- #1

joypav

- 151

- 0

**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!