Limit of the average of a sequence of functions
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Let $B(epsilon)$ be the ball of radius $epsilon>0$ in $ mathbb{R}^{n} $ centered at the origin. Let $phi_{epsilon}$ be a test function, $geq0 $ and $leq 1$ with support in $B(epsilon)$ that is equal to 1 on the closure of $B(epsilon/2)$, and let $f$ be a continuous function at a neighborhood of the origin. Can we say that
$$lim_{epsilonto0}frac{1}{m(B(epsilon))}int_{B(epsilon)}f(x) phi_{epsilon}(x)dx=f(0)? $$ ($m(B(epsilon))$ is the measure of the ball)
real-analysis
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
$endgroup$
add a comment |
$begingroup$
Let $B(epsilon)$ be the ball of radius $epsilon>0$ in $ mathbb{R}^{n} $ centered at the origin. Let $phi_{epsilon}$ be a test function, $geq0 $ and $leq 1$ with support in $B(epsilon)$ that is equal to 1 on the closure of $B(epsilon/2)$, and let $f$ be a continuous function at a neighborhood of the origin. Can we say that
$$lim_{epsilonto0}frac{1}{m(B(epsilon))}int_{B(epsilon)}f(x) phi_{epsilon}(x)dx=f(0)? $$ ($m(B(epsilon))$ is the measure of the ball)
real-analysis
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
$endgroup$
$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
1
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11
add a comment |
$begingroup$
Let $B(epsilon)$ be the ball of radius $epsilon>0$ in $ mathbb{R}^{n} $ centered at the origin. Let $phi_{epsilon}$ be a test function, $geq0 $ and $leq 1$ with support in $B(epsilon)$ that is equal to 1 on the closure of $B(epsilon/2)$, and let $f$ be a continuous function at a neighborhood of the origin. Can we say that
$$lim_{epsilonto0}frac{1}{m(B(epsilon))}int_{B(epsilon)}f(x) phi_{epsilon}(x)dx=f(0)? $$ ($m(B(epsilon))$ is the measure of the ball)
real-analysis
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
$endgroup$
Let $B(epsilon)$ be the ball of radius $epsilon>0$ in $ mathbb{R}^{n} $ centered at the origin. Let $phi_{epsilon}$ be a test function, $geq0 $ and $leq 1$ with support in $B(epsilon)$ that is equal to 1 on the closure of $B(epsilon/2)$, and let $f$ be a continuous function at a neighborhood of the origin. Can we say that
$$lim_{epsilonto0}frac{1}{m(B(epsilon))}int_{B(epsilon)}f(x) phi_{epsilon}(x)dx=f(0)? $$ ($m(B(epsilon))$ is the measure of the ball)
real-analysis
real-analysis
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
edited Jan 13 at 4:51
M. Rahmat
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
asked Jan 13 at 4:09
M. RahmatM. Rahmat
291212
291212
$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
1
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11
add a comment |
$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
1
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11
$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
1
1
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11
add a comment |
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$begingroup$
You have probably copied the result wrongly. If $phi_{epsilon}=1$ on $B(x,epsilon)$ then this is true. But this is not true the way it is stated. It is necessary that $phi_{epsilon}=1$ on a large enough neighborhood of $0$ for thus to be true.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:48
$begingroup$
I edited. What do you think now?
$endgroup$
– M. Rahmat
Jan 13 at 4:52
$begingroup$
It is still false.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 4:53
1
$begingroup$
The basic idea is: under the given hypothesis $phi_{epsilon}$ can be too small in, say $B(epsilon)setminus B(3epsilon/4)$ in which case the limit on the left can be much smaller then $f(0)$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:50
$begingroup$
Is it possible to construct a test function $phi_{epsilon}$ so that $phi_{epsilon}$ is in $ B(epsilon)$ and the above equation holds?
$endgroup$
– M. Rahmat
Jan 13 at 22:11