Real Analysis - Convergence to Essential Supremum

[joypav]

Problem:
Let $\left(X, M, \mu\right)$ be a probability space. Suppose $f \in L^\infty\left(\mu\right)$ and $\left| \left| f \right| \right|_\infty > 0$. Prove that
$lim_{n \rightarrow \infty} \frac{\int_{X}^{}\left| f \right|^{n+1} \,d\mu}{\int_{X}^{}\left| f \right|^{n} \,d\mu} = \left| \left| f \right| \right|_\infty$

Proof:
So... I'm not really sure how to approach this problem.
$\left(X, M, \mu\right)$ is a probability space $\implies \mu\left(X\right) = 1$
Which then, $\left| \left| f \right| \right|_p \leq \left| \left| f \right| \right|_\infty$ for all $p$.

I thought maybe to just use our usual definition for convergence? Meaning,
$\forall \epsilon>0, \exists \delta>0, n>\delta \implies \left|\frac{\int_{X}^{}\left| f \right|^{n+1} \,d\mu}{\int_{X}^{}\left| f \right|^{n} \,d\mu} - \left| \left| f \right| \right|_\infty \right| \leq \epsilon$

But again, I'm not seeing what the next step would be... if someone could give some help I would really appreciate it!

 

[Euge]

Hello joypav,

If $\alpha_n = \int_X \lvert f\rvert^n\, d\mu$, then using the fact $\lvert f(x)\rvert \le \|f\|_\infty$ a.e., one finds $\alpha_{n+1} \le \|f\|_\infty \alpha_n$, forcing $\limsup_{n\to \infty} \alpha_{n+1}/\alpha_n \le \|f\|_\infty$. On the other hand, if $\varepsilon \in (0, \|f\|_\infty)$, the set $A := \{x\in X : |f(x)| > \|f\|_\infty - \varepsilon\}$ has finite positive measure. Using Hölder's inequality, show that $$\left(\frac{1}{\mu(A)} \int_A \lvert f\vert^{n+1}\, d\mu\right)^{1/(n+1)} \ge \left(\frac{1}{\mu(A)}\int_A \lvert f\vert^n\, d\mu\right)^{1/n}$$ and consequently $$\alpha_{n+1} \ge \int_A \lvert f\rvert^{n+1}\, d\mu \ge \mu(A)^{-1/n}(\|f\|_\infty - \varepsilon)^{n+1}$$ Since $\alpha_n \le \|f\|_\infty^n$, then $\alpha_{n+1}/\alpha_n \ge \mu(A)^{-1/n}(\|f\|_\infty - \varepsilon)$. Taking inferior limits as $n \to \infty$ results in $$\liminf_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \ge \|f\|_\infty-\varepsilon$$As $\epsilon$ was arbitrary, $\liminf_{n\to \infty} \alpha_{n+1}/\alpha_n \ge \|f\|_\infty$. Hence $\alpha_{n+1}/\alpha_n$ converges to $\|f\|_\infty$.

 

[math771]

Hi there! I think you're on the right track with using the definition of convergence. Here's how I would approach this problem:

We know that $\left| \left| f \right| \right|_\infty > 0$, which means that there exists some $x_0 \in X$ such that $\left| f\left(x_0\right) \right| > 0$. Now, let's consider the sequence of functions $g_n = \left| f \right|^n$. We have:
$$
\lim_{n \rightarrow \infty} g_n\left(x_0\right) = \lim_{n \rightarrow \infty} \left| f\left(x_0\right) \right|^n = 0
$$
since $\left| f\left(x_0\right) \right| < 1$. This means that $g_n$ converges pointwise to the function $g = 0$.

Now, we can use the dominated convergence theorem to show that the sequence $\frac{\int_{X}^{} g_{n+1} \,d\mu}{\int_{X}^{} g_{n} \,d\mu}$ converges to $\frac{\int_{X}^{} g \,d\mu}{\int_{X}^{} g \,d\mu} = 0$. Since $\left| \left| f \right| \right|_\infty > 0$, we know that $\left| f \right|^n$ is dominated by $g_n$ for all $n$, so we can use the theorem.

This means that for any $\epsilon > 0$, there exists some $N$ such that for all $n > N$, we have:
$$
\left| \frac{\int_{X}^{} \left| f \right|^{n+1} \,d\mu}{\int_{X}^{} \left| f \right|^{n} \,d\mu} - 0 \right| < \epsilon
$$
which is equivalent to:
$$
\left| \frac{\int_{X}^{} \left| f \right|^{n+1} \,d\mu}{\int_{X}^{} \left| f \right|^{n} \,