13

Suppose that $\|f\|_p &lt \infty$ for all $1\leq p &lt p'$, 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.

J. M. ain't a mathematician
  • 75,051
Hawii
  • 1,279

1 Answers1

17

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 $|f|$).

Zarrax
  • 44,950
  • 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 06:46
  • 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 03 '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 03 '13 at 20:25