When $lim_{r to 0}frac{1}{|B_r(bar x)|}int_0^aint_{B_r(bar x)} u(y,x) dx dy = 0$ implies $u(y,bar x)=0$ for...
1
$begingroup$
Let $f$ be a nonnegative $L^infty([0,a], L^1(mathbb{R}^N))$ function. Fix any $bar x in mathbb{R}^N$ where $f$ is defined. If
$$lim_{r to 0}frac{1}{|B_r(bar x)|}int_0^aint_{B_r(bar x)} f(y,x) dx dy = 0,$$
does this imply $$f(y,bar x)=0$$ for a.e. $y in [0,a]$?
real-analysis calculus functional-analysis measure-theory
asked Dec 29 '18 at 19:39
DalDal
1,55322473
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add a comment |
1
$begingroup$
Let $f$ be a nonnegative $L^infty([0,a], L^1(mathbb{R}^N))$ function. Fix any $bar x in mathbb{R}^N$ where $f$ is defined. If
$$lim_{r to 0}frac{1}{|B_r(bar x)|}int_0^aint_{B_r(bar x)} f(y,x) dx dy = 0,$$
does this imply $$f(y,bar x)=0$$ for a.e. $y in [0,a]$?
real-analysis calculus functional-analysis measure-theory
asked Dec 29 '18 at 19:39
DalDal
1,55322473
$endgroup$
$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59
add a comment |
1
1
1
$begingroup$
Let $f$ be a nonnegative $L^infty([0,a], L^1(mathbb{R}^N))$ function. Fix any $bar x in mathbb{R}^N$ where $f$ is defined. If
$$lim_{r to 0}frac{1}{|B_r(bar x)|}int_0^aint_{B_r(bar x)} f(y,x) dx dy = 0,$$
does this imply $$f(y,bar x)=0$$ for a.e. $y in [0,a]$?
real-analysis calculus functional-analysis measure-theory
asked Dec 29 '18 at 19:39
DalDal
1,55322473
$endgroup$
Let $f$ be a nonnegative $L^infty([0,a], L^1(mathbb{R}^N))$ function. Fix any $bar x in mathbb{R}^N$ where $f$ is defined. If
$$lim_{r to 0}frac{1}{|B_r(bar x)|}int_0^aint_{B_r(bar x)} f(y,x) dx dy = 0,$$
does this imply $$f(y,bar x)=0$$ for a.e. $y in [0,a]$?
real-analysis calculus functional-analysis measure-theory
real-analysis calculus functional-analysis measure-theory
asked Dec 29 '18 at 19:39
DalDal
1,55322473
asked Dec 29 '18 at 19:39
DalDal
1,55322473
edited Dec 29 '18 at 20:58
Dal
asked Dec 29 '18 at 19:39
DalDal
1,55322473
asked Dec 29 '18 at 19:39
DalDal
1,55322473
asked Dec 29 '18 at 19:39
DalDal
1,55322473
1,55322473
$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59
add a comment |
$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59
$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59
add a comment |
1 Answer
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No. Simply let $f(y, overline{x}) = 1$ for all $y$, and $f(y, x) = 0$ for all $x, y$ where $x neq overline{x}$.
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
$endgroup$
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
add a comment |
Your Answer
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1 Answer
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1 Answer
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No. Simply let $f(y, overline{x}) = 1$ for all $y$, and $f(y, x) = 0$ for all $x, y$ where $x neq overline{x}$.
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
$endgroup$
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
add a comment |
1
$begingroup$
No. Simply let $f(y, overline{x}) = 1$ for all $y$, and $f(y, x) = 0$ for all $x, y$ where $x neq overline{x}$.
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
$endgroup$
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
add a comment |
1
1
1
$begingroup$
No. Simply let $f(y, overline{x}) = 1$ for all $y$, and $f(y, x) = 0$ for all $x, y$ where $x neq overline{x}$.
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
$endgroup$
No. Simply let $f(y, overline{x}) = 1$ for all $y$, and $f(y, x) = 0$ for all $x, y$ where $x neq overline{x}$.
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
answered Dec 29 '18 at 21:29
John JiangJohn Jiang
369312
369312
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
add a comment |
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
Thank you. Then what is the minimal assumption that makes the statement true?
$endgroup$
– Dal
Dec 29 '18 at 21:31
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
$begingroup$
@Dal not sure about minimal, but continuity in x for instance would make it true. Even weaker conditions include upper/lower continuity.
$endgroup$
– John Jiang
Dec 29 '18 at 21:35
add a comment |
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$begingroup$
Idea: Suppose not. Then, use Fubini and the Lebesgue Density theorem to get a contradiction.
$endgroup$
– Matematleta
Dec 29 '18 at 19:54
$begingroup$
what does $L^infty([0,a], L^1(mathbb{R}^N))$ mean? Also, did $f$ turn into $u?$
$endgroup$
– zhw.
Dec 29 '18 at 20:36
$begingroup$
@zhw. Thanks for the correction. It means that $sup_{y in [0,a]} int_{mathbb{R}^N} |f| dx < infty$
$endgroup$
– Dal
Dec 29 '18 at 20:59