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

$$\lim\limits_{n\to \infty}\int_X n \log\left(1+\left(\frac{f}{n} \right)^{\alpha} \right)d\mu$$

depending on positive $\alpha$.

(1). When $f>0$, calculate $\lim\limits_{n\to \infty}n \log\left(1+\left(\frac{f}{n} \right)^{\alpha} \right)$.

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

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

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

My Solution:

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

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

(3). When $\alpha=1$, $|n\log(1+f/n)|\le n(f/n)=f$, using domainted convergence theorem. $\lim\limits_{n\to \infty}\int_X n\log(1+(\frac{f}{n})^{\alpha})d\mu=\int_X fd\mu=c$

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

Then $\lim\limits_{n\to \infty}\int_X n\log(1+(\frac{f}{n})^{\alpha})d\mu=\int_X \alpha fd\mu=\alpha c$.

  • 1
    $\begingroup$ Your first equality in (1) isn't correct. The exponent $n^\alpha$ should be $n\alpha$. What about using L'Hospital's rule? $\endgroup$ – Umberto P. May 7 '13 at 13:52

Fatou's Lemma says that, if $\left(f_{n}\right)$ is a sequence of positive measurable functions, then $$ \int_{X}\liminf f_{n}d\mu\leq\liminf\int_{X}f_{n}d\mu. $$ 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 $n\log\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)\rightarrow+\infty$. Hence, by Fatou's lemma, $\int_{X}n\log\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)d\mu\rightarrow+\infty$.


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