The stated problem is as follows: Let $f$ be non-negative measurable function defined on a finite measure space. Let ess sup $f = M$ (hence, $f \in L^{\infty})$. Prove that $$ \lim_{n \rightarrow \infty} \frac{\int_X f^{n+1} \ d\mu}{\int_X f^n \ d\mu} = M $$
Now, my original thought (and what prompted this question) was to use the fact that $\lim_{n \rightarrow \infty} ||f||_n = ||f||_{\infty} = M$. Then, treating the numerator as $||f||_{n+1}^{n+1}$ and the denominator as $||f||_n^n$, I was $\textit{hoping}$ that I could take the limit just as that and arrive at something like
$$ \lim_{n \rightarrow \infty} \frac{\int_X f^{n+1} \ d\mu}{\int_X f^n \ d\mu} = \lim_{n \rightarrow \infty} \frac{||f||_{\infty}^{n+1}}{||f||_{\infty}^n} = \lim_{n \rightarrow \infty} \frac{M^{n+1}}{M^n} = M $$ Now, the problem, I think, is taking the limit of the norm while leaving the exponent as $n+1$ and $n$ respectively. What I imagine should be happening is that I would have to take both limits simultaneously. In my soul, I feel like that is correct, but I'm not sure why. My first question here isn't related to the question prompt specifically, but in general, why am I not allowed to take the limit of just a part of the function? Would I ever be able to write something like $$ \lim_{n \rightarrow \infty} \ \lim_{k \rightarrow \infty} \ ||f||_n^k $$ and have it make sense in this context? Or in any context?
My second question is about the problem itself. Assuming that I am correct in that I can't break up the limit like that, my second step was to consider L'Hospital. However, just because $f \in L^{\infty}$ doesn't tell me that it is differentiable. I'm not sure there is really anything else I can do with this idea.
The last thing I tried to to show dual inequalities. The first way is easy, we have
$$ \lim_{n \rightarrow \infty} \frac{\int_X f^{n+1} \ d\mu}{\int_X f^n \ d\mu} \leq \frac{\int_X M^{n+1} d\mu}{\int_X M^n \ d\mu} = \frac{M^{n+1} \mu(X)}{M^n \mu(X)} = M $$ This one takes into account the fact that $\mu(X) < \infty$ (and I always feel better when I use every part of the prompt. Why would it be there if it wasn't necessary? Unfortunately, showing the other direction has been quite the hassle. Any tips or suggestions on where to go, relevant theorems, etc? I'm working out of Rudin and Royden, though this question is not from either of them (I don't think). Thanks in advance!
Let $I_n = \int_X f^n \, d\mu$. Then it is clear that $I_{n+1} \leq M I_n$ and thus $\limsup_{n\to\infty} (I_{n+1}/I_n) \leq M$. So it suffices to prove that $\liminf_{n\to\infty} (I_{n+1}/I_n) \geq M$.
To this end, fix $m \in (0, M)$ and write $E = \{ x \in X : f(x) \geq m \}$. By the assumption, $\mu(E) > 0$. Moreover, since $f \geq m \mathbf{1}_E$, we have
\begin{align*} I_{n+1} &\geq \int_{E} m f^n \, d\mu \\ &= \int_{X} m f^n \, d\mu - \int_{X\setminus E} m f^n \, d\mu \\ &\geq m I_n - m^{n+1}\mu(X\setminus E) \end{align*}
Since we know that $I_n^{1/n} \to M$ as $n\to\infty$, it follows that $m^{n+1}/I_n \to 0$ as $n \to \infty$. Consequently
$$ \liminf_{n\to\infty} \frac{I_{n+1}}{I_n} \geq \lim_{n\to\infty} \left( m - \frac{m^{n+1}\mu(X\setminus E)}{I_n} \right) = m. $$
Since this is true for all $m \in (0,M)$, we can take $m \uparrow M$ and the conclusion follows.