I am just wondering whether following is true?
Let $p>1$ If $f_n\to f$ in $L^p(\mu)$, is it then true that $f^p\to f^p$ in $L^1(\mu)$?
This is true if $||x|^p-|y|^p|\le |x-y|^p$. I know this is true for $p=1$. Is the inequality true for $p>1$?
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$|x|^p=\Big( \sum_{i=1}{\vert x_i\vert}^{p}\Big)^{1/p} $ ? – R.N Sep 15 '15 at 06:28
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@Razieh Noori: Yes. – Ashok Sep 15 '15 at 07:05
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@Razieh Noori: Sorry, $|x|^p$ doesn't mean norm. It's just the $p$th power of the scalar $|x|$. – Ashok Sep 15 '15 at 12:09
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It is ok, dear :) – R.N Sep 15 '15 at 12:10
2 Answers
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This is not true since $$2\leq \left(u+\frac{1}{u}\right)^2 -u^2 $$
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$||x|^p-|y|^p|\le |x-y|^p$ is just a result of Triangle inequality. for first you need only wright definision $
$|x+y|^p\leq |x|^p +|y|^p $ for every $x$ and $y$. now replace $y$ with $y-x$ $$|x+y-x|^p\leq |x|^p+|x-y|^p \Rightarrow |y|^p\leq |x|^p+|x-y|^p$$ hence $$|y|^p - |x|^p\leq |x-y|^p.$$ this time replace $x$ with $x-y$ and then you will have $$-(|y|^p - |x|^p) =|x|^p - |y|^p\leq |x-y|^p.$$ Now use the fact that if $-a\leq b\leq a$ then $|b|\leq |a|$.
For first part try to wright if you had problem just say.
R.N
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No. $|x+y|^p \le |x|^p+|y|^p$ isn't correct. Take for example, $ x=1,y=1$. – Ashok Sep 15 '15 at 12:03
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then $|1+1|^p= (2^p)^\dfrac{1}{p}=2\leq (1^p)^\dfrac{1}{p}+(1^p)^\dfrac{1}{p}=1+1=2$ – R.N Sep 15 '15 at 12:05
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