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If $X_n$ converges to $X$ in $L^p$, do we have $X_n^p$ converges to $X^p$ in $L^1$?

We can prove that it is true when p=1,2 easily. I am curious whether this is true for all $p>0$.

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You should consider $|X|^p$ and so on, as $X^p$ does not make sense for all $X$ and all $p>0$. –  Mariano Suárez-Alvarez Nov 8 '12 at 3:05
    
Thanks for your comment. When I consider Lp convergence, I will take modulus. I don't think we need to add modulus in the beginning. –  XXX11235 Nov 8 '12 at 23:53

3 Answers 3

up vote 6 down vote accepted

Note that $$|x^p - y^p| = \left|\int_y^x p t^{p-1}\ dt \right| \le p (|x|+|y|)^{p-1} |x - y|$$ Thus if $1/p + 1/q = 1$ (where $1 < p,q < \infty$) $$\eqalign{\|X^p - Y^p\|_1 &\le \int p (|X| + |Y|)^{p-1} |X - Y| \cr &\le p \|(|X| + |Y|)^{p-1}\|_q \|X - Y\|_p \cr &= p \left(\int (|X| + |Y|)^p\right)^{1/q} \|X - Y\|_p \cr &= p \||X|+|Y|\|_p^{p/q} \|X - Y\|_p \cr &\le p (\|X\|_p + \|Y\|_p)^{p/q} \|X - Y\|_p\cr}$$ If $X_n \to X$ in $L^p$, $\|X_n\|_p$ is bounded, and so we get $\|X_n^p - X^p\|_1 \to 0$.

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Get it. Thank you very much!! –  XXX11235 Nov 8 '12 at 23:48
    
@Robert Israel: Dear Sir. I have an interesting question in the following math.stackexchange.com/questions/346937/… I would like to ask your comments? –  blindman Apr 12 '13 at 2:18

It's not correct for $0 < p < 1$. Take $X_n: [0,1] \to \mathbb{R}$ as $X_n(x) = n \cdot I_{[0,1/n)}(x)$, where $I_A$ = indicator function of a set $A$. Then $\|X_n\|_p = n^{1-1/p}$ and therefore $X_n \to 0$ in $L^p$ for $p < 1$, but $X_n$ does not converge in $L^1$.

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I don't understand you solution. But many thanks anyway. –  XXX11235 Nov 8 '12 at 23:53

It is solved by Robert Israel.

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If you cite an author, give book, edition and page number. Be complete. –  ncmathsadist Nov 9 '12 at 2:38
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It is actually the first answer of this page. I should have said "It is answered by Robert" –  XXX11235 Nov 10 '12 at 3:23

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