$\lim_{\epsilon \searrow 0}\frac{1}{\epsilon} \int_{[0,1]}(f^{\epsilon}-1)dm=\int_{[0,1]}log f dm$

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!

• Is the integrand on the left hand side $(f^\varepsilon-1)/\varepsilon$ or why do you consider that function? – Brandon Jan 20 '14 at 4:29
• Sorry, I'm correcting it – user120005 Jan 20 '14 at 4:30
• If there is no restriction on $f$ being $L^1$ then there's no way you can find a bounding function. It's got to be, somehow, by the monotone convergence theorem that you pass the limit through. – abnry Jan 20 '14 at 5:02
• f is assumed to be in $L^1([0,1],m)$. – user120005 Jan 20 '14 at 5:08

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.)
• Isn't the derivative $(\frac{log(f)}{\epsilon}-\frac{1}{\epsilon^2})f^{\epsilon}+\frac{1}{\epsilon^2}$? In this case I couldn't show it is decreasing. If I can show this I see what you mean for the rest. Thanks a lot! – user120005 Jan 20 '14 at 21:11
• Whoops, you are right. In this case it is increasing as $\epsilon \to 0$ and you should straight up be able to use the MCT. – abnry Jan 20 '14 at 21:16