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Willie Wong
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Let $\Omega$ be the unit ball in $\mathbb{R}^2$.

Let $u_k(x,y) = \tan^{-1}(k^3 x)$.

Let $v_k(x,y)$ be a function that agrees with $u_k$ on $\partial\Omega$, and is constant on the level sets of $\{ (x- k)^2 + y^2\}$.

So $\nabla u_k / |\nabla u_k| = \partial_x$, and $ \nabla v_k / |\nabla v_k| = \partial_x + O(1/k) $, so the directions of their gradients differ only by a little.

Let $B_k$ be the ball of radius $1/k^3$ centered at $(k - \sqrt{k^2 + 1} \approx -1/(2k), 0)$.

For large $k$, we have that $\nabla u_k \approx 1/k$ on $B_k$.

On the other hand, $\nabla v_k \approx k^3$ on $B_k$.

This implies that $\|u_k - v_k\|_{H^1_0(\Omega)} \geq \|\nabla u_k - \nabla v_k\|_{L^2(B_k)} \approx 1$ is bounded uniformly away from zero.


This shows that, under the assumptions that

  • $\Omega$ is simply connected with connected boundary, and bounded
  • $u,v\in C^1(\Omega)$ are both functions bounded by $M$
  • $u = v$ on $\partial\Omega$.
  • and that $\nabla u, \nabla v \neq 0$ on $\Omega$

there does not exist a constant $C$ such that $$ \|u - v\|_{H^1_0(\Omega)} \leq C \| \frac{\nabla u}{|\nabla u|} - \frac{\nabla v}{|\nabla v|} \|_{L^2(\Omega)}$$

Willie Wong
  • 39.1k
  • 4
  • 94
  • 176