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Let$(X,mathcal{M},mu)$ be measure space, $f$ is non-negative integrable function on $X$, and $int_X fdmu=c,0<c<infty$.Then calculate the following:

$$limlimits_{nto infty}int_X n logleft(1+left(frac{f}{n} right)^{alpha} right)dmu$$

depending on positive $alpha$.

(1). When $f>0$, calculate $limlimits_{nto infty}n logleft(1+left(frac{f}{n} right)^{alpha} right)$.

(2). When $0<alpha<1$, use Fatou's Lemma.

(3). When $alpha=1$, Using $xge 0$, then $log(1+x)le x$, then use Lebesgue convergence theorem.

(4). When $alpha>1$, Using $xge 0$, then $1+x^{alpha} le (1+x)^{alpha}$.

My Solution:

(1). $nlog left(1+left(frac{f}{n}right)^{alpha} right) = log(1+frac{f^alpha}{n^alpha})^n=logsqrt[alpha]{left(1+dfrac{f^alpha}{n^alpha}right)^{n^alpha}} to logsqrt[alpha]{e^{f^alpha}}=f$ as $nto infty$

(2). I don't know the answer.

(3). When $alpha=1$, $|nlog(1+f/n)|le n(f/n)=f$, using domainted convergence theorem.
$limlimits_{nto infty}int_X nlog(1+(frac{f}{n})^{alpha})dmu=int_X fdmu=c$

(4). As (3), $|nlog(1+(f/n)^alpha)|<nlog(1+f/n)^{alpha}=log(1+f/n)^{nalpha}le falpha$, as $nto infty$.

Then $limlimits_{nto infty}int_X nlog(1+(frac{f}{n})^{alpha})dmu=int_X alpha fdmu=alpha c$.

Fatou's Lemma says that, if $left(f_{n}right)$ is a sequence of
positive measurable functions, then
$$
int_{X}liminf f_{n}dmuleqliminfint_{X}f_{n}dmu.
$$
It is not an equality, but it does not require monotonicity nor domination.
Hence if $lim f_{n}=+infty$, then the two members of the inequality
are equal to $+infty$.

Here, If $alpha<1$, then $nlogleft(1+left(frac{f}{n}right)^{alpha}right)rightarrow+infty$.
Hence, by Fatou's lemma, $int_{X}nlogleft(1+left(frac{f}{n}right)^{alpha}right)dmurightarrow+infty$.