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 $\infty$.


  • 1
    $\begingroup$ 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. $\endgroup$ – kkc Mar 22 at 21:28
  • $\begingroup$ @kkc why $\frac{f}{1+ |f|}$ is Lebesgue integrable? $\endgroup$ – user 242964 Mar 22 at 21:39
  • 6
    $\begingroup$ Because $$\left|\frac {f^n} {1+n|f|}\right|\leq |f|^n\leq|f|$$ In fact, you could take $f$ as your $g$. $\endgroup$ – 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 $\left\lvert f_n(x)\right\rvert\leqslant \left\lvert f(x)\right\rvert/n$.


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