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Suppose that $\|f\|_p < \infty$ for all $1\leq p < p'$, I want to know if the the following is true and in that case how to show it

$p \mapsto \|f\|_p$ is continuous on $[1,p')$

Or maybe we need to impose some more constraints such as finite measure space. In case of finite measure space, I tried to use Egoroff's theorem.

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1 Answer 1

up vote 7 down vote accepted

It suffices to show $g(p) = \int_X |f|^p$ is continuous, because then $||f||_p = g(p)^{1 \over p} = e^{\log(g(p)) \over p}$ is continuous.

Note that $g(p) = \int_{|f| > 1} |f|^p + \int_{|f| \leq 1}|f|^p$. To show continuity of each term at some $p_0$, you can use the dominated convergence theorem; for some $\epsilon > 0$, $|f|^{p_0 + \epsilon}$ will serve as a dominating function for the first integral, and $|f|^{p_0 - \epsilon}$ will serve as a dominating function for the second (unless $p_0 = 1$ in which case just use $|f|$).

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Nice! I was thinking about using DCT, but I got stuck on finding a dominating function. The splitting part is great. –  Hawii Apr 19 '12 at 6:46
    
What if $\int_X |f|^{p_0+\epsilon} d\mu=\infty$ ? In this case $|f|^{p_0+\epsilon}$ is not a dominating function –  Amr Oct 14 '13 at 14:30
    
If $\epsilon$ is small enough it will finite; recall that $f \in L^p$ for all $1 \leq p \leq p'$ and $p = p_0$ is in this range. –  Zarrax Oct 14 '13 at 20:54
    
@Zarrax How do we show that if $\epsilon$ is small enough the integral of $|f|^{p_0+\epsilon}$ will be finite? –  Amr Dec 3 '13 at 19:41
    
Aaaah. I misread the question and didn't see the condition $\|f\|_p < \infty$ for all $1\leq p < p'$. Now it makes sense. +1 –  Amr Dec 3 '13 at 20:25

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