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...
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
asked Dec 6 at 10:27
J.Guo
1769
add a comment |
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
asked Dec 6 at 10:27
J.Guo
1769
2
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19
add a comment |
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
asked Dec 6 at 10:27
J.Guo
1769
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
asked Dec 6 at 10:27
J.Guo
1769
asked Dec 6 at 10:27
J.Guo
1769
asked Dec 6 at 10:27
J.Guo
1769
asked Dec 6 at 10:27
J.Guo
1769
asked Dec 6 at 10:27
J.Guo
1769
1769
2
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19
add a comment |
2
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19
2
2
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19
add a comment |
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Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know.
– Kavi Rama Murthy
Dec 6 at 10:29
Thank you ! I see the point now.
– J.Guo
Dec 6 at 10:46
By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $lim_{|x|to infty}f(x)=0$.
– MaoWao
Dec 6 at 12:45
@ 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 $int _{R_d} vert f(x) vert ^p , dx to infty$
– J.Guo
Dec 6 at 13:19