If $,f$ and $g$ are continuous with compact support then $f*g$ (convolution ) is also continuous with compact...
up vote
1
down vote
favorite
I do not know how to find a compact that satisfies $f$ and $g$ support.
Could someone explain how find this new compact ?
How $f$ and $g$ are continuous in a compact then are bounded in this compact. But,
Are $f$ and $g$ bounded in all real line?
Could someone starts the proofs?
real-analysis definite-integrals continuity convolution uniform-continuity
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
closed as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cameron Williams, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
1
down vote
favorite
I do not know how to find a compact that satisfies $f$ and $g$ support.
Could someone explain how find this new compact ?
How $f$ and $g$ are continuous in a compact then are bounded in this compact. But,
Are $f$ and $g$ bounded in all real line?
Could someone starts the proofs?
real-analysis definite-integrals continuity convolution uniform-continuity
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
closed as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cameron Williams, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I do not know how to find a compact that satisfies $f$ and $g$ support.
Could someone explain how find this new compact ?
How $f$ and $g$ are continuous in a compact then are bounded in this compact. But,
Are $f$ and $g$ bounded in all real line?
Could someone starts the proofs?
real-analysis definite-integrals continuity convolution uniform-continuity
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
I do not know how to find a compact that satisfies $f$ and $g$ support.
Could someone explain how find this new compact ?
How $f$ and $g$ are continuous in a compact then are bounded in this compact. But,
Are $f$ and $g$ bounded in all real line?
Could someone starts the proofs?
real-analysis definite-integrals continuity convolution uniform-continuity
real-analysis definite-integrals continuity convolution uniform-continuity
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
edited Nov 19 at 19:02
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
asked Nov 19 at 16:56
Gabriel Corrêa
93
asked Nov 19 at 16:56
Gabriel Corrêa
93
asked Nov 19 at 16:56
Gabriel Corrêa
93
93
closed as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cameron Williams, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cameron Williams, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03
add a comment |
1
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03
1
1
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
Compact support of $f*g$. Assume that $,f,gin C_0(mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.
Then for $x>2M$,
$$
(f*g)(x)=int_{-infty}^infty f(t),g(x-t),dt=int_{-M}^M f(t),g(x-t),dt=0,
$$
since $x-t>M$, when $|t|le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.
Continuity of $f*g$.
Observe that
$$
(f*g)(x+h)-(f*g)(x)=int_{-infty}^infty f(t),big(g(x+h-t)-g(x-t)big),dt
$$
and hence
$$
|(f*g)(x+h)-(f*g)(x)|le max |f|cdot int_{-infty}^infty |g(x+h-t)-g(x-t)|,dt
$$
Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Compact support of $f*g$. Assume that $,f,gin C_0(mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.
Then for $x>2M$,
$$
(f*g)(x)=int_{-infty}^infty f(t),g(x-t),dt=int_{-M}^M f(t),g(x-t),dt=0,
$$
since $x-t>M$, when $|t|le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.
Continuity of $f*g$.
Observe that
$$
(f*g)(x+h)-(f*g)(x)=int_{-infty}^infty f(t),big(g(x+h-t)-g(x-t)big),dt
$$
and hence
$$
|(f*g)(x+h)-(f*g)(x)|le max |f|cdot int_{-infty}^infty |g(x+h-t)-g(x-t)|,dt
$$
Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
add a comment |
up vote
3
down vote
Compact support of $f*g$. Assume that $,f,gin C_0(mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.
Then for $x>2M$,
$$
(f*g)(x)=int_{-infty}^infty f(t),g(x-t),dt=int_{-M}^M f(t),g(x-t),dt=0,
$$
since $x-t>M$, when $|t|le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.
Continuity of $f*g$.
Observe that
$$
(f*g)(x+h)-(f*g)(x)=int_{-infty}^infty f(t),big(g(x+h-t)-g(x-t)big),dt
$$
and hence
$$
|(f*g)(x+h)-(f*g)(x)|le max |f|cdot int_{-infty}^infty |g(x+h-t)-g(x-t)|,dt
$$
Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
add a comment |
up vote
3
down vote
up vote
3
down vote
Compact support of $f*g$. Assume that $,f,gin C_0(mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.
Then for $x>2M$,
$$
(f*g)(x)=int_{-infty}^infty f(t),g(x-t),dt=int_{-M}^M f(t),g(x-t),dt=0,
$$
since $x-t>M$, when $|t|le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.
Continuity of $f*g$.
Observe that
$$
(f*g)(x+h)-(f*g)(x)=int_{-infty}^infty f(t),big(g(x+h-t)-g(x-t)big),dt
$$
and hence
$$
|(f*g)(x+h)-(f*g)(x)|le max |f|cdot int_{-infty}^infty |g(x+h-t)-g(x-t)|,dt
$$
Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
Compact support of $f*g$. Assume that $,f,gin C_0(mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.
Then for $x>2M$,
$$
(f*g)(x)=int_{-infty}^infty f(t),g(x-t),dt=int_{-M}^M f(t),g(x-t),dt=0,
$$
since $x-t>M$, when $|t|le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.
Continuity of $f*g$.
Observe that
$$
(f*g)(x+h)-(f*g)(x)=int_{-infty}^infty f(t),big(g(x+h-t)-g(x-t)big),dt
$$
and hence
$$
|(f*g)(x+h)-(f*g)(x)|le max |f|cdot int_{-infty}^infty |g(x+h-t)-g(x-t)|,dt
$$
Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 17:41
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
add a comment |
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
And the way this compact support idea works is that we could have several such intervals that we might have to consider but they all basically act the same way? Or there is really just one interval to consider?
– Mason
Nov 19 at 17:47
1
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
Nov 19 at 18:53
add a comment |
1
Loosely speaking, if $f*g$ is non-zero at $x=a$, then there must be $d,c$ such that $b+c=a$ and $f(b),g(c)$ are both non-zero.
– Arthur
Nov 19 at 17:03