# Inequality for dense subset implies inequality for whole set? (PDE)

Suppose I have an inequality that holds for all $f \in C^\infty(\Omega)$. Then since $C^\infty(\Omega)$ is dense in, say, $H^1(\Omega)$ under the latter norm, does the inequality hold for all $f \in H^1(\Omega)$ too?

(Suppose the inequality involves norms in $L^2$ space)

-

Let $X$ be a topological space. Let $F,G$ be continuous maps from $X$ to $\mathbb{R}$. Let $Y\subset X$ be a dense subspace.

Then $$F|_Y \leq G|_Y \iff F \leq G$$

The key is continuity. (Actually, semi-continuity of the appropriate direction is enough.) Continuity guarantees for $x\in X\setminus Y$ and $x_\alpha \in Y$ such that $x_\alpha \to x$ you have

$$F(x) \leq \lim F(x_\alpha) \leq \lim G(x_\alpha) \leq G(x)$$

So the question you need to ask yourself is: are the two functionals on the two sides of your inequality continuous functionals on $H^1(\Omega)$? Just knowing that they involve norms in $L^2$ space is not (generally) enough (imagine $f\mapsto \|\triangle f\|_{L^2(\Omega)}$).

For a slightly silly example: Let $G:H^1(\Omega)\to\mathbb{R}$ be identically 1, and let $F:C^\infty(\Omega)\to \mathbb{R}$ be identically zero, but let $F:H^1(\Omega) \setminus C^\infty(\Omega) \to \mathbb{R}$ be equal to $\|f\|_{L^2(\Omega)}$. Then clearly $F|_{C^\infty} \leq G|_{C^\infty}$ but the extension to $H^1$ is false in general.

-
Thanks very much – pde_lover Jan 7 '13 at 12:21