How Does the Limit lim. $\Phi_u(p)$ Indicate $u \in L^{\infty}(\Omega)$?

Question

Question: Let $\Omega \subset \mathbb{R}^d$ be measurable with $|\Omega|<\infty$, and let $u: \Omega \rightarrow \mathbb{R}$ be measurable. Define $$ \Phi_u:[1, \infty) \rightarrow[0, \infty], \quad \Phi_u(p):=|\Omega|^{-\frac{1}{p}}\|u\|_{L^p(\Omega)} . $$

Prove the following:

It holds $u \in L^{\infty}(\Omega)$ if and only if the $\operatorname{limit} \lim _{p \rightarrow \infty} \Phi_u(p)$ is finite. In this case $$ \lim _{p \rightarrow \infty} \Phi_u(p)=\|u\|_{L^{\infty}(\Omega)} . $$

My solution

Part 1: $u \in L^{\infty}(\Omega) \Rightarrow \lim _{p \rightarrow \infty} \Phi_u(p)=\|u\|_{L^{\infty}(\Omega)}$

1. Suppose $u \in L^{\infty}(\Omega)$. This means there exists a bound $M$ such that $|u(x)| \leq M$ almost everywhere in $\Omega$.
2. Consider $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$. Since $|u(x)| \leq M$, we have $\|u\|_{L^p(\Omega)} \leq$ $\left(\int_{\Omega} M^p d x\right)^{1 / p}$, which simplifies to $M|\Omega|^{1 / p}$.
3. Thus, $\Phi_u(p) \leq|\Omega|^{-1 / p} \cdot M|\Omega|^{1 / p}=M$.
4. As $p \rightarrow \infty,|\Omega|^{1 / p}$ approaches 1, so $\lim _{p \rightarrow \infty} \Phi_u(p) \leq M$.
5. Conversely, for any $\epsilon>0$, there exists a set $E \subset \Omega$ with $|E|>0$ where $|u(x)| \geq$ $M-\epsilon$ (since $\|u\|_{L^{\infty}(\Omega)}=M$ ).
6. For such $E,\|u\|_{L^p(\Omega)}$ must be greater than $(M-\epsilon)|\Omega|^{1 / p}$, so $\Phi_u(p) \geq(M-\epsilon)$.
7. Thus, $\lim _{p \rightarrow \infty} \Phi_u(p) \geq M-\epsilon$. Since $\epsilon$ is arbitrary, $\lim _{p \rightarrow \infty} \Phi_u(p) \geq M$.
8. Combining, we have $\lim _{p \rightarrow \infty} \Phi_u(p)=M=\|u\|_{L^{\infty}(\Omega)}$.

Part 2: $\lim _{p \rightarrow \infty} \Phi_u(p)$ finite $\Rightarrow u \in L^{\infty}(\Omega)$

1. Assume $\lim _{p \rightarrow \infty} \Phi_u(p)=L$ is finite.
2. By definition, $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$.
3. For any $p,\|u\|_{L^p(\Omega)} \leq L|\Omega|^{1 / p}$.
4. As $p \rightarrow \infty,|\Omega|^{1 / p}$ approaches 1, and so $\|u\|_{L^p(\Omega)} \leq L$ for all $p$.
5. This implies that $|u(x)|$ cannot exceed $L$ almost everywhere in $\Omega$, otherwise $\|u\|_{L^p(\Omega)}$ would exceed $L$ for some $p$.
6. Therefore, $u \in L^{\infty}(\Omega)$ with $\|u\|_{L^{\infty}(\Omega)} \leq L$.

-- I would love your proofreading and suggestion! :)