Hint Prove $||b|^p-|a|^p-|a-b|^p|leq C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$
$begingroup$
I have looked at this inequality in several ways, but cannot find the correct path:
Show for $a,b in mathbb R$ and $C_{p} > 0$ with $p in ]1,infty[$
$||b|^p-|a|^p-|a-b|^p|leq C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$
My attempt:
In this first case
Looking at LHS: $|b|^p-|a|^p-|a-b|^pleq|b-a|^{p}-|a-b|^{p}=|a-b|^{p}-|a-b|^{p}=0$
and the RHS is: $C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$ and note $(|a|^{p-1}|a-b|+|a||a-b|^{p-1}) > 0$, so we could easily find $C_{p}$ (e.g. $1$) so that LHS $leq$ RHS
In the second case, looking at LHS: $|a|^p+|a-b|^p-|b|^p$ and attempting to prove LHS $geq$ RHS, all I can say is $|a|^p+|a-b|^p-|b|^pgeq |a|^p+|a|^p-|b|^p-|b|^p=2(|a|^p-|b|^p)...$ but this does not lead anywhere, does it?
Any ideas?
real-analysis measure-theory inequality
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
$endgroup$
add a comment |
$begingroup$
I have looked at this inequality in several ways, but cannot find the correct path:
Show for $a,b in mathbb R$ and $C_{p} > 0$ with $p in ]1,infty[$
$||b|^p-|a|^p-|a-b|^p|leq C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$
My attempt:
In this first case
Looking at LHS: $|b|^p-|a|^p-|a-b|^pleq|b-a|^{p}-|a-b|^{p}=|a-b|^{p}-|a-b|^{p}=0$
and the RHS is: $C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$ and note $(|a|^{p-1}|a-b|+|a||a-b|^{p-1}) > 0$, so we could easily find $C_{p}$ (e.g. $1$) so that LHS $leq$ RHS
In the second case, looking at LHS: $|a|^p+|a-b|^p-|b|^p$ and attempting to prove LHS $geq$ RHS, all I can say is $|a|^p+|a-b|^p-|b|^pgeq |a|^p+|a|^p-|b|^p-|b|^p=2(|a|^p-|b|^p)...$ but this does not lead anywhere, does it?
Any ideas?
real-analysis measure-theory inequality
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
$endgroup$
add a comment |
$begingroup$
I have looked at this inequality in several ways, but cannot find the correct path:
Show for $a,b in mathbb R$ and $C_{p} > 0$ with $p in ]1,infty[$
$||b|^p-|a|^p-|a-b|^p|leq C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$
My attempt:
In this first case
Looking at LHS: $|b|^p-|a|^p-|a-b|^pleq|b-a|^{p}-|a-b|^{p}=|a-b|^{p}-|a-b|^{p}=0$
and the RHS is: $C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$ and note $(|a|^{p-1}|a-b|+|a||a-b|^{p-1}) > 0$, so we could easily find $C_{p}$ (e.g. $1$) so that LHS $leq$ RHS
In the second case, looking at LHS: $|a|^p+|a-b|^p-|b|^p$ and attempting to prove LHS $geq$ RHS, all I can say is $|a|^p+|a-b|^p-|b|^pgeq |a|^p+|a|^p-|b|^p-|b|^p=2(|a|^p-|b|^p)...$ but this does not lead anywhere, does it?
Any ideas?
real-analysis measure-theory inequality
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
$endgroup$
I have looked at this inequality in several ways, but cannot find the correct path:
Show for $a,b in mathbb R$ and $C_{p} > 0$ with $p in ]1,infty[$
$||b|^p-|a|^p-|a-b|^p|leq C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$
My attempt:
In this first case
Looking at LHS: $|b|^p-|a|^p-|a-b|^pleq|b-a|^{p}-|a-b|^{p}=|a-b|^{p}-|a-b|^{p}=0$
and the RHS is: $C_{p}(|a|^{p-1}|a-b|+|a||a-b|^{p-1})$ and note $(|a|^{p-1}|a-b|+|a||a-b|^{p-1}) > 0$, so we could easily find $C_{p}$ (e.g. $1$) so that LHS $leq$ RHS
In the second case, looking at LHS: $|a|^p+|a-b|^p-|b|^p$ and attempting to prove LHS $geq$ RHS, all I can say is $|a|^p+|a-b|^p-|b|^pgeq |a|^p+|a|^p-|b|^p-|b|^p=2(|a|^p-|b|^p)...$ but this does not lead anywhere, does it?
Any ideas?
real-analysis measure-theory inequality
real-analysis measure-theory inequality
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
asked Dec 11 '18 at 12:47
SABOYSABOY
710311
710311
add a comment |
add a comment |
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