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non-decreasing by young's inequality
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Suppose that $int_0^inftyf(x)dx=1 $ and $int_0^inftye^kxf(x)dx<infty$ for some $k>0 $. Prove $(int_0^inftyx^tf(x)dx)^frac1t$ is non-decreasing in t (Hint: Use young's inequality).I want to follow the hint,but i don't know how to apply young's inequality.Can someone give more hint or a complete proof?
calculus analysis
asked Jul 26 at 0:48
kai
94
add a comment |Â
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Suppose that $int_0^inftyf(x)dx=1 $ and $int_0^inftye^kxf(x)dx<infty$ for some $k>0 $. Prove $(int_0^inftyx^tf(x)dx)^frac1t$ is non-decreasing in t (Hint: Use young's inequality).I want to follow the hint,but i don't know how to apply young's inequality.Can someone give more hint or a complete proof?
calculus analysis
asked Jul 26 at 0:48
kai
94
There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
Is it also another possible approach?
– kai
Jul 26 at 7:22
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $int_0^inftyf(x)dx=1 $ and $int_0^inftye^kxf(x)dx<infty$ for some $k>0 $. Prove $(int_0^inftyx^tf(x)dx)^frac1t$ is non-decreasing in t (Hint: Use young's inequality).I want to follow the hint,but i don't know how to apply young's inequality.Can someone give more hint or a complete proof?
calculus analysis
asked Jul 26 at 0:48
kai
94
Suppose that $int_0^inftyf(x)dx=1 $ and $int_0^inftye^kxf(x)dx<infty$ for some $k>0 $. Prove $(int_0^inftyx^tf(x)dx)^frac1t$ is non-decreasing in t (Hint: Use young's inequality).I want to follow the hint,but i don't know how to apply young's inequality.Can someone give more hint or a complete proof?
calculus analysis
asked Jul 26 at 0:48
kai
94
edited Jul 26 at 1:08
asked Jul 26 at 0:48
kai
94
asked Jul 26 at 0:48
kai
94
asked Jul 26 at 0:48
kai
94
94
There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
Is it also another possible approach?
– kai
Jul 26 at 7:22
add a comment |Â
There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
Is it also another possible approach?
– kai
Jul 26 at 7:22
There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
Is it also another possible approach?
– kai
Jul 26 at 7:22
Is it also another possible approach?
– kai
Jul 26 at 7:22
add a comment |Â
1 Answer
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1
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Young's inequality says $ab leq frac a^p p + frac b^p q$ if $a,b geq 0$,$1<p<infty $ and $frac 1 p + frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(int_0^infty x^sf(x), dx)^t/s$. Let $p=frac s t$, $q=frac s s-t$. Apply Young's inequality with $a=frac x^t c$ and $b=1$. You will get $frac x^t c leq frac x^s pc^p+frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ frac int_0^inftyx^tf(x), dx c leq frac 1 p +frac 1 q =1$ (where we used the fact that $c^p$ in the denominator cancels with $int_0^infty x^s f(x), dx$). Hence $ int_0^inftyx^tf(x), dx leq c=(int_0^infty x^sf(x), dx)^t/s$. This gives $ (int_0^inftyx^tf(x), dx )^1/t leq c=(int_0^infty x^sf(x), dx)^1/s$ which is what we want to prove.
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Young's inequality says $ab leq frac a^p p + frac b^p q$ if $a,b geq 0$,$1<p<infty $ and $frac 1 p + frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(int_0^infty x^sf(x), dx)^t/s$. Let $p=frac s t$, $q=frac s s-t$. Apply Young's inequality with $a=frac x^t c$ and $b=1$. You will get $frac x^t c leq frac x^s pc^p+frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ frac int_0^inftyx^tf(x), dx c leq frac 1 p +frac 1 q =1$ (where we used the fact that $c^p$ in the denominator cancels with $int_0^infty x^s f(x), dx$). Hence $ int_0^inftyx^tf(x), dx leq c=(int_0^infty x^sf(x), dx)^t/s$. This gives $ (int_0^inftyx^tf(x), dx )^1/t leq c=(int_0^infty x^sf(x), dx)^1/s$ which is what we want to prove.
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
add a comment |Â
up vote
1
down vote
Young's inequality says $ab leq frac a^p p + frac b^p q$ if $a,b geq 0$,$1<p<infty $ and $frac 1 p + frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(int_0^infty x^sf(x), dx)^t/s$. Let $p=frac s t$, $q=frac s s-t$. Apply Young's inequality with $a=frac x^t c$ and $b=1$. You will get $frac x^t c leq frac x^s pc^p+frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ frac int_0^inftyx^tf(x), dx c leq frac 1 p +frac 1 q =1$ (where we used the fact that $c^p$ in the denominator cancels with $int_0^infty x^s f(x), dx$). Hence $ int_0^inftyx^tf(x), dx leq c=(int_0^infty x^sf(x), dx)^t/s$. This gives $ (int_0^inftyx^tf(x), dx )^1/t leq c=(int_0^infty x^sf(x), dx)^1/s$ which is what we want to prove.
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Young's inequality says $ab leq frac a^p p + frac b^p q$ if $a,b geq 0$,$1<p<infty $ and $frac 1 p + frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(int_0^infty x^sf(x), dx)^t/s$. Let $p=frac s t$, $q=frac s s-t$. Apply Young's inequality with $a=frac x^t c$ and $b=1$. You will get $frac x^t c leq frac x^s pc^p+frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ frac int_0^inftyx^tf(x), dx c leq frac 1 p +frac 1 q =1$ (where we used the fact that $c^p$ in the denominator cancels with $int_0^infty x^s f(x), dx$). Hence $ int_0^inftyx^tf(x), dx leq c=(int_0^infty x^sf(x), dx)^t/s$. This gives $ (int_0^inftyx^tf(x), dx )^1/t leq c=(int_0^infty x^sf(x), dx)^1/s$ which is what we want to prove.
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
Young's inequality says $ab leq frac a^p p + frac b^p q$ if $a,b geq 0$,$1<p<infty $ and $frac 1 p + frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(int_0^infty x^sf(x), dx)^t/s$. Let $p=frac s t$, $q=frac s s-t$. Apply Young's inequality with $a=frac x^t c$ and $b=1$. You will get $frac x^t c leq frac x^s pc^p+frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ frac int_0^inftyx^tf(x), dx c leq frac 1 p +frac 1 q =1$ (where we used the fact that $c^p$ in the denominator cancels with $int_0^infty x^s f(x), dx$). Hence $ int_0^inftyx^tf(x), dx leq c=(int_0^infty x^sf(x), dx)^t/s$. This gives $ (int_0^inftyx^tf(x), dx )^1/t leq c=(int_0^infty x^sf(x), dx)^1/s$ which is what we want to prove.
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
answered Jul 26 at 9:32
Kavi Rama Murthy
20k2829
20k2829
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
add a comment |Â
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
Thank you for the answer. Now, I realize it.
– kai
Jul 26 at 9:42
add a comment |Â
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There are a lot of $f$ that satisfy your hypothesis for which $int_0^infty x^t f (x)dx $ is negative, so it doesn't make sense in general to take the $1/t $-power of this quantity. Are you sure that you didn't miss any hypothesis?
– Bob
Jul 26 at 5:46
I think it should be added that f is non-negative
– kai
Jul 26 at 5:54
I can't see how this follows from Young's inequality (by the way, which one of the three?). However, it is a simple consequence of Jensen's inequality (and you can also get rid of the useless condition $int_0^infty e^kxf(x)dx<infty$)
– Bob
Jul 26 at 6:05
I think that we can prove Holder's inequality by young's inequality.For $t_1<t_2$ we take $p=fract_2t_1$ and $q=fract_2t_2-t_1$. By using Holder's inequality ,we will have $(int_0^inftyx^t_1f(x)dx)^frac1t_1<(int_0^inftyx^t_2f(x)dx)^frac1t_2$
– kai
Jul 26 at 7:18
Is it also another possible approach?
– kai
Jul 26 at 7:22