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does weak derivative zero imply independence?
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If $fcolonmathbbR^2 to mathbbR$ is a continuous function, and $partial_x f = 0$ in the weak sense, i.e.
$$ iint f(x,y) partial_x phi( x,y) dx dy =0 $$
for all $phi in C^infty_c(mathbbR^2)$, does it follow that $f(x,y)=c(y)$?
Many thanks in advance for any suggestions!
real-analysis pde distribution-theory
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
add a comment |Â
up vote
2
down vote
favorite
If $fcolonmathbbR^2 to mathbbR$ is a continuous function, and $partial_x f = 0$ in the weak sense, i.e.
$$ iint f(x,y) partial_x phi( x,y) dx dy =0 $$
for all $phi in C^infty_c(mathbbR^2)$, does it follow that $f(x,y)=c(y)$?
Many thanks in advance for any suggestions!
real-analysis pde distribution-theory
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $fcolonmathbbR^2 to mathbbR$ is a continuous function, and $partial_x f = 0$ in the weak sense, i.e.
$$ iint f(x,y) partial_x phi( x,y) dx dy =0 $$
for all $phi in C^infty_c(mathbbR^2)$, does it follow that $f(x,y)=c(y)$?
Many thanks in advance for any suggestions!
real-analysis pde distribution-theory
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
If $fcolonmathbbR^2 to mathbbR$ is a continuous function, and $partial_x f = 0$ in the weak sense, i.e.
$$ iint f(x,y) partial_x phi( x,y) dx dy =0 $$
for all $phi in C^infty_c(mathbbR^2)$, does it follow that $f(x,y)=c(y)$?
Many thanks in advance for any suggestions!
real-analysis pde distribution-theory
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
asked Jul 16 at 13:54
Aerinmund Fagelson
1,6491614
1,6491614
Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17
add a comment |Â
Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17
Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17
Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17
add a comment |Â
1 Answer
1
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votes
up vote
2
down vote
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You can use functions $phi in C_mathrmc^infty (mathbbR^2)$ of the form $phi(x,y) = chi(x) psi(y)$ with $chi, psi in C_mathrmc^infty (mathbbR)$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that
$$ int limits_mathbbR psi(y) int limits_mathbbR f(x,y) chi' (x) , mathrmd x , mathrmd y = 0 $$
holds for every $psi in C_mathrmc^infty (mathbbR)$ . Thus the continuous function $y mapsto int_mathbbR f(x,y) chi'(x) , mathrmd x $ vanishes identically for every $chi in C_mathrmc^infty (mathbbR)$ by the fundamental theorem of the calculus of variations.
But then for fixed $y in mathbbR$ the function $x mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) in mathbbR^2$ with $c in C^0 (mathbbR)$ .
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can use functions $phi in C_mathrmc^infty (mathbbR^2)$ of the form $phi(x,y) = chi(x) psi(y)$ with $chi, psi in C_mathrmc^infty (mathbbR)$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that
$$ int limits_mathbbR psi(y) int limits_mathbbR f(x,y) chi' (x) , mathrmd x , mathrmd y = 0 $$
holds for every $psi in C_mathrmc^infty (mathbbR)$ . Thus the continuous function $y mapsto int_mathbbR f(x,y) chi'(x) , mathrmd x $ vanishes identically for every $chi in C_mathrmc^infty (mathbbR)$ by the fundamental theorem of the calculus of variations.
But then for fixed $y in mathbbR$ the function $x mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) in mathbbR^2$ with $c in C^0 (mathbbR)$ .
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
add a comment |Â
up vote
2
down vote
accepted
You can use functions $phi in C_mathrmc^infty (mathbbR^2)$ of the form $phi(x,y) = chi(x) psi(y)$ with $chi, psi in C_mathrmc^infty (mathbbR)$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that
$$ int limits_mathbbR psi(y) int limits_mathbbR f(x,y) chi' (x) , mathrmd x , mathrmd y = 0 $$
holds for every $psi in C_mathrmc^infty (mathbbR)$ . Thus the continuous function $y mapsto int_mathbbR f(x,y) chi'(x) , mathrmd x $ vanishes identically for every $chi in C_mathrmc^infty (mathbbR)$ by the fundamental theorem of the calculus of variations.
But then for fixed $y in mathbbR$ the function $x mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) in mathbbR^2$ with $c in C^0 (mathbbR)$ .
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can use functions $phi in C_mathrmc^infty (mathbbR^2)$ of the form $phi(x,y) = chi(x) psi(y)$ with $chi, psi in C_mathrmc^infty (mathbbR)$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that
$$ int limits_mathbbR psi(y) int limits_mathbbR f(x,y) chi' (x) , mathrmd x , mathrmd y = 0 $$
holds for every $psi in C_mathrmc^infty (mathbbR)$ . Thus the continuous function $y mapsto int_mathbbR f(x,y) chi'(x) , mathrmd x $ vanishes identically for every $chi in C_mathrmc^infty (mathbbR)$ by the fundamental theorem of the calculus of variations.
But then for fixed $y in mathbbR$ the function $x mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) in mathbbR^2$ with $c in C^0 (mathbbR)$ .
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
You can use functions $phi in C_mathrmc^infty (mathbbR^2)$ of the form $phi(x,y) = chi(x) psi(y)$ with $chi, psi in C_mathrmc^infty (mathbbR)$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that
$$ int limits_mathbbR psi(y) int limits_mathbbR f(x,y) chi' (x) , mathrmd x , mathrmd y = 0 $$
holds for every $psi in C_mathrmc^infty (mathbbR)$ . Thus the continuous function $y mapsto int_mathbbR f(x,y) chi'(x) , mathrmd x $ vanishes identically for every $chi in C_mathrmc^infty (mathbbR)$ by the fundamental theorem of the calculus of variations.
But then for fixed $y in mathbbR$ the function $x mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) in mathbbR^2$ with $c in C^0 (mathbbR)$ .
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
answered Jul 16 at 14:27
ComplexYetTrivial
2,702624
2,702624
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
add a comment |Â
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
Thanks for the great answer :) I had been struggling to reduce this to the 1D case, and actually I think I managed a proof in the end after posting using a mollification argument, but your method is much cleaner and would work on more general domains :)
– Aerinmund Fagelson
Jul 16 at 15:13
add a comment |Â
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Yes, it does follow. You can modify this argument to work in higher dimensions.
– Daniel Fischer♦
Jul 16 at 14:17