Background
Let $(\Omega, \mathcal{F}, P)$ be a probability space (or a finite measure space). Then, it is well-known that, for a fixed function $f$ with $\lVert f \rVert_\infty := \mathrm{ess.sup}_{x} \lvert f(x) \rvert < \infty$, we have $$ \lVert f \rVert_p := \left( \int |f|^p \, dP \right)^{1/p} \ \overset{p \to \infty}{\rightarrow} \ \lVert f \rVert_\infty \text{.} $$ (See, e.g., this question.)
I'm curious what happens if $f$ depends on $p$. More precisely, suppose we have a sequence of functions $\{f_p\}_{p=1}^\infty \subset L^\infty$ that converges to $f \in L^\infty$ a.s. Do we always have $ \lim_{p \to \infty} \lVert f_p \rVert_p = \lVert f \rVert_\infty $? If not always, under what conditions?
What I've tried so far
Maybe one of the possible sufficient conditions is that $|f_p| \nearrow |f|$ almost surely. Under this condition, we can prove $\lim_{p \to \infty} \lVert f_p \rVert_p = \lVert f \rVert_\infty$ as follows.
By the assumption, for any $p = 1, 2, \ldots$, we have $|f_p| \leq |f| \leq \lVert f \rVert_\infty$ a.s. So the monotonicity of integral implies $\int |f_p|^p \, dP \leq \lVert f \rVert_\infty^p$, which proves $\limsup_{p \to \infty} \lVert f_p \rVert_p \leq \lVert f \rVert_\infty$.
Now we are going to prove $\liminf_{p \to \infty} \lVert f_p \rVert_p \geq \lVert f \rVert_\infty$. To this end, fix any $\epsilon > 0$ and define the events $A_p(\epsilon) := \{|f_p| > \lVert f \rVert_\infty - \epsilon\}$ for $p = 1, 2, \ldots$. Then, we have \begin{align} \int_\Omega |f_p|^p \, dP \geq \int_{A_p(\epsilon)} |f_p|^p \, dP \geq \int_{A_p(\epsilon)} (\lVert f \rVert_\infty - \epsilon)^p \, dP = P(A_p(\epsilon)) (\lVert f \rVert_\infty - \epsilon)^p \end{align} and therefore $$ \lVert f_p \rVert_p \geq P(A_p(\epsilon))^{1/p} (\lVert f \rVert_\infty - \epsilon) \text{.} $$ Now, $|f_p| \nearrow |f|$ implies $A_p(\epsilon) \nearrow A(\epsilon) := \{|f| > \lVert f \rVert_\infty - \epsilon\}$. Noting that $A(\epsilon)$ is a positive constant (by the definition of $\lVert f \rVert_\infty$), by the continuity of measure, we have $$ \lim_{p \to \infty} P(A_p(\epsilon))^{1/p} = P(A(\epsilon))^{0} = 1 \text{.} $$ Therefore, we have proven that $\liminf_{p \to \infty} \lVert f_p \rVert_p \geq \lVert f \rVert_\infty - \epsilon$. Since $\epsilon > 0$ can be taken arbitrarily small, we obtain $\liminf_{p \to \infty} \lVert f_p \rVert_p \geq \lVert f \rVert_\infty$ as desired.
Questions
So my questions are:
- Is the proof that I wrote above correct?
- Are there any other conditions under which we have $ \lim_{p \to \infty} \lVert f_p \rVert_p = \lVert f \rVert_\infty $?
- Are there any counterexamples?
