at 15:56
C. BishopC. Bishop
398
$endgroup$
add a comment |
$begingroup$
I'm trying to understand why this inequality is true
$$ Vert fVert_L^2^pcdot hleft(Vert fVert_inftyright) leq c Vert fVert_k+1^p cdot hleft(Vert fVert_k+1right), $$
where $ Vert cdotVert_k+1 $ denotes the usual norm of the Sobolev Space $ H^k+1(R^n) $, $ h(y) $ denotes a nondecreasing nonnegative and continuous function, c is a costant, $k geq 0 $ and $ p> 1$.
Maybe i can use an embedding theorem? which one? Can anyone please help me?
real-analysis inequality sobolev-spaces
asked Mar 28 at 15:56
C. BishopC. Bishop
398
$endgroup$
1
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
2
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25
add a comment |
$begingroup$
I'm trying to understand why this inequality is true
$$ Vert fVert_L^2^pcdot hleft(Vert fVert_inftyright) leq c Vert fVert_k+1^p cdot hleft(Vert fVert_k+1right), $$
where $ Vert cdotVert_k+1 $ denotes the usual norm of the Sobolev Space $ H^k+1(R^n) $, $ h(y) $ denotes a nondecreasing nonnegative and continuous function, c is a costant, $k geq 0 $ and $ p> 1$.
Maybe i can use an embedding theorem? which one? Can anyone please help me?
real-analysis inequality sobolev-spaces
asked Mar 28 at 15:56
C. BishopC. Bishop
398
$endgroup$
I'm trying to understand why this inequality is true
$$ Vert fVert_L^2^pcdot hleft(Vert fVert_inftyright) leq c Vert fVert_k+1^p cdot hleft(Vert fVert_k+1right), $$
where $ Vert cdotVert_k+1 $ denotes the usual norm of the Sobolev Space $ H^k+1(R^n) $, $ h(y) $ denotes a nondecreasing nonnegative and continuous function, c is a costant, $k geq 0 $ and $ p> 1$.
Maybe i can use an embedding theorem? which one? Can anyone please help me?
real-analysis inequality sobolev-spaces
real-analysis inequality sobolev-spaces
asked Mar 28 at 15:56
C. BishopC. Bishop
398
asked Mar 28 at 15:56
C. BishopC. Bishop
398
asked Mar 28 at 15:56
C. BishopC. Bishop
398
asked Mar 28 at 15:56
C. BishopC. Bishop
398
asked Mar 28 at 15:56
C. BishopC. Bishop
398
398
1
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
2
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25
add a comment |
1
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
2
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25
1
1
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
2
2
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using a bit of Fourier analysis, one can show that, for $s>fracd2$,
$$H^s(mathbbR^d) hookrightarrow L^infty(mathbbR^d).$$
In particular, consider
$$|f|_L^infty leq |hatf|_L^1 leq |langle xirangle ^-s|_L^2 |langle xirangle ^shatf|_L^2 lesssim |f|_H^s,
$$
where the first inequality follows, pretty much directly, from the Fourier inversion formula.
Now, $|f|_L^2 leq|f|_H^k+1$ for every $kgeq 0$, so that part is fine, too. But, you still have some work to do. In particular, you have
$$h(|f|_L^infty) leq h(c|f|_H^s),$$
but you need to get to
$$h(|f|_L^infty) lesssim h(|f|_H^s).$$
You may need some additional assumptions on $h$ (particularly on how $h$ scales) to do this.
As a bit of a tangent, it's actually the case that for $s > fracd2 + k$, we have $H^s(mathbbR^d) hookrightarrow C^k(mathbbR^d) cap L^infty(mathbbR^d)$.
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Using a bit of Fourier analysis, one can show that, for $s>fracd2$,
$$H^s(mathbbR^d) hookrightarrow L^infty(mathbbR^d).$$
In particular, consider
$$|f|_L^infty leq |hatf|_L^1 leq |langle xirangle ^-s|_L^2 |langle xirangle ^shatf|_L^2 lesssim |f|_H^s,
$$
where the first inequality follows, pretty much directly, from the Fourier inversion formula.
Now, $|f|_L^2 leq|f|_H^k+1$ for every $kgeq 0$, so that part is fine, too. But, you still have some work to do. In particular, you have
$$h(|f|_L^infty) leq h(c|f|_H^s),$$
but you need to get to
$$h(|f|_L^infty) lesssim h(|f|_H^s).$$
You may need some additional assumptions on $h$ (particularly on how $h$ scales) to do this.
As a bit of a tangent, it's actually the case that for $s > fracd2 + k$, we have $H^s(mathbbR^d) hookrightarrow C^k(mathbbR^d) cap L^infty(mathbbR^d)$.
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
$endgroup$
add a comment |
$begingroup$
Using a bit of Fourier analysis, one can show that, for $s>fracd2$,
$$H^s(mathbbR^d) hookrightarrow L^infty(mathbbR^d).$$
In particular, consider
$$|f|_L^infty leq |hatf|_L^1 leq |langle xirangle ^-s|_L^2 |langle xirangle ^shatf|_L^2 lesssim |f|_H^s,
$$
where the first inequality follows, pretty much directly, from the Fourier inversion formula.
Now, $|f|_L^2 leq|f|_H^k+1$ for every $kgeq 0$, so that part is fine, too. But, you still have some work to do. In particular, you have
$$h(|f|_L^infty) leq h(c|f|_H^s),$$
but you need to get to
$$h(|f|_L^infty) lesssim h(|f|_H^s).$$
You may need some additional assumptions on $h$ (particularly on how $h$ scales) to do this.
As a bit of a tangent, it's actually the case that for $s > fracd2 + k$, we have $H^s(mathbbR^d) hookrightarrow C^k(mathbbR^d) cap L^infty(mathbbR^d)$.
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
$endgroup$
add a comment |
$begingroup$
Using a bit of Fourier analysis, one can show that, for $s>fracd2$,
$$H^s(mathbbR^d) hookrightarrow L^infty(mathbbR^d).$$
In particular, consider
$$|f|_L^infty leq |hatf|_L^1 leq |langle xirangle ^-s|_L^2 |langle xirangle ^shatf|_L^2 lesssim |f|_H^s,
$$
where the first inequality follows, pretty much directly, from the Fourier inversion formula.
Now, $|f|_L^2 leq|f|_H^k+1$ for every $kgeq 0$, so that part is fine, too. But, you still have some work to do. In particular, you have
$$h(|f|_L^infty) leq h(c|f|_H^s),$$
but you need to get to
$$h(|f|_L^infty) lesssim h(|f|_H^s).$$
You may need some additional assumptions on $h$ (particularly on how $h$ scales) to do this.
As a bit of a tangent, it's actually the case that for $s > fracd2 + k$, we have $H^s(mathbbR^d) hookrightarrow C^k(mathbbR^d) cap L^infty(mathbbR^d)$.
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
$endgroup$
Using a bit of Fourier analysis, one can show that, for $s>fracd2$,
$$H^s(mathbbR^d) hookrightarrow L^infty(mathbbR^d).$$
In particular, consider
$$|f|_L^infty leq |hatf|_L^1 leq |langle xirangle ^-s|_L^2 |langle xirangle ^shatf|_L^2 lesssim |f|_H^s,
$$
where the first inequality follows, pretty much directly, from the Fourier inversion formula.
Now, $|f|_L^2 leq|f|_H^k+1$ for every $kgeq 0$, so that part is fine, too. But, you still have some work to do. In particular, you have
$$h(|f|_L^infty) leq h(c|f|_H^s),$$
but you need to get to
$$h(|f|_L^infty) lesssim h(|f|_H^s).$$
You may need some additional assumptions on $h$ (particularly on how $h$ scales) to do this.
As a bit of a tangent, it's actually the case that for $s > fracd2 + k$, we have $H^s(mathbbR^d) hookrightarrow C^k(mathbbR^d) cap L^infty(mathbbR^d)$.
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
edited Mar 28 at 19:40
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
answered Mar 28 at 18:58
Gary MoonGary Moon
92127
92127
add a comment |
add a comment |
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1
$begingroup$
Looks like a key comparison needed here is between $||f||_infty$ and $||f||_k+1$. This should allow a comparison between $h ( ||f||_infty) $ and (a constant times) $h ($||f||_k+1)$. Although I am not sure why $h$ needs to be continuous. Cursorily (haven't checked the details), a candidate embedding theorem is Rellich-Kondrashov.
$endgroup$
– avs
Mar 28 at 16:29
2
$begingroup$
This can only be correct if $k+1 > n/2$ since otherwise $H^k+1$ cannot be embedded into $L^infty$. Even then it can only be true if $h$ has polynomial growth, since otherwise there are scaling problems.
$endgroup$
– Hans Engler
Mar 28 at 19:25