Mixing
Integration by parts in $mathbb{R}^n$
$begingroup$
Let $U subset mathbb{R}^n$ be open. Let $f colon U to mathbb{R}$ be a function of class $C^1$, and let $phicolon U to mathbb{R}$ be a function in $C_c^infty (U)$. Is it true that $$intlimits_U f(x)partial_iphi(x) dx = -intlimits_U partial_i f(x) phi(x) dx$$ for every $i$?
real-analysis calculus sobolev-spaces
asked Jan 24 at 22:20
RenRen
368
$endgroup$
add a comment |
$begingroup$
Let $U subset mathbb{R}^n$ be open. Let $f colon U to mathbb{R}$ be a function of class $C^1$, and let $phicolon U to mathbb{R}$ be a function in $C_c^infty (U)$. Is it true that $$intlimits_U f(x)partial_iphi(x) dx = -intlimits_U partial_i f(x) phi(x) dx$$ for every $i$?
real-analysis calculus sobolev-spaces
asked Jan 24 at 22:20
RenRen
368
$endgroup$
$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30
add a comment |
$begingroup$
Let $U subset mathbb{R}^n$ be open. Let $f colon U to mathbb{R}$ be a function of class $C^1$, and let $phicolon U to mathbb{R}$ be a function in $C_c^infty (U)$. Is it true that $$intlimits_U f(x)partial_iphi(x) dx = -intlimits_U partial_i f(x) phi(x) dx$$ for every $i$?
real-analysis calculus sobolev-spaces
asked Jan 24 at 22:20
RenRen
368
$endgroup$
Let $U subset mathbb{R}^n$ be open. Let $f colon U to mathbb{R}$ be a function of class $C^1$, and let $phicolon U to mathbb{R}$ be a function in $C_c^infty (U)$. Is it true that $$intlimits_U f(x)partial_iphi(x) dx = -intlimits_U partial_i f(x) phi(x) dx$$ for every $i$?
real-analysis calculus sobolev-spaces
real-analysis calculus sobolev-spaces
asked Jan 24 at 22:20
RenRen
368
asked Jan 24 at 22:20
RenRen
368
edited Jan 25 at 6:29
Ren
asked Jan 24 at 22:20
RenRen
368
asked Jan 24 at 22:20
RenRen
368
asked Jan 24 at 22:20
RenRen
368
368
$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30
add a comment |
$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30
$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a consequence of Fubini's theorem and the usual integration by parts in $mathbb R.$ To simplify notation assume $i = n,$ and write $x = (x',x_n) in mathbb R^n = mathbb R^{n-1} times mathbb R$ (the general case follows by interchanging coordinates using Fubini).
Observe that as $varphi in C^{infty}_c(U),$ we have $f partial_{x_n}varphi in C^1_c(mathbb R^n)$ extending by zero (this is okay as $varphi$ vanishes in a neighbourhood of $partial U$). Hence we have,
begin{align*}
int_U f(x)partial_{x_n}varphi(x) ,mathrm{d} x &= int_{mathbb R^n} f(x)partial_{x_n}varphi(x) ,mathrm{d}x \
&= int_{mathbb R^{n-1}} int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n mathrm{d}x' \
&= int_{mathbb R^{n-1}} left( - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_nright)mathrm{d}x' \
&= - int_U partial_{x_n}f(x) varphi(x),mathrm{d}x.
end{align*}
Here we have used Fubini in the second line and the fact that,
$$ int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n = - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_n $$
for all $x' in mathbb R^{n-1}$ by integrating by parts, noting the boundary term vanish since $varphi$ has compact support.
answered Jan 28 at 20:51
ktoiktoi
2,4161618
$endgroup$
add a comment |
$begingroup$
You can also apply the divergence theorem to the vector field
$$
u(x)=e_i f(x)phi(x)
$$
Then, the divergence becomes
$$
nabla cdot u(x)=phi(x) partial_i f(x)+phi(x)partial_i phi(x)
$$
Then, for any compact set $K subset U subset mathbb{R}^n$ such that $supp(phi) subseteq K$ :
$$
0=int_{partial U}u(x)cdot n; dS=int_U nabla cdot u(x)dx=int_U phi(x) partial_i f(x)+f(x)partial_i phi(x) dx
$$
Since $u$ has zero boundary values since $phi$ has 0 boundary values. Rearranging gives the result.
However, you need your boundary to be somewhat smooth. Else the theorem wont hold. I finished this answer before noting this missing requirement; I will just leave it up for now.
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a consequence of Fubini's theorem and the usual integration by parts in $mathbb R.$ To simplify notation assume $i = n,$ and write $x = (x',x_n) in mathbb R^n = mathbb R^{n-1} times mathbb R$ (the general case follows by interchanging coordinates using Fubini).
Observe that as $varphi in C^{infty}_c(U),$ we have $f partial_{x_n}varphi in C^1_c(mathbb R^n)$ extending by zero (this is okay as $varphi$ vanishes in a neighbourhood of $partial U$). Hence we have,
begin{align*}
int_U f(x)partial_{x_n}varphi(x) ,mathrm{d} x &= int_{mathbb R^n} f(x)partial_{x_n}varphi(x) ,mathrm{d}x \
&= int_{mathbb R^{n-1}} int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n mathrm{d}x' \
&= int_{mathbb R^{n-1}} left( - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_nright)mathrm{d}x' \
&= - int_U partial_{x_n}f(x) varphi(x),mathrm{d}x.
end{align*}
Here we have used Fubini in the second line and the fact that,
$$ int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n = - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_n $$
for all $x' in mathbb R^{n-1}$ by integrating by parts, noting the boundary term vanish since $varphi$ has compact support.
answered Jan 28 at 20:51
ktoiktoi
2,4161618
$endgroup$
add a comment |
$begingroup$
This is a consequence of Fubini's theorem and the usual integration by parts in $mathbb R.$ To simplify notation assume $i = n,$ and write $x = (x',x_n) in mathbb R^n = mathbb R^{n-1} times mathbb R$ (the general case follows by interchanging coordinates using Fubini).
Observe that as $varphi in C^{infty}_c(U),$ we have $f partial_{x_n}varphi in C^1_c(mathbb R^n)$ extending by zero (this is okay as $varphi$ vanishes in a neighbourhood of $partial U$). Hence we have,
begin{align*}
int_U f(x)partial_{x_n}varphi(x) ,mathrm{d} x &= int_{mathbb R^n} f(x)partial_{x_n}varphi(x) ,mathrm{d}x \
&= int_{mathbb R^{n-1}} int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n mathrm{d}x' \
&= int_{mathbb R^{n-1}} left( - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_nright)mathrm{d}x' \
&= - int_U partial_{x_n}f(x) varphi(x),mathrm{d}x.
end{align*}
Here we have used Fubini in the second line and the fact that,
$$ int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n = - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_n $$
for all $x' in mathbb R^{n-1}$ by integrating by parts, noting the boundary term vanish since $varphi$ has compact support.
answered Jan 28 at 20:51
ktoiktoi
2,4161618
$endgroup$
add a comment |
$begingroup$
This is a consequence of Fubini's theorem and the usual integration by parts in $mathbb R.$ To simplify notation assume $i = n,$ and write $x = (x',x_n) in mathbb R^n = mathbb R^{n-1} times mathbb R$ (the general case follows by interchanging coordinates using Fubini).
Observe that as $varphi in C^{infty}_c(U),$ we have $f partial_{x_n}varphi in C^1_c(mathbb R^n)$ extending by zero (this is okay as $varphi$ vanishes in a neighbourhood of $partial U$). Hence we have,
begin{align*}
int_U f(x)partial_{x_n}varphi(x) ,mathrm{d} x &= int_{mathbb R^n} f(x)partial_{x_n}varphi(x) ,mathrm{d}x \
&= int_{mathbb R^{n-1}} int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n mathrm{d}x' \
&= int_{mathbb R^{n-1}} left( - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_nright)mathrm{d}x' \
&= - int_U partial_{x_n}f(x) varphi(x),mathrm{d}x.
end{align*}
Here we have used Fubini in the second line and the fact that,
$$ int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n = - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_n $$
for all $x' in mathbb R^{n-1}$ by integrating by parts, noting the boundary term vanish since $varphi$ has compact support.
answered Jan 28 at 20:51
ktoiktoi
2,4161618
$endgroup$
This is a consequence of Fubini's theorem and the usual integration by parts in $mathbb R.$ To simplify notation assume $i = n,$ and write $x = (x',x_n) in mathbb R^n = mathbb R^{n-1} times mathbb R$ (the general case follows by interchanging coordinates using Fubini).
Observe that as $varphi in C^{infty}_c(U),$ we have $f partial_{x_n}varphi in C^1_c(mathbb R^n)$ extending by zero (this is okay as $varphi$ vanishes in a neighbourhood of $partial U$). Hence we have,
begin{align*}
int_U f(x)partial_{x_n}varphi(x) ,mathrm{d} x &= int_{mathbb R^n} f(x)partial_{x_n}varphi(x) ,mathrm{d}x \
&= int_{mathbb R^{n-1}} int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n mathrm{d}x' \
&= int_{mathbb R^{n-1}} left( - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_nright)mathrm{d}x' \
&= - int_U partial_{x_n}f(x) varphi(x),mathrm{d}x.
end{align*}
Here we have used Fubini in the second line and the fact that,
$$ int_{mathbb R} f(x',x_n) partial_{x_n}varphi(x',x_n) ,mathrm{d}x_n = - int_{mathbb R} partial_{x_n}f(x',x_n)varphi(x',x_n) ,mathrm{d}x_n $$
for all $x' in mathbb R^{n-1}$ by integrating by parts, noting the boundary term vanish since $varphi$ has compact support.
answered Jan 28 at 20:51
ktoiktoi
2,4161618
answered Jan 28 at 20:51
ktoiktoi
2,4161618
answered Jan 28 at 20:51
ktoiktoi
2,4161618
answered Jan 28 at 20:51
ktoiktoi
2,4161618
2,4161618
add a comment |
add a comment |
$begingroup$
You can also apply the divergence theorem to the vector field
$$
u(x)=e_i f(x)phi(x)
$$
Then, the divergence becomes
$$
nabla cdot u(x)=phi(x) partial_i f(x)+phi(x)partial_i phi(x)
$$
Then, for any compact set $K subset U subset mathbb{R}^n$ such that $supp(phi) subseteq K$ :
$$
0=int_{partial U}u(x)cdot n; dS=int_U nabla cdot u(x)dx=int_U phi(x) partial_i f(x)+f(x)partial_i phi(x) dx
$$
Since $u$ has zero boundary values since $phi$ has 0 boundary values. Rearranging gives the result.
However, you need your boundary to be somewhat smooth. Else the theorem wont hold. I finished this answer before noting this missing requirement; I will just leave it up for now.
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
$endgroup$
add a comment |
$begingroup$
You can also apply the divergence theorem to the vector field
$$
u(x)=e_i f(x)phi(x)
$$
Then, the divergence becomes
$$
nabla cdot u(x)=phi(x) partial_i f(x)+phi(x)partial_i phi(x)
$$
Then, for any compact set $K subset U subset mathbb{R}^n$ such that $supp(phi) subseteq K$ :
$$
0=int_{partial U}u(x)cdot n; dS=int_U nabla cdot u(x)dx=int_U phi(x) partial_i f(x)+f(x)partial_i phi(x) dx
$$
Since $u$ has zero boundary values since $phi$ has 0 boundary values. Rearranging gives the result.
However, you need your boundary to be somewhat smooth. Else the theorem wont hold. I finished this answer before noting this missing requirement; I will just leave it up for now.
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
$endgroup$
add a comment |
$begingroup$
You can also apply the divergence theorem to the vector field
$$
u(x)=e_i f(x)phi(x)
$$
Then, the divergence becomes
$$
nabla cdot u(x)=phi(x) partial_i f(x)+phi(x)partial_i phi(x)
$$
Then, for any compact set $K subset U subset mathbb{R}^n$ such that $supp(phi) subseteq K$ :
$$
0=int_{partial U}u(x)cdot n; dS=int_U nabla cdot u(x)dx=int_U phi(x) partial_i f(x)+f(x)partial_i phi(x) dx
$$
Since $u$ has zero boundary values since $phi$ has 0 boundary values. Rearranging gives the result.
However, you need your boundary to be somewhat smooth. Else the theorem wont hold. I finished this answer before noting this missing requirement; I will just leave it up for now.
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
$endgroup$
You can also apply the divergence theorem to the vector field
$$
u(x)=e_i f(x)phi(x)
$$
Then, the divergence becomes
$$
nabla cdot u(x)=phi(x) partial_i f(x)+phi(x)partial_i phi(x)
$$
Then, for any compact set $K subset U subset mathbb{R}^n$ such that $supp(phi) subseteq K$ :
$$
0=int_{partial U}u(x)cdot n; dS=int_U nabla cdot u(x)dx=int_U phi(x) partial_i f(x)+f(x)partial_i phi(x) dx
$$
Since $u$ has zero boundary values since $phi$ has 0 boundary values. Rearranging gives the result.
However, you need your boundary to be somewhat smooth. Else the theorem wont hold. I finished this answer before noting this missing requirement; I will just leave it up for now.
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
edited Jan 28 at 22:22
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
answered Jan 28 at 22:07
F. ConradF. Conrad
1,296412
1,296412
add a comment |
add a comment |
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$begingroup$
I think you may missed a negative sign. Also, the domain $Usubsetmathbb{R}^n$ is not sufficient, as far as I know, this result holds for domain with Lipschitiz continuous boundary. Maybe there are counterexamples for $C^{alpha}$ domains. ;)
$endgroup$
– Zixiao_Liu
Jan 25 at 0:49
$begingroup$
@Zixiao_Liu Since your function has compact support inside $U$, no conditions are required on its boundary. $U$ just has to be measurable, and it is.
$endgroup$
– GReyes
Jan 25 at 2:46
$begingroup$
@GReyes Yes, you are right. I forgot this condition here. Thanks for pointing out!
$endgroup$
– Zixiao_Liu
Jan 25 at 3:25
$begingroup$
@Zixiao_Liu Yes I have corrected it. Thanks ;) Could you explain how I can prove it?
$endgroup$
– Ren
Jan 25 at 6:29
$begingroup$
@GReyes Do you know how to show this equation?
$endgroup$
– Ren
Jan 25 at 6:30