I'm trying to solve an exercise in Brezis' Functional Analysis
Let $(\Omega, \mathcal F, \mu)$ be a measure space with $\mu (\Omega) < \infty$. Fix $p \in [1, \infty)$ and let $E$ be a closed subspace of $L^p (\Omega)$ such that $E \subset L^\infty (\Omega)$.
- Prove that there is a constant $C$ such that $$ \|u\|_\infty \le C \|u\|_p \quad \forall u \in E. $$
- Prove that there is a constant $M$ such that $$ \|u\|_\infty \le M \|u\|_2 \quad \forall u \in E. $$
Could you have a check on my attempt?
- Fix $q \in [p, \infty]$. Because $\mu (\Omega) < \infty$, we get $$ \| u \|_p \le (\mu (\Omega))^{\frac{1}{p} - \frac{1}{q}} \|u\|_q \quad \forall u \in L^q (\Omega). \qquad \qquad (\star) $$
We will prove that $E$ is closed in $L^q (\Omega)$. Let $(u_n) \subset E$ and $u \in L^q (\Omega)$ such that $\|u_n-u\|_q \to 0$. By $(\star)$, $\|u_n-u\|_p \to 0$. On the other hand, $E$ is closed in $L^p (\Omega)$. So $u \in E$. It follows that $(E, \| \cdot\|_p)$ and $(E, \| \cdot\|_q)$ are Banach spaces. It follows from $(\star)$ and Corollary 2.8 in the same book that $\| \cdot\|_p$ and $\| \cdot\|_q$ are equivalent norms on $E$. The claim then follows by picking $q = \infty$.
Corollary 2.8. Let $E$ be a vector space provided with two norms, ||$_1$ and ||$_2$. Assume that $E$ is a Banach space for both norms and that there exists a constant $C \geq 0$ such that $$ \|x\|_2 \leq C\|x\|_1 \quad \forall x \in E . $$ Then the two norms are equivalent, i.e., there is a constant $c>0$ such that $$ \|x\|_1 \leq c\|x\|_2 \quad \forall x \in E . $$
- If $2 \ge p$, then the claim follows from (1.). Now we assume $2 < p$. We fix some $q \in (p, \infty)$. Then $2 < p < q < \infty$. Then there is $\alpha \in (0, 1)$ such that $\frac{1}{p} = \frac{\alpha}{2} + \frac{1-\alpha}{q}$. By interpolation inequality, $$ \|u\|_p \le \|u\|_2^\alpha \|u\|_q^{1-\alpha} \quad \forall u \in E. $$
By $(\star)$, $\|u\|_q \le (\mu (\Omega))^{\frac{1}{q}} \|u\|_\infty$ for all $u \in E$. By (1.), $\|u\|_\infty \le C \|u\|_p$ for all $u \in E$. Then $$ \|u\|_p \le \|u\|_2^\alpha (C(\mu (\Omega))^{\frac{1}{q}} \|u\|_p)^{1-\alpha} \quad \forall u \in E. $$
Then $$ \|u\|^\alpha_p \le \|u\|_2^\alpha (C(\mu (\Omega))^{\frac{1}{q}})^{1-\alpha} \quad \forall u \in E. $$
Then $$ \|u\|_p \le \|u\|_2 (C(\mu (\Omega))^{\frac{1}{q}})^{(1-\alpha)/\alpha} \quad \forall u \in E. $$
Then $$ \|u\|_\infty \le C \|u\|_p \le C (C(\mu (\Omega))^{\frac{1}{q}})^{(1-\alpha)/\alpha} \|u\|_2 \quad \forall u \in E. $$
This completes the proof.
