The Poincaré inequality for $H_0^1(\Omega)$

Question

The following is the well known Poincaré inequality for $H_0^1(\Omega)$:

Suppose that $\Omega$ is an open set in $\mathbb{R}^n$ that is bounded in some direction. Then there is a constant $C$ such that $$ \int_\Omega u^2\ dx\leq C\int_\Omega|Du|^2\ dx\quad \textrm{ for all }\ \ \color{red}{u\in H_0^1(\Omega)}. $$

Here are my questions:

• Could anyone come up with an example that $f\in H^1(\Omega)\setminus H_0^1(\Omega)$?
• Is the statement above true if one replaces $H_0^1(\Omega)$ with $H^1(\Omega)$?

Answer 1

Take $u \equiv 1$. Then, $u \in H^1(\Omega) \setminus H_0^1(\Omega)$ and your inequality fails (as long as $\Omega \ne \emptyset$).

Answer 2

Let $u \in H^1_0(\Omega)$ be any function, then $u + c \in H^1(\Omega) \setminus H^1_0(\Omega)$ (provided $\Omega$ is bounded)

And yes, you need the function in $H^1_0$ for this to work. The intuition is easy (picture one dimension): if you know how big the derivative is (how fast the function is growing), you can know more or less which level will the function reach after some point, provided that you know where it started.

So if $u(0) = 0$ and $u'(x) \le 1$, you can say "I know $u(1) = u(0) + \int_0^1 u'(x)dx \le 1$". If you don't know $u(0)$ though you can make no such assertion.

Now clearly this analogy is not perfect (there are different kinds of norms in play) but it boils down to this