Question. Let $(X, M, m)$ be an arbitrary measure space. Let $f \in L^1_m(X)$. Then $$\lim_{n \rightarrow \infty} \int_X n \log \left( 1 + \frac{|f(x)|^2}{n} \right) \,dm(x) = 0.$$
Solution.
Observe that $$n \log \left( 1 + \frac{|f(x)|^2}{n} \right) \leq n \log \left( 1 + \frac{|f(x)|}{n} \right)^2 = 2n \log \left( 1 + \frac{|f(x)|}{n} \right) \leq 2n \frac{|f(x)|}{n} = 2|f(x)|.$$
Then $n \log \left( 1 + \frac{|f(x)|^2}{n} \right)$ is dominated by $2|f(x)|$.
Moreover, we have
$$\lim_{n \rightarrow \infty} \frac{\log \left( 1 + \frac{|f(x)|^2}{n} \right)}{1/n} = \frac{0}{0}.$$
So we apply L'Hopital's Rule and we obtain:
$$\lim_{n \rightarrow \infty} \frac{\log \left( 1 + \frac{|f(x)|^2}{n} \right)}{1/n} = \lim_{n \rightarrow \infty} \frac{\frac{-|f(x)|^2}{n^2} (-n^2)}{ 1 + \frac{|f(x)|^2}{n} } = |f(x)|^2.$$
Then by Lebesgue Dominated Convergence Theorem, we conclude that
$$\lim_{n \rightarrow \infty} \int_X n \log \left( 1 + \frac{|f(x)|^2}{n} \right) \,dm(x) = \int_X |f(x)|^2 \,dm(x).$$
Where is the proof wrong? Thanks.
Added Later. I found the mistake: $n \log \left( 1 + \frac{|f(x)|^2}{n} \right) \leq n \log \left( 1 + \frac{|f(x)|}{n} \right)^2$ is not true for all $n$. As @did suggested, dominating function should be $|f|^2$.
Let $g_n=n\log(1+|f|^2/n)$, then $\log(1+t)\leqslant t$ for every $t$, hence $g_n\leqslant|f|^2$. If $f$ is in $L^2$, this shows that $(g_n)$ is dominated. Since $g_n\to|f|^2$ pointwise, one gets $\int\limits_Xg_n\to\int\limits_X|f|^2$.
If $f$ is not in $L^2$, let $A(C)=\{|f|\leqslant C\}$ for some $C\gt0$, then $g_n\geqslant n\log(1+|f|^2/n)\mathbf 1_{A(C)}$. Furthermore, $\log(1+t)\geqslant t\log2$ for every $t\leqslant1$ hence, for every $n\geqslant C^2$, $g_n\geqslant(\log2)|f|^2\mathbf 1_{A(C)}$ and $\liminf\limits_{n\to\infty}\int\limits_Xg_n\geqslant\int\limits_{A(C)}(\log2)|f|^2$. This holds for every $C$ and $\int\limits_{A(C)}|f|^2$ can be made as large as one wants hence $\int\limits_Xg_n\to+\infty$.