Let $I=(0,1)$ and $u\in W^{1,2}(I)$. It is not difficult to see that there is a constant $C>0$ such that $$\|u\|_{2}\le C(|u(0)|+\|u'\|_2).\tag{1}$$

If we restrict the inequality $(1)$ to the set $A_c=\{u\in W^{1,2}(I):\ u(0)=c\}$, where $c$ is a fixed constant, we conclude that $$\|u\|_2\le C(c+\|u'\|_2),\ \forall\ u\in A_c.$$

In particular, we conclude that if $\|u\|_2\to \infty$ then, $\|u'\|_2\to \infty$. Now, let $\Omega\subset \mathbb{R}^N$ be a bounded smooth domain, $w\in H^{1/2}(\partial \Omega)$ and $T$ the usual trace operator. Define $B_w$ by$$B_w=\{u\in W^{1,2}(\Omega):\ T(u)=w\}.$$

Are there constants $C_1,C_2$ such that $$\|u\|_2\le C_1(C_2+\|\nabla u\|_2),\ u\in B_w?$$

If the above is not true:

Can we conclude that $\|\nabla u\|_2\to \infty$ if $\|u\|_2\to \infty$ with $u\in B_w$?


Fix a function $u_0\in W^{1,2}$ such that $T(u_0)=w$. The Poincaré inequality, applied to $u-u_0$ (which is in $W^{1,2}_0(\Omega)$) yields $$ \|u-u_0\|_2\le C\|\nabla (u-u_0)\|_2 \le C\|\nabla u\|_2 + C\|\nabla u_0\|_2 $$ Hence $$ \|u\|_2 \le (\|u_0\|_2 + C\|\nabla u_0\|_2) + C\|\nabla u\|_2 $$ which is an estimate of the desired type. The additive constant depends on $u_0$, which depends only on $w$.

  • $\begingroup$ It was just a simple idea and I'm here trying to prove it by plenty of calculations... $\endgroup$ – Tomás Mar 18 '16 at 22:46

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