Let $f>0$ be a Lebesgue integrable function on [0,1]. Show that
$\lim_{\epsilon \searrow 0}\frac{1}{\epsilon} \int_{[0,1]}(f^{\epsilon}-1)dm=\int_{[0,1]}log f dm$
Here $m$ denotes Lebesgue measure. HINT: Decompose f (or log f) into two parts.
I can show that $\frac{f^\epsilon -1}{\epsilon}$ is Lebesgue integrable on [0,1] for every $\epsilon\in(0,1)$. But I cannot bound $\frac{f^\epsilon -1}{\epsilon}$ by an integrable function on [0,1] to pass the limit inside. If I can show that I can pass the limit inside then I'll be done since $\frac{f^\epsilon -1}{\epsilon}\rightarrow log f$ as $\epsilon$ decreasing to 0.
Thanks for any help!
With the condition that $f$ is $L^1$ I have a solution.
Consider the set $f \geq 1$: Let $\epsilon < 1/2$. By the MVT there is a $0<c<\epsilon$ such that $$|(f^\epsilon-1)/\epsilon| = |\log(f)f^c| \leq C |f^{1/2} f^{\epsilon}| \leq |f|$$ on our set. This is because $|log(f)| \leq C|f|^{1/2}$ for some $C$ when $|f|\geq 1$. Thus the integrand is dominated and we can pass the limit through.
Consider the set $f < 1$: Take a derivative of the integrand getting $$\left( \frac{\log(f)}{\epsilon} - \frac{1}{\epsilon^2} \right)f^\epsilon - \frac{1}{\epsilon^2},$$ and conclude it is decreasing. Now use the function $g_\epsilon = (f^1-1)/1-(f^\epsilon-1)/\epsilon$ to apply the monotone convergence test and pass the limit through. (I am omitting details here.)