If $|f| < 1$, compute $\lim_{n\to \infty} \int \left ( \frac{f ^n}{1 + n |f|} \right )\, d\mu.$

Let $$f: X \to \mathbb C$$ be integrable in a measure space $$(X, \mathfrak M, \mu)$$, i.e. $$\int |f| \, d\mu < \infty$$. Suppose that $$|f(x)| \leq 1$$ for all $$x \in X$$. How can one compute the limit $$\lim_{n\to \infty} \int \left ( \frac{f ^n}{1 + n |f|} \right )\, d\mu \quad ?$$

My attempt:

I want to find a Lebesgue integrable $$g$$ that dominates the sequence $$f_n = \frac{f ^n}{1 + n |f|}$$ and, then, I would conclude that

$$\lim_{n\to \infty} \int \left ( \frac{f ^n}{1 + n |f|} \right ) d\mu = \int \left ( \lim \frac{f ^n}{1 + n |f|} \right ) d\mu = 0,$$ since $$|f| < 1$$ implies $$\lim \frac{f ^n}{1 + n |f|} = 0$$.

My problem is in find such function $$g$$, I can see that $$|f_n| < 1$$ for each $$n$$, however the function $$g = 1$$ does not need to be Lebesgue integrable since $$\mu(X)$$ maybe $$\infty$$.

Help?

• Isn't your sequence monotonically decreasing? In that case, you could use $\frac{f^1}{1+|f|}$ as your $g$ and then apply dominated convergence theorem. – kkc Mar 22 at 21:28
• @kkc why $\frac{f}{1+ |f|}$ is Lebesgue integrable? – user 242964 Mar 22 at 21:39
• Because $$\left|\frac {f^n} {1+n|f|}\right|\leq |f|^n\leq|f|$$ In fact, you could take $f$ as your $g$. – Stefan Lafon Mar 22 at 21:49

For all $$x\in [0,1]$$ and $$n\geqslant 2$$, $$\frac{t^n}{1+nt}=\frac{t}{1+nt}t\cdot t^{n-2}\leqslant \frac{t}{1+nt}t\leqslant \frac 1nt$$ hence applying the previous inequality with $$t=\left\lvert f(x)\right\rvert$$ gives $$\left\lvert f_n(x)\right\rvert\leqslant \left\lvert f(x)\right\rvert/n$$.