Let $I=(a, b)$ be an open interval in $\mathbb R$. I would like to verify Remark 11 at page 214 in Brezis' Functional Analysis
Let $I$ be bounded and $p,q \in [1, \infty]$. By Theorems 8.2 and 8.8, $$ [u] := \|u'\|_p + \|u\|_q $$ is an equivalent norm of $W^{1, p} (I)$.
In below attempt, I don't use Theorems 8.2. There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
We have $\|u\|_{W^{1,p}} := \|u'\|_p + \|u\|_p$. By Theorem 8.8 in the same book, there exists a constant $C_1$ (depending only on $|I|$) such that $$ \|u\|_{\infty} \leq C_1\|u\|_{W^{1, p}} \quad \forall u \in W^{1, p}(I). $$
- We consider the case $1 \le p \le q \le \infty$. By interpolation inequality, there exist constant $C_2$ (depending only on $p,q,|I|$) such that for all $u \in W^{1, p}(I)$ we have $$ \|u\|_p \le C_2 \|u\|_q \quad \text{and} \quad \|u\|_q \le \|u\|_p^{p/q} \|u\|_\infty^{(q-p)/q}. $$
Then for all $u \in W^{1, p}(I)$ we have $$ \begin{align*} \|u\|_q &\le \|u\|_p^{p/q} (C_1\|u\|_{W^{1, p}})^{(q-p)/q} \\ & \le C_1^{(q-p)/q} \|u\|_{W^{1,p}}, \end{align*} $$ and thus $$ [u] \le (1+C_1^{(q-p)/q}) \|u\|_{W^{1,p}}. $$
On the other hand, $$ \begin{align*} \|u\|_{W^{1,p}} &\le \|u'\|_p + C_2 \|u\|_q \\ &\le (1+C_2) [u]. \end{align*} $$
- We consider the case $1 \le q < p \le \infty$. By interpolation inequality, there exist constant $C_3$ (depending only on $p,q,|I|$) such that for all $u \in W^{1, p}(I)$ we have $$ \|u\|_q \le C_2 \|u\|_p \quad \text{and} \quad \|u\|_p \le \|u\|_q^{q/p} \|u\|_\infty^{(p-q)/p}. $$
We precede as previously and complete the proof.