The Convergence of Convolution of $f$ and $g$












0












$begingroup$


I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.



I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.



Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$



The last equality is by translation invariance and reflection invariance.



Therefore, $|f*g(x)|$ is defined everywhere and bounded.



Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$



Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$



It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.



I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.



Any idea will be really appreciated!






measure-theory convolution holder-inequality







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.



    I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.



    Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$



    The last equality is by translation invariance and reflection invariance.



    Therefore, $|f*g(x)|$ is defined everywhere and bounded.



    Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$



    Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$



    It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.



    I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.



    Any idea will be really appreciated!






    measure-theory convolution holder-inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.



      I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.



      Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$



      The last equality is by translation invariance and reflection invariance.



      Therefore, $|f*g(x)|$ is defined everywhere and bounded.



      Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$



      Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$



      It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.



      I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.



      Any idea will be really appreciated!






      measure-theory convolution holder-inequality







      share|cite|improve this question









      $endgroup$




      I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.



      I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.



      Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$



      The last equality is by translation invariance and reflection invariance.



      Therefore, $|f*g(x)|$ is defined everywhere and bounded.



      Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$



      Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$



      It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.



      I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.



      Any idea will be really appreciated!






      measure-theory convolution holder-inequality




      measure-theory convolution holder-inequality






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 18 '18 at 23:12









      JacobsonRadicalJacobsonRadical

      502111




      502111






















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          $begingroup$

          There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
            $endgroup$
            – JacobsonRadical
            Dec 18 '18 at 23:52










          • $begingroup$
            @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:53










          • $begingroup$
            Oh! I got it, I was thinking too much. Thank you!
            $endgroup$
            – JacobsonRadical
            Dec 19 '18 at 0:36











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          active

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          2












          $begingroup$

          There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
            $endgroup$
            – JacobsonRadical
            Dec 18 '18 at 23:52










          • $begingroup$
            @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:53










          • $begingroup$
            Oh! I got it, I was thinking too much. Thank you!
            $endgroup$
            – JacobsonRadical
            Dec 19 '18 at 0:36
















          2












          $begingroup$

          There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
            $endgroup$
            – JacobsonRadical
            Dec 18 '18 at 23:52










          • $begingroup$
            @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:53










          • $begingroup$
            Oh! I got it, I was thinking too much. Thank you!
            $endgroup$
            – JacobsonRadical
            Dec 19 '18 at 0:36














          2












          2








          2





          $begingroup$

          There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].






          share|cite|improve this answer









          $endgroup$



          There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 23:21









          Kavi Rama MurthyKavi Rama Murthy

          60.3k42161




          60.3k42161












          • $begingroup$
            Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
            $endgroup$
            – JacobsonRadical
            Dec 18 '18 at 23:52










          • $begingroup$
            @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:53










          • $begingroup$
            Oh! I got it, I was thinking too much. Thank you!
            $endgroup$
            – JacobsonRadical
            Dec 19 '18 at 0:36


















          • $begingroup$
            Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
            $endgroup$
            – JacobsonRadical
            Dec 18 '18 at 23:52










          • $begingroup$
            @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:53










          • $begingroup$
            Oh! I got it, I was thinking too much. Thank you!
            $endgroup$
            – JacobsonRadical
            Dec 19 '18 at 0:36
















          $begingroup$
          Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
          $endgroup$
          – JacobsonRadical
          Dec 18 '18 at 23:52




          $begingroup$
          Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
          $endgroup$
          – JacobsonRadical
          Dec 18 '18 at 23:52












          $begingroup$
          @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
          $endgroup$
          – Kavi Rama Murthy
          Dec 18 '18 at 23:53




          $begingroup$
          @JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
          $endgroup$
          – Kavi Rama Murthy
          Dec 18 '18 at 23:53












          $begingroup$
          Oh! I got it, I was thinking too much. Thank you!
          $endgroup$
          – JacobsonRadical
          Dec 19 '18 at 0:36




          $begingroup$
          Oh! I got it, I was thinking too much. Thank you!
          $endgroup$
          – JacobsonRadical
          Dec 19 '18 at 0:36


















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