Brezis excercise 4.12: $L_p$ is uniformly convex for $1<pleq 2$
1
$begingroup$
- Let $1<p<infty$. Prove that there is a constant $C$(depending only on $p$) such that $$|a-b|^pleq C(|a|^p+|b|^p)^{1-s}(|a|^p+|b|^p-2|dfrac{a+b}{2}|^p)^s$$ for all $a,bin mathbb{R}$ and $s=dfrac{p}{2}$.
2.Deduce that $L_p$ is uniformly convex for $1<pleq 2$.
I have an idea on how to prove 2, assuming 1 and using Holder's inequality I think I can do it, however I dont see a way to prove 1. any hints? Thanks in advance.
functional-analysis lp-spaces
asked Jan 10 at 18:42
AlfdavAlfdav
787
$endgroup$
add a comment |
1
$begingroup$
- Let $1<p<infty$. Prove that there is a constant $C$(depending only on $p$) such that $$|a-b|^pleq C(|a|^p+|b|^p)^{1-s}(|a|^p+|b|^p-2|dfrac{a+b}{2}|^p)^s$$ for all $a,bin mathbb{R}$ and $s=dfrac{p}{2}$.
2.Deduce that $L_p$ is uniformly convex for $1<pleq 2$.
I have an idea on how to prove 2, assuming 1 and using Holder's inequality I think I can do it, however I dont see a way to prove 1. any hints? Thanks in advance.
functional-analysis lp-spaces
asked Jan 10 at 18:42
AlfdavAlfdav
787
$endgroup$
$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30
add a comment |
1
1
1
$begingroup$
- Let $1<p<infty$. Prove that there is a constant $C$(depending only on $p$) such that $$|a-b|^pleq C(|a|^p+|b|^p)^{1-s}(|a|^p+|b|^p-2|dfrac{a+b}{2}|^p)^s$$ for all $a,bin mathbb{R}$ and $s=dfrac{p}{2}$.
2.Deduce that $L_p$ is uniformly convex for $1<pleq 2$.
I have an idea on how to prove 2, assuming 1 and using Holder's inequality I think I can do it, however I dont see a way to prove 1. any hints? Thanks in advance.
functional-analysis lp-spaces
asked Jan 10 at 18:42
AlfdavAlfdav
787
$endgroup$
- Let $1<p<infty$. Prove that there is a constant $C$(depending only on $p$) such that $$|a-b|^pleq C(|a|^p+|b|^p)^{1-s}(|a|^p+|b|^p-2|dfrac{a+b}{2}|^p)^s$$ for all $a,bin mathbb{R}$ and $s=dfrac{p}{2}$.
2.Deduce that $L_p$ is uniformly convex for $1<pleq 2$.
I have an idea on how to prove 2, assuming 1 and using Holder's inequality I think I can do it, however I dont see a way to prove 1. any hints? Thanks in advance.
functional-analysis lp-spaces
functional-analysis lp-spaces
asked Jan 10 at 18:42
AlfdavAlfdav
787
asked Jan 10 at 18:42
AlfdavAlfdav
787
edited Jan 10 at 18:58
Alfdav
asked Jan 10 at 18:42
AlfdavAlfdav
787
asked Jan 10 at 18:42
AlfdavAlfdav
787
asked Jan 10 at 18:42
AlfdavAlfdav
787
787
$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30
add a comment |
$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30
$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30
add a comment |
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$begingroup$
This does not look homogenous. I would say $1-s$ instead of $s-1$.
$endgroup$
– Mindlack
Jan 10 at 18:44
$begingroup$
Right. edited, thanks.
$endgroup$
– Alfdav
Jan 10 at 18:48
$begingroup$
You may assume wlog that $a = 1$, due to homogeneity. Then it's essentially a one variable calculus problem.
$endgroup$
– Hans Engler
Jan 10 at 20:30