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On page 65 of "Sobolev Spaces (Adams ed2)", it is said that to prove $V=\{L_v: v \in L^{p'}(\Omega)\}$ is dense in $(W_0^{m,p}(\Omega))'$, it is sufficient to prove that if $F \in (W_0^{m,p}(\Omega))''$ satisfies $F(L_v) = 0\;\forall L_v \in V$, then $F=0$, hence the denseness of $V$ is proved. I would like to know why there is such equivalence between the denseness of $V$ and the zero operator $F$. Many thanks!

Ire Shaw
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TJH
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1 Answers1

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This is an easy corollary from Hahn-Banach:

Let $X$ be a normed vector space and $V \subset X$ a subspace. Then $V$ is dense, if and only if $$\forall x' \in X': \Big( \big( \forall v \in V : x'(v) = 0 \big)\Rightarrow x' = 0\Big).$$

Hint for the proof: If $V$ is not dense, you can separate the closure of $V$ from any point not belonging to this closure.

gerw
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  • Thanks for your help! Then I understand. If $V$ is not dense, we can find a point $x_0 \in X-\bar{V}$, then there exists a $y' \in X'$ such that $y'(x_0) = 1$ and $y'(v) = 0; \forall v \in V$. – TJH Jan 12 '16 at 04:08
  • Yes, this is correct. – gerw Jan 12 '16 at 08:20