is there a simple way to prove that $X_n \rightarrow_{L^p} X$ implies that $\mathrm{E}(X^p_n) \rightarrow \mathrm{E}(X^p)$? the proof for $p=1$ is easy. but what about the case $p>1$? I would appreciate any comments. many thanks!
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
If $0<a<b$ then, by the Mean Value Theorem, there exists $c\in[a,b]$ such that $$ |a^p-b^p|=p\,c^{p-1}\,|a-b|\implies |a^p-b^p|\le p\,b^{p-1}\,|a-b|\le p\,(a^{p-1}+b^{p-1})\,|a-b|. $$ Then $$ |E(X_n^p)-E(X^p)|\le p\,E((X_n^{p-1}+X^{p-1})|X_n-X|). $$ Now use Hölder's inequality an the boundedness of $E(X_n^p)$ to deduce that $E(X_n^p)\to E(X^p)$.
answered Oct 10, 2013 at 16:05
-
$\begingroup$ thanks for your help! great idea, I just don't see how the boundedness of E(X_n^p) kicks in (the conjugate Hölder-norm is the essential supremum, right?). so I would have to establish that the essential supremum of X_n^(p-1) is finite - but I don't know how to do that? any further hints? many thanks! $\endgroup$– s_2Commented Oct 10, 2013 at 19:42
-
$\begingroup$ Let $q=p/(p-1)$ be the conjugate exponent of $p$. The $L^q$ norm of $X_n^{p-1}$ (which appears when you use Hölder's inequality) is precisely the $L^p$ norm of $X_n$, and this is bounded. $\endgroup$ Commented Oct 10, 2013 at 21:20
-
$\begingroup$ of course - silly me that I did not see it! many thanks for your help, I really appreciate it! $\endgroup$– s_2Commented Oct 10, 2013 at 21:44