5
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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 $\left\lvert f_n(x)\right\rvert\leqslant \left\lvert f(x)\right\rvert/n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.