Locally closed in the sense of distributions implies closed?
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Let $F in L^{p}(mathbb{(-1,1)^{n}}; Lambda^{2}mathbb{R}^{n})$ be an $L^{p}$ $2$-form on the open cube $(-1,1)^{n}.$ Suppose we know that for every $x in (-1,1)^{n},$ that there exists $0 < r = r(x) < operatorname{dist}(x, partial (-1,1)^{n})$ such that $$ dF = 0 quad text{ in the sense of distributions on } B_{r}(x). $$
Does it imply $$ dF = 0 quad text{ in the sense of distributions on } (-1,1)^{n}? $$
I think it must be true but somehow I can not figure out a way to prove it.
differential-forms weak-derivatives
asked Nov 20 at 11:06
Swarnendu Sil
416
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up vote
1
down vote
favorite
Let $F in L^{p}(mathbb{(-1,1)^{n}}; Lambda^{2}mathbb{R}^{n})$ be an $L^{p}$ $2$-form on the open cube $(-1,1)^{n}.$ Suppose we know that for every $x in (-1,1)^{n},$ that there exists $0 < r = r(x) < operatorname{dist}(x, partial (-1,1)^{n})$ such that $$ dF = 0 quad text{ in the sense of distributions on } B_{r}(x). $$
Does it imply $$ dF = 0 quad text{ in the sense of distributions on } (-1,1)^{n}? $$
I think it must be true but somehow I can not figure out a way to prove it.
differential-forms weak-derivatives
asked Nov 20 at 11:06
Swarnendu Sil
416
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $F in L^{p}(mathbb{(-1,1)^{n}}; Lambda^{2}mathbb{R}^{n})$ be an $L^{p}$ $2$-form on the open cube $(-1,1)^{n}.$ Suppose we know that for every $x in (-1,1)^{n},$ that there exists $0 < r = r(x) < operatorname{dist}(x, partial (-1,1)^{n})$ such that $$ dF = 0 quad text{ in the sense of distributions on } B_{r}(x). $$
Does it imply $$ dF = 0 quad text{ in the sense of distributions on } (-1,1)^{n}? $$
I think it must be true but somehow I can not figure out a way to prove it.
differential-forms weak-derivatives
asked Nov 20 at 11:06
Swarnendu Sil
416
Let $F in L^{p}(mathbb{(-1,1)^{n}}; Lambda^{2}mathbb{R}^{n})$ be an $L^{p}$ $2$-form on the open cube $(-1,1)^{n}.$ Suppose we know that for every $x in (-1,1)^{n},$ that there exists $0 < r = r(x) < operatorname{dist}(x, partial (-1,1)^{n})$ such that $$ dF = 0 quad text{ in the sense of distributions on } B_{r}(x). $$
Does it imply $$ dF = 0 quad text{ in the sense of distributions on } (-1,1)^{n}? $$
I think it must be true but somehow I can not figure out a way to prove it.
differential-forms weak-derivatives
differential-forms weak-derivatives
asked Nov 20 at 11:06
Swarnendu Sil
416
asked Nov 20 at 11:06
Swarnendu Sil
416
asked Nov 20 at 11:06
Swarnendu Sil
416
asked Nov 20 at 11:06
Swarnendu Sil
416
asked Nov 20 at 11:06
Swarnendu Sil
416
416
add a comment |
add a comment |
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