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I regularly see proofs involving Sobolev spaces where the proof states it will show some result holds for, say, $u \in H_0^1(\Omega)$ where $\Omega$ is a smooth bounded domain.

Then right away it will say that it suffices to the result holds for $u \in C^\infty(\Omega)$. So we have proved that the result holds for $u \in C^\infty(\Omega)$, how then can we show rigorously that this result also holds for $u \in H_0^1(\Omega)$?

sonicboom
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    In most case they use the fact that $C_0^\infty(\Omega)$ is dense in $H^1_0(\Omega)$. –  Jul 19 '18 at 11:34

1 Answers1

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The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that

  1. $F$ and $G$ are continuous with respect to the Sobolev norm
  2. $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space

then the conclusion $F\equiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$

The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.