Let $(X,\Sigma,\mu)$ a measure space and $f: X \to [0,+\infty]$ and integrable function with positive integral and $a>1$
Prove that $\int_X n\log{(1+\frac{f^a(x)}{n^a})}d\mu(x) \to^{n \to +\infty}0$
Clearly we are interested at $x \in X$ where $f(x)>0$
Let $f(x)>0$. Then it is easy to see that $f_n(x):=n\log{(1+\frac{f^a(x)}{n^a})} \to 0$
The problem is that i did not come up with something to dominate $f_n(x)$ except $f^a$ which does not tell so much for $a>1$
We do not know if $f^a$ is integrable.
Also i believe that this statement may not hold in general but i could not come up with a counterexample.
Can some give me a hint for this problem or giving a counterexample in case the statement is wrong?
Thank you in advance.
For any $\alpha>1$ there exists a constant $C>0$ such that
$$1+x^{\alpha} \leq e^{Cx} \quad \text{for all $x \geq 0$}.$$
This implies
$$\log(1+x^{\alpha}) \leq Cx, \qquad x \geq 0.$$
Hence,
$$0 \leq n \log \left(1 +\frac{f(x)^{\alpha}}{n^{\alpha}} \right)\leq Cn \frac{f(x)}{n}=C f(x) \in L^1(\mu).$$
Since
$$n \log \left(1 +\frac{f(x)^{\alpha}}{n^{\alpha}} \right) \to 0$$
it follows from the dominated convergence theorem that
$$\int_X n \log \left(1 +\frac{f(x)^{\alpha}}{n^{\alpha}}\right) \, d\mu(x) \xrightarrow[]{n \to \infty} 0.$$