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$ fin C_0^infty Rightarrow fin L^m $?
The 2019 Stack Overflow Developer Survey Results Are InIs $C_0^infty(Omega)cap H^m,p(Omega)$ dense in $H^m,p(Omega)$?Approximation of Lipschitz functions on Riemannian manifoldsAbout the property of Littlewood-Paley partition of unity.$fin L^p(mathbb R)cap C_0(mathbb R); (1<p<infty), gin C^infty_c(mathbb R) implies fast g in C^k(mathbb R)$?Help understanding $limsup_ntoinfty(a_n+b_n)leq limsup_ntoinfty a_n + limsup_ntoinfty b_n$Range of $fin C_c(X)$ is compact subset of complex planeInequality for the Laplace operatorHow to prove this Holder- type inequality?Supremum of product $L^infty$ functions
$begingroup$
Let $ C_0^infty $ be the subspace of $ C^infty $ functions with compact support in $ R^n $. It's true that if $ f in C_0^infty, $ then $ f in L^m cap H^s $ where $ 1leq m leq 2$ and $ sgeq 0 $ is an integer number?
Can anyone can me explain why, please?
real-analysis lebesgue-measure sobolev-spaces
asked Mar 30 at 9:44
C. BishopC. Bishop
419
$endgroup$
add a comment |
$begingroup$
Let $ C_0^infty $ be the subspace of $ C^infty $ functions with compact support in $ R^n $. It's true that if $ f in C_0^infty, $ then $ f in L^m cap H^s $ where $ 1leq m leq 2$ and $ sgeq 0 $ is an integer number?
Can anyone can me explain why, please?
real-analysis lebesgue-measure sobolev-spaces
asked Mar 30 at 9:44
C. BishopC. Bishop
419
$endgroup$
$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
2
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
2
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49
add a comment |
$begingroup$
Let $ C_0^infty $ be the subspace of $ C^infty $ functions with compact support in $ R^n $. It's true that if $ f in C_0^infty, $ then $ f in L^m cap H^s $ where $ 1leq m leq 2$ and $ sgeq 0 $ is an integer number?
Can anyone can me explain why, please?
real-analysis lebesgue-measure sobolev-spaces
asked Mar 30 at 9:44
C. BishopC. Bishop
419
$endgroup$
Let $ C_0^infty $ be the subspace of $ C^infty $ functions with compact support in $ R^n $. It's true that if $ f in C_0^infty, $ then $ f in L^m cap H^s $ where $ 1leq m leq 2$ and $ sgeq 0 $ is an integer number?
Can anyone can me explain why, please?
real-analysis lebesgue-measure sobolev-spaces
real-analysis lebesgue-measure sobolev-spaces
asked Mar 30 at 9:44
C. BishopC. Bishop
419
asked Mar 30 at 9:44
C. BishopC. Bishop
419
edited 12 hours ago
C. Bishop
asked Mar 30 at 9:44
C. BishopC. Bishop
419
asked Mar 30 at 9:44
C. BishopC. Bishop
419
asked Mar 30 at 9:44
C. BishopC. Bishop
419
419
$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
2
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
2
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49
add a comment |
$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
2
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
2
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49
$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
2
2
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
2
2
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me first prove that $C_0^infty subset L^p$. Let $fin C_0^infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ Vert f Vert_p = left( int vert f (x) vert^p dx right)^frac1p
= left( int_B(R) vert f (x) vert^p dx right)^frac1p
leq Vert f Vert_infty vert B(R)vert <infty$$
where $vert B(R)vert$ denote the measure of the ball $B(R)$. Thus, $fin L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$. The main trick here is the following: Let $mgeq s$ be an integer. Then we get
$$ (1+ vert x vert)^s vert hatf(x) vert
leq frac1(1+vert x vert)^m cdot (1+ vert x vert)^2m vert hatf(x) vert leq fracC_m(1+vert x vert)^m $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ fracC_m(1+vert x vert)^m $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s hatf$ is in $L^2$ and hence $fin H^s$.
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Let me first prove that $C_0^infty subset L^p$. Let $fin C_0^infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ Vert f Vert_p = left( int vert f (x) vert^p dx right)^frac1p
= left( int_B(R) vert f (x) vert^p dx right)^frac1p
leq Vert f Vert_infty vert B(R)vert <infty$$
where $vert B(R)vert$ denote the measure of the ball $B(R)$. Thus, $fin L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$. The main trick here is the following: Let $mgeq s$ be an integer. Then we get
$$ (1+ vert x vert)^s vert hatf(x) vert
leq frac1(1+vert x vert)^m cdot (1+ vert x vert)^2m vert hatf(x) vert leq fracC_m(1+vert x vert)^m $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ fracC_m(1+vert x vert)^m $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s hatf$ is in $L^2$ and hence $fin H^s$.
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
add a comment |
$begingroup$
Let me first prove that $C_0^infty subset L^p$. Let $fin C_0^infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ Vert f Vert_p = left( int vert f (x) vert^p dx right)^frac1p
= left( int_B(R) vert f (x) vert^p dx right)^frac1p
leq Vert f Vert_infty vert B(R)vert <infty$$
where $vert B(R)vert$ denote the measure of the ball $B(R)$. Thus, $fin L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$. The main trick here is the following: Let $mgeq s$ be an integer. Then we get
$$ (1+ vert x vert)^s vert hatf(x) vert
leq frac1(1+vert x vert)^m cdot (1+ vert x vert)^2m vert hatf(x) vert leq fracC_m(1+vert x vert)^m $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ fracC_m(1+vert x vert)^m $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s hatf$ is in $L^2$ and hence $fin H^s$.
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
add a comment |
$begingroup$
Let me first prove that $C_0^infty subset L^p$. Let $fin C_0^infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ Vert f Vert_p = left( int vert f (x) vert^p dx right)^frac1p
= left( int_B(R) vert f (x) vert^p dx right)^frac1p
leq Vert f Vert_infty vert B(R)vert <infty$$
where $vert B(R)vert$ denote the measure of the ball $B(R)$. Thus, $fin L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$. The main trick here is the following: Let $mgeq s$ be an integer. Then we get
$$ (1+ vert x vert)^s vert hatf(x) vert
leq frac1(1+vert x vert)^m cdot (1+ vert x vert)^2m vert hatf(x) vert leq fracC_m(1+vert x vert)^m $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ fracC_m(1+vert x vert)^m $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s hatf$ is in $L^2$ and hence $fin H^s$.
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
Let me first prove that $C_0^infty subset L^p$. Let $fin C_0^infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ Vert f Vert_p = left( int vert f (x) vert^p dx right)^frac1p
= left( int_B(R) vert f (x) vert^p dx right)^frac1p
leq Vert f Vert_infty vert B(R)vert <infty$$
where $vert B(R)vert$ denote the measure of the ball $B(R)$. Thus, $fin L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$. The main trick here is the following: Let $mgeq s$ be an integer. Then we get
$$ (1+ vert x vert)^s vert hatf(x) vert
leq frac1(1+vert x vert)^m cdot (1+ vert x vert)^2m vert hatf(x) vert leq fracC_m(1+vert x vert)^m $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ fracC_m(1+vert x vert)^m $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s hatf$ is in $L^2$ and hence $fin H^s$.
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
edited Mar 30 at 11:07
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
answered Mar 30 at 10:53
Severin SchravenSeverin Schraven
6,7802936
6,7802936
add a comment |
add a comment |
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$begingroup$
You haven't stated what $alpha$ is. Also giving the context could help with people being able to answer the question. This looks like a very specific choice of numbers, so you either suspect it is true as a corollary or as something you encounered.
$endgroup$
– Keen-ameteur
Mar 30 at 9:53
2
$begingroup$
Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+vert cdot vert)^s hatf$. However, using that $hatf$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $sgeq 0$.
$endgroup$
– Severin Schraven
Mar 30 at 9:55
2
$begingroup$
Also any continuous compactly supported function on $mathbbR^n$, is in $L^q$ for all $qgeq1$.
$endgroup$
– Keen-ameteur
Mar 30 at 9:56
$begingroup$
I understood the reasoning of Severin Schraven. Thank you. But there is a “simple” way to prove that every $ fin C_0^infty $ is also in $ L^p$? I only know that $ C_0$ is dense in $L^p$ for $ 1leq p < infty$.
$endgroup$
– C. Bishop
Mar 30 at 10:49