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Sequence of continuous functions with discontinuous limits
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A sequence of smooth functions such as $h_n (x) = arctan(nx)$ has discontinuous limit that depends on x. When $n rightarrow infty$, $h_n (x)$ converges to
$$left{beginarrayccc
-pi/2, & quad & x < 0\
0, & quad & x = 0\
pi/2, & quad & x > 0
endarrayright.$$
In order to show that we would evaluate three limits when $x<0, x=0$ and $x>0$.
We would do it similarly for a continuous sequence of piecewise linear functions such as
$$f_n (x)= begincases
0, xleq0 \[2ex]
nx, 0<x<frac1n \[2ex]
1, xgeq frac1n
endcases$$
My question is: In these cases, it is easy to find limit function. What happens when we know that limit is discontinuous but we do not know precisely how it looks like? How would we show any convergence? Here we can't separate on cases based on the value of $x$.
These examples are just for ilustration. Problem in PDEs I am dealing with comes down to this. Also, just to mention: I work in Sobolev spaces. So maybe I could use test functions?
limits pde continuity discontinuous-functions
asked Jul 19 at 11:21
Mark
316
add a comment |Â
up vote
1
down vote
favorite
A sequence of smooth functions such as $h_n (x) = arctan(nx)$ has discontinuous limit that depends on x. When $n rightarrow infty$, $h_n (x)$ converges to
$$left{beginarrayccc
-pi/2, & quad & x < 0\
0, & quad & x = 0\
pi/2, & quad & x > 0
endarrayright.$$
In order to show that we would evaluate three limits when $x<0, x=0$ and $x>0$.
We would do it similarly for a continuous sequence of piecewise linear functions such as
$$f_n (x)= begincases
0, xleq0 \[2ex]
nx, 0<x<frac1n \[2ex]
1, xgeq frac1n
endcases$$
My question is: In these cases, it is easy to find limit function. What happens when we know that limit is discontinuous but we do not know precisely how it looks like? How would we show any convergence? Here we can't separate on cases based on the value of $x$.
These examples are just for ilustration. Problem in PDEs I am dealing with comes down to this. Also, just to mention: I work in Sobolev spaces. So maybe I could use test functions?
limits pde continuity discontinuous-functions
asked Jul 19 at 11:21
Mark
316
For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A sequence of smooth functions such as $h_n (x) = arctan(nx)$ has discontinuous limit that depends on x. When $n rightarrow infty$, $h_n (x)$ converges to
$$left{beginarrayccc
-pi/2, & quad & x < 0\
0, & quad & x = 0\
pi/2, & quad & x > 0
endarrayright.$$
In order to show that we would evaluate three limits when $x<0, x=0$ and $x>0$.
We would do it similarly for a continuous sequence of piecewise linear functions such as
$$f_n (x)= begincases
0, xleq0 \[2ex]
nx, 0<x<frac1n \[2ex]
1, xgeq frac1n
endcases$$
My question is: In these cases, it is easy to find limit function. What happens when we know that limit is discontinuous but we do not know precisely how it looks like? How would we show any convergence? Here we can't separate on cases based on the value of $x$.
These examples are just for ilustration. Problem in PDEs I am dealing with comes down to this. Also, just to mention: I work in Sobolev spaces. So maybe I could use test functions?
limits pde continuity discontinuous-functions
asked Jul 19 at 11:21
Mark
316
A sequence of smooth functions such as $h_n (x) = arctan(nx)$ has discontinuous limit that depends on x. When $n rightarrow infty$, $h_n (x)$ converges to
$$left{beginarrayccc
-pi/2, & quad & x < 0\
0, & quad & x = 0\
pi/2, & quad & x > 0
endarrayright.$$
In order to show that we would evaluate three limits when $x<0, x=0$ and $x>0$.
We would do it similarly for a continuous sequence of piecewise linear functions such as
$$f_n (x)= begincases
0, xleq0 \[2ex]
nx, 0<x<frac1n \[2ex]
1, xgeq frac1n
endcases$$
My question is: In these cases, it is easy to find limit function. What happens when we know that limit is discontinuous but we do not know precisely how it looks like? How would we show any convergence? Here we can't separate on cases based on the value of $x$.
These examples are just for ilustration. Problem in PDEs I am dealing with comes down to this. Also, just to mention: I work in Sobolev spaces. So maybe I could use test functions?
limits pde continuity discontinuous-functions
asked Jul 19 at 11:21
Mark
316
asked Jul 19 at 11:21
Mark
316
asked Jul 19 at 11:21
Mark
316
asked Jul 19 at 11:21
Mark
316
316
For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55
add a comment |Â
For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55
For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55
add a comment |Â
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For your first question, you should find examples of discontinuous limits in the Fourier sums. You can prove the sum converges by dominating the terms, you can be sure the limit is not continuous if the series decay is slow enough, but still have no idea of what your function looks like.
– Benoit Gaudeul
Jul 19 at 12:11
For @BenoitGaudeul: Thank you. I wasn't thinking about Fourier sums but because I am working in Sobolev spaces that sounds interesting to me. Could you explain it a little bit more or send me a title of the paper that deals with that?
– Mark
Jul 19 at 19:44
You can consider convergence in the $L^p$ norm for $1leq p <infty$ instead.
– Jeff
Jul 19 at 23:15
The general answer to your question is that you have to use what you do know about the functions. If that isn't enough to deduce the limit, then there is no limit to deduce. Information does not appear out of nowhere - it can only be deduced from other information.
– Paul Sinclair
Jul 20 at 0:55