Limit of $L^p$ norm when $p\to0$

Let ($\Omega$, $\cal{F}$, $\mu$) be a probability space and $f\in L^1(\Omega)$. Prove that

$$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu \right],$$

where $\exp[-\infty]=0$. To simplify the problem, we may assume $\log|f|\in L^1(\Omega).$

-
–  Byron Schmuland Jan 30 '13 at 23:31

Let $f(p):=\frac 1p\log\int_{\Omega}|f|^pd\mu-\int_{\Omega}\log|f|d\mu$.
Since $t\mapsto \log t$ is concave, by Jensen inequality we get $f(p)\geqslant 0$. Using the inequality $\ln(1+t)\leqslant t$ we have $$0\leqslant f(p)\leqslant \frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu.$$ Now the problem reduces to show that $\lim_{p\to 0}\frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu=0$. To see that, take a sequence $\{p_n\}$ which converges to $0$ and put $f_n(x):=\frac{|f(x)|^{p_n}-1}{p_n}-\log |f(x)|$. The sequence $\{f_n\}$ converges almost everywhere to $0$ and we have, if $t\geq 1$, $0<p<1$ $$\left|\frac{t^p-1}p\right|=\int_1^t s^{p-1}ds\leqslant t-1$$ since the map $s\mapsto s^{p-1}$ is decreasing, and if $0<t<1$ $$\left|\frac{t^p-1}p\right|=\int_t^1s^{p-1}ds\leqslant \int_t^1s^{-1}ds=-\log t$$ so denoting $A=\{x, |f(x)|\geqslant 1\}$, $$|f_n(x)|\leqslant (|f(x)|-1)\mathbf 1_A(x)-\log|f(x)|\mathbf 1_B(x)+\log|f(x)|,$$ which is integrable. We can conclude by the dominated convergence theorem.
A log is missing in $f(p)$. More importantly, the inequality $p^{-1}(t^p-1)-\log t\leqslant p(t-1-\log t)$ does not hold for $t$ in $(0,1)$. –  Did Mar 3 '12 at 11:06
@DidierPiau You are right, the fact that $\log t$ is negative led me to wrong bounds. I've done this exercise using dominated convergence theorem, and I thought it was possible to work without it. I believed I found a simpler solution, but I confused "simple" and "wrong". –  Davide Giraudo Mar 3 '12 at 11:42