# How to prove $\lim_{n \rightarrow \infty } \frac{1}{n} \int_0^1 \ln \big( 1 + e^{nf(t)} \big) dt = \int_0^1 f^+(t) dt$

Let $f \in L^1[0,1]$ be a real valued Lebesgue integrable function on $[0,1]$. Prove that

$$\lim_{n \rightarrow \infty } \frac{1}{n} \int_0^1 \ln \big( 1 + e^{nf(t)} \big) dt = \int_0^1 f^+(t) dt \, ,$$ where $f^+(t) =\max (0,f)$.

I feel like I should apply the dominated convergence theorem somehow but I can't see a substitution that makes this work. Any suggestions, hints would be greatly appreciated.

• Consider that for $t$ where $f(t)$ is not zero but is positive, you can essentially ignore the $1+$ in the log. Where $f\leq0$, you can see that you get zero in the numerator. This is not rigorous, but maybe provides some helpful insight. – jdods Jul 21 '16 at 15:58
• Very helpful! Thank you! – user209663 Jul 21 '16 at 17:03

I leave to you the proof that $g_n(t)=\frac{1}{n}\log (1+\exp(nf(t)) \to f^+(t)$ for fixed $t$.

Now $\exp(nf(t))\leq \exp(n|f(t)|)$ and then $$0\leq g_n(t)\leq \frac{1}{n}\log (1+\exp(n|f(t)|)=|f(t)|+\frac{1}{n}\log (1+\exp(-n|f(t)|))$$ and then $$0\leq g_n(t) \leq |f(t)|+\log(2)=h(t)\in L^1$$

Now you can use the DCT.

Write $[0,1]= ([0,1]\cap\{f>0\})\cup ([0,1]\cap\{f\leq 0\})=A\cup B$.

Define $g_n(t)=ln(1+(e^{f(t)})^n)/n$. Over A, $g_n\to f(t)$ and over B $g_n\to 0$.

Apply BeppoLevy and mayorated, respectly.

• I think that over $A$, $g_n(t)\to f(t)$ – Svetoslav Jul 21 '16 at 16:32
• @Svetoslav yes! Tks – Martín Vacas Vignolo Jul 21 '16 at 16:36