Sobolev inequality involving summing from $j = 0$ to $m - 2$, exists constant

Question

Let $I = (0, 1)$ and $1 \le q < \infty$. For all $\epsilon > 0$, does there exist $C = C(\epsilon, q)$ such that$$\|D^{(m - 1)}u\|_{L^q(I)} + \sum_{j = 0}^{m - 2} \|D^ju\|_{L^\infty(I)} \le \epsilon\|D^mu\|_{L^1(I)} + C\|u\|_{L^1(I)}$$for all $u \in W^{m, 1}(I)$?

Answer 1

We argue by contradiction. After scaling, we have a sequence $(u_n)_{n=1}^\infty \subset W^{m,1}(I)$ with $$ \| D^{(m-1)} u_n \|_q + \sum_{j=0}^{m-2} \| D^j u_n \|_\infty = 1 > \varepsilon \| D^m u_n \|_1 + n \| u_n \|_1. $$ We have $(D^{m-1}u_n)$ is a bounded subset of $W^{1,1}(I)$ from $$ \| D^{m-1} u_n \|_1 \leq \| D^{m-1} u_n \|_q \leq 1, \quad \| D^{m} u_n \|_1 < \varepsilon^{-1}. $$ Furthermore, $(D^j u_n)$ is a bounded subsequence of $W^{1,1}(I)$ for $0 \leq j \leq m-2$ from $$ \| D^{j} u_n \|_1 \leq \| D^{j} u_n \|_\infty \leq 1, \quad \| D^{m-1} u_n \|_1 \leq \| D^{m-1} u_n \|_q \leq 1. $$ By the compact embedding $W^{1,1}(I) \subset L^q(I)$, we may assume, after passing to subsequences, that $D^j u_n \to g_j$ in $L^q(I)$ for $0 \leq j \leq m-1$, for some $g_j \in L^q(I)$. It is clear that $u_n \to 0$ in $L^1(I)$. Thus $$ \| u_n \|_q \leq \| u_n \|_\infty^{(q-1)/q} \| u_n \|_1^{1/q} \leq \| u_n \|_1^{1/q} \to 0. $$ Thus $(u_n)$ converges to $0$ in $W^{m-1,q}(I)$. Since $I$ is finite, there is some $C$ for which $$ \| D^j u_n \|_\infty \leq C \| D^j u_n \|_{W^{1,q}} \quad \forall 0 \leq j \leq m-2. $$ Thus $\| D^j u_n \|_\infty \to 0$ for all $0 \leq j \leq m-2$. We have also shown $\| D^{(m-1)} u_n \|_q \to 0$, but by assumption $$ \| D^{(m-1)} u_n \|_q + \sum_{j=0}^{m-2} \| D^j u_n \|_\infty = 1, $$ contradiction.