0
$\begingroup$

Let be $f:[0, 1] \rightarrow \mathbb{R}$ a function such that $\int_{[0,1]}|f|<+\infty$. Calculate: $$\lim_{n \rightarrow +\infty}\frac1n\int_0^1log(1+e^{nf(x)})dx$$ I wanted to either use the dominated convergence theorem, but I dont know how to find a function that is upperbound Lebesgue integrable function, or the monotony convergence theorem, but I dont know how to prove the monotony. Any help?

$\endgroup$
2
  • 1
    $\begingroup$ I think that it really depend on $f$. For example, if $f\equiv 0$, then the limit is 0, however, if $f\equiv 1$, the limit will be $\infty $. $\endgroup$
    – Surb
    Mar 11, 2022 at 14:29
  • $\begingroup$ How can I calculate the integral as function of $f$? $\endgroup$ Mar 11, 2022 at 15:00

1 Answer 1

0
$\begingroup$

First, consider the following:

Where $f(x)>0$, the expression $\ln(1+e^{nf(x)})$ is almost $\;nf(x)$.

Where $f(x)=0$, you get $\ln(1+e^{nf(x)})=\ln(2)$.

Where $f(x)\leq0$ instead, $\ln(1+e^{nf(x)})$ gets close to $0$.

This helps us to estimate $\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x$.

$\\$

Let $S=\{t\in[0,1]:f(t)>0\}$.

If $S$ has positive measure, then automatically $\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x=+\infty$.

So we may as well now assume $S$ is a null set.

$\\$

Let $T=\{t\in[0,1]:f(t)=0\}$.

You can see that $\displaystyle\int_T\ln(1+e^{nf(x)})\;\text{d}x=\ln(2)\mu(T)$ for all $n\in\mathbb{Z}^+$.

$\\$

Let $U=\{t\in[0,1]:f(t)<0\}$

As for $\displaystyle\int_U\ln(1+e^{nf(x)})\;\text{d}x$, how do we know that it tends to $0$ as $n$ tends to infinity?

This is an application of a standard trick: look at $U_r=\{t\in[0,1]:f(t)<-r\}$ where $r>0$.

Obviously for a fixed $r$,

$\displaystyle\int_{U_r}\ln(1+e^{nf(x)})\;\text{d}x \leq\mu(U_r)\ln(1+e^{-rn})\xrightarrow[n \to \infty]{} 0$.

So, given that $\displaystyle\int_U\ln(1+e^{nf(x)})\;\text{d}x=\int_{U_r}\ln(1+e^{nf(x)})\;\text{d}x+\int_{U\setminus U_r}\ln(1+e^{nf(x)})\;\text{d}x$,

by varying $r$, we can make the second summand arbitrarily small, because $\ln(1+e^{nf(x)})<1$ on $x\in U$, and $\displaystyle\lim_{r\to0^+}\mu(U\setminus U_r)=0$;

varying $n$ after we vary $r$, we can make the first summand tend to $0$.

So $\displaystyle\lim_{n\to\infty}\int_U\ln(1+e^{nf(x)})\;\text{d}x=0$.

$\\$

In conclusion,

$\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x=\begin{cases} +\infty & \text{if } \mu(S)>0 \\ \ln(2)\mu(T) & \text{if } \mu(S)=0 \end{cases}$

$\endgroup$
2
  • $\begingroup$ So, there is no need to use the monotone convergence theorem or the dominated convergence theorem? Also can you please explain with more details the first part? Why the fact that $log(1+e^{nf(x)})$ is almost $nf(x)$ implies that the integral of $log(1+e^{nf(x)})$ diverges to $+\infty$? $\endgroup$ Mar 11, 2022 at 15:26
  • $\begingroup$ If $S$ has positive measure, then there is some $r>0$ such that $S_r=\{t\in[0,1]:f(t)>r\}$ has positive measure. For such a value of $r$, you get $\mu(S_r)\ln(1+e^{nr})\to\infty$ $\endgroup$ Mar 11, 2022 at 15:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .