Prove that if $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $lim_{vert x vert to...












0












$begingroup$



The convolution of $f$ and $g$ on $R^d$ equipped with the lebsgue measure is defined by $$(f*g)(x)=int_{R_d} f(x-y)g(y) , dy$$ Prove that if $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $$lim_{vert x vert to infty}(f*g)(x)=0$$




There's something I have already proved:



(a) If $f in L^p$ and $g in L^1$ , then $f*g in L^p$ with $$vertvert f*g vertvert _{L^p} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^1}$$
(b) If $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $f*g in L^{infty}$ with $$vertvert f*g vertvert _{L^infty} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^q}$$ Moreover , the convolution $f*g$ is uniformly continuous on $R^d$



I want to show that $f*g in L^a$ for some $a lt infty$ , then by the uniform ontinuous I can get the desired conclution. However , can I find the desired $a$ ？






real-analysis functional-analysis integral-inequality







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 10:29










  • $begingroup$
    Thank you ! I see the point now.
    $endgroup$
    – J.Guo
    Dec 6 '18 at 10:46










  • $begingroup$
    By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 12:45










  • $begingroup$
    @ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
    $endgroup$
    – J.Guo
    Dec 6 '18 at 13:19
















0












$begingroup$



The convolution of $f$ and $g$ on $R^d$ equipped with the lebsgue measure is defined by $$(f*g)(x)=int_{R_d} f(x-y)g(y) , dy$$ Prove that if $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $$lim_{vert x vert to infty}(f*g)(x)=0$$




There's something I have already proved:



(a) If $f in L^p$ and $g in L^1$ , then $f*g in L^p$ with $$vertvert f*g vertvert _{L^p} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^1}$$
(b) If $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $f*g in L^{infty}$ with $$vertvert f*g vertvert _{L^infty} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^q}$$ Moreover , the convolution $f*g$ is uniformly continuous on $R^d$



I want to show that $f*g in L^a$ for some $a lt infty$ , then by the uniform ontinuous I can get the desired conclution. However , can I find the desired $a$ ？






real-analysis functional-analysis integral-inequality







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 10:29










  • $begingroup$
    Thank you ! I see the point now.
    $endgroup$
    – J.Guo
    Dec 6 '18 at 10:46










  • $begingroup$
    By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 12:45










  • $begingroup$
    @ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
    $endgroup$
    – J.Guo
    Dec 6 '18 at 13:19














0












0








0





$begingroup$



The convolution of $f$ and $g$ on $R^d$ equipped with the lebsgue measure is defined by $$(f*g)(x)=int_{R_d} f(x-y)g(y) , dy$$ Prove that if $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $$lim_{vert x vert to infty}(f*g)(x)=0$$




There's something I have already proved:



(a) If $f in L^p$ and $g in L^1$ , then $f*g in L^p$ with $$vertvert f*g vertvert _{L^p} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^1}$$
(b) If $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $f*g in L^{infty}$ with $$vertvert f*g vertvert _{L^infty} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^q}$$ Moreover , the convolution $f*g$ is uniformly continuous on $R^d$



I want to show that $f*g in L^a$ for some $a lt infty$ , then by the uniform ontinuous I can get the desired conclution. However , can I find the desired $a$ ？






real-analysis functional-analysis integral-inequality







share|cite|improve this question









$endgroup$





The convolution of $f$ and $g$ on $R^d$ equipped with the lebsgue measure is defined by $$(f*g)(x)=int_{R_d} f(x-y)g(y) , dy$$ Prove that if $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $$lim_{vert x vert to infty}(f*g)(x)=0$$




There's something I have already proved:



(a) If $f in L^p$ and $g in L^1$ , then $f*g in L^p$ with $$vertvert f*g vertvert _{L^p} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^1}$$
(b) If $f in L^p$ and $g in L^q$ where $p$ and $q$ are conjugate exponents , then $f*g in L^{infty}$ with $$vertvert f*g vertvert _{L^infty} le vertvert f vertvert _{L^p} vertvert g vertvert _{L^q}$$ Moreover , the convolution $f*g$ is uniformly continuous on $R^d$



I want to show that $f*g in L^a$ for some $a lt infty$ , then by the uniform ontinuous I can get the desired conclution. However , can I find the desired $a$ ？






real-analysis functional-analysis integral-inequality




real-analysis functional-analysis integral-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 10:27









J.GuoJ.Guo

2599




2599








  • 2




    $begingroup$
    Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 10:29










  • $begingroup$
    Thank you ! I see the point now.
    $endgroup$
    – J.Guo
    Dec 6 '18 at 10:46










  • $begingroup$
    By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 12:45










  • $begingroup$
    @ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
    $endgroup$
    – J.Guo
    Dec 6 '18 at 13:19














  • 2




    $begingroup$
    Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 10:29










  • $begingroup$
    Thank you ! I see the point now.
    $endgroup$
    – J.Guo
    Dec 6 '18 at 10:46










  • $begingroup$
    By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 12:45










  • $begingroup$
    @ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
    $endgroup$
    – J.Guo
    Dec 6 '18 at 13:19








2




2




$begingroup$
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 10:29




$begingroup$
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 10:29












$begingroup$
Thank you ! I see the point now.
$endgroup$
– J.Guo
Dec 6 '18 at 10:46




$begingroup$
Thank you ! I see the point now.
$endgroup$
– J.Guo
Dec 6 '18 at 10:46












$begingroup$
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
$endgroup$
– MaoWao
Dec 6 '18 at 12:45




$begingroup$
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
$endgroup$
– MaoWao
Dec 6 '18 at 12:45












$begingroup$
@ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
$endgroup$
– J.Guo
Dec 6 '18 at 13:19




$begingroup$
@ MaoWao If not , then there exist $a gt0 ,, delta gt 0$ such that for every $M ge 0$ ,there exist $x_0 gt M$ , $vert f(x) vert gt a$ whenever $vert x-x_0 vert lt delta$ , so we have $int _{R_d} vert f(x) vert ^p , dx to infty$
$endgroup$
– J.Guo
Dec 6 '18 at 13:19










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