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If $,f$ and $g$ are continuous with compact support then $f*g$ (convolution ) is also continuous with compact...
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1
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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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
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Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 yesterday
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 yesterday
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
2 days ago
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
edited 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
edited 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
61.5k1383161
asked 2 days ago
Gabriel Corrêa
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Gabriel Corrêa
93
asked 2 days ago
Gabriel Corrêa
93
93
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 yesterday
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.
put on hold as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 yesterday
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
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
2 days ago
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
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
2 days ago
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
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
2 days ago
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
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 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
answered 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
answered 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
answered 2 days ago
Yiorgos S. Smyrlis
61.5k1383161
61.5k1383161
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
2 days ago
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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
2 days ago
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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
2 days ago
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
2 days ago
1
1
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
One interval, or one closed ball, in higher dimensions.
– Yiorgos S. Smyrlis
2 days ago
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
2 days ago