If $,f$ and $g$ are continuous with compact support then $f*g$ (convolution ) is also continuous with compact...











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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







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put on hold as off-topic by Cameron Williams, amWhy, Rebellos, jgon, user10354138 yesterday


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  • "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

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    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















up vote
1
down vote

favorite
1












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







share|cite|improve this question









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













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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Gabriel Corrêa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 2 days ago









Yiorgos S. Smyrlis

61.5k1383161




61.5k1383161






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asked 2 days ago









Gabriel Corrêa

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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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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














  • 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










1 Answer
1






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oldest

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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.






share|cite|improve this answer





















  • 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


















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.






share|cite|improve this answer





















  • 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















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.






share|cite|improve this answer





















  • 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













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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



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