I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 4.5 Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. Let $p\in [1, \infty)$ and $q \in [1, \infty]$.
- Prove that the set $$ F:=\{f \in L^p (\Omega) \cap L^q (\Omega) : \|f\|_q \le 1\} $$ is closed in $L^p (\Omega)$.
- Let $f_n \in L^p (\Omega) \cap L^q (\Omega)$ and $f \in L^p (\Omega)$. Assume $f_n \to f$ in $L^p$ and $\|f_n\|_q \le C$ for all $n$. Then $f \in L^r (\Omega)$ for all $r$ between $p$ and $q$ with $r \neq q$.
Could you confirm if my below attempt is correct?
Proof
Let $f_n \in F$ and $f \in L^p (\Omega)$ such that $f_n \to f$ in $L^p$. There is a sub-sequence $\varphi$ of $\mathbb N$ such that $f_{\varphi (n)} \to f$ $\mu$-a.e. First, we consider the case $q=\infty$. Then $|f_{\varphi (n)}| \le 1$ $\mu$-a.e. for all $n$. By point-wise convergence, $|f| \le 1$ $\mu$-a.e. Hence $\|f\|_{L^\infty} \le 1$ and thus $f \in F$. Second, we consider the case $p \neq \infty$. By Fatou's lemma, $$ \int |f|^{q} \le \liminf_n \int |f_{\varphi (n)}|^q \le 1. $$ Hence $\|f\|_q \le 1$ and thus $f \in F$. This completes the proof.
Let $r$ be between $p$ and $q$ such that $r \neq q$. There is $\alpha \in (0, 1]$ such that $$ \frac{1}{r} = \frac{\alpha}{p} + \frac{1-\alpha}{q}. $$ By interpolation inequality, $$ \|f_n-f_m\|_r \le \|f_n-f_m\|_p^\alpha \cdot \|f_n-f_m\|_q^{1-\alpha} \le (2C)^{1-\alpha} \|f_n-f_m\|_p^\alpha. \tag{1} $$ There is a sub-sequence $\varphi$ of $\mathbb N$ such that $f_{\varphi (n)} \to f$ $\mu$-a.e. By $(1)$, the sequence $(f_{\varphi (n)})$ is Cauchy in $L^r (\Omega)$. By definition of Bochner integral, $f_{\varphi (n)} \to f$ in $L^r$. Because $L^r (\Omega)$ is complete, $f_n \to f$ in $L^r$. Notice that Bochner integral coincides with Lebesgue integral. This completes the proof.
