I'm reading the book Real and Complex Analysis recently. And I encountered a question in the exercises of Chapter 3 that I can not solve for a long time. This question is about what the $L^p$ norm(not a norm when $p<1$) will become if we send p to zero.
Explicitly, Suppose $X$ is a measure space, $\mu$ is a probability measure on $X$. Define $$||f||_p = \left(\int_X|f|^pd\mu\right)^{1/p}$$for all $0<p<+\infty$ and complex measurable function $f$.
Assume that $||f||_r < \infty$ for some $r > 0$, and prove that $$\lim_{p\rightarrow0}||f||_p = \exp\left\{\int_X\log|f|d\mu\right\}$$if $\exp\{-\infty\}$ is defined to be $0$.
By using Jensen's Inequality and a conclusion that $|f|_r \leq |f|_s$ if $0<r<s<\infty$, I get half part of the conclusion:$$\lim_{p\rightarrow0}||f||_p \geq \exp\left\{\int_X\log|f|d\mu\right\}$$ Then it remains to prove $$\lim_{p\rightarrow0}||f||_p \leq \exp\left\{\int_X\log|f|d\mu\right\}$$ Or another form $$\lim_{p\rightarrow 0}\frac{1}{p}\left(\int_X\log|f|^pd\mu - \log\int_X|f|^pd\mu\right) \geq 0$$ Then by inequality $\log(1+x) < x$, I only need to prove $$\lim_{p\rightarrow 0}\frac{1}{p}\left(\int_X\log|f|^p-|f|^p + 1d\mu\right) \geq 0$$But the inequality seems to be wrong. Can this way work? Or are there any other solutions for this question?