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Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $H^{-1}(\Lambda):=H_0^1(\Lambda)'$ denote the topological dual sapce of $H_0^1(\Lambda)$

Since $$\langle u,f\rangle_{H_0^1,\:H^{-1}}:=\langle u,f\rangle_{L^2(\Lambda)}\;\;\;\text{for }u\in H_0^1(\Lambda)$$ is a bounded linear functional on $H_0^1(\Lambda)$ for all $f\in L^2(\Lambda)$, $$\iota:L^2(\Lambda)\to H^{-1}(\Lambda)\;,\;\;\;f\mapsto\langle\;\cdot\;,f\rangle_{H_0^1,\:H^{-1}}$$ is well-defined. Note that $\iota$ is injective and linear.

How can we show that $\iota L^2(\Lambda)$ is a dense in $H^{-1}(\Lambda)$?

I guess we somehow need to use the Hahn-Banach theorem or the Stone–Weierstrass theorem. For the first one, we would need to show that $\left.\eta\right|_{\iota L^2(\Lambda)}=0$ implies $\eta=0$, for all $\eta\in H^{-1}(\Lambda)'$.

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    Let $\Lambda= [0,1]$. For $f \in L^2$, how do you choose a sequence $g_n \in L^2$ such that $G_n(u) = \int_0^1 u(x) g_n(x)dx$ approximates $F(u) = \int_0^1 u'(x) f(x)dx$ ? – 2017-01-22

1 Answers 1

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Let $F \in H^{-1}$. If $\Lambda$ is bounded take $\kappa \ge 0$ and if $\Lambda$ is unbounded take $\kappa >0$. We will take the inner-product on $H^1_0$ to be $$ (u,v)_{H^1_0} = \int_\Lambda\left( \nabla u \cdot \nabla v + \kappa uv\right). $$

Using Riesz / the weak-solvability of $-\Delta u +\kappa u =F$ with Dirichlet BCs, we can find a unique $u \in H^1_0$ such that $$ \langle F,v \rangle = \int_\Lambda \left(\nabla u \cdot \nabla v + \kappa uv \right) \text{ for all }v \in H^1_0. $$

Let $\epsilon >0$. Since $C^\infty_c(\Lambda)$ is dense in $H^1_0$ we can find $u_\epsilon \in C^\infty_c$ such that $$ \Vert u - u_\epsilon \Vert_{H^1_0} = \left( \int_\Lambda |\nabla u - \nabla u_\epsilon|^2 + \kappa|u-u_\epsilon|^2\right)^{1/2} < \epsilon. $$ Then $$ \langle F,v \rangle = \int_{\Lambda}\left[ (\nabla u -\nabla u_\epsilon) \cdot \nabla v + \kappa (u-u_\epsilon)v\right] + \int_\Lambda \left(\nabla u_\epsilon \cdot \nabla v + \kappa u_\epsilon v \right) $$ for $v \in H^1_0$. Since $u_\epsilon \in C^\infty_c$ we can rewrite the last term as $$ \int_\Lambda \left(\nabla u_\epsilon \cdot \nabla v + \kappa u_\epsilon v\right)= \int_\Lambda (- \Delta u_\epsilon + \kappa u_\epsilon) v = (-\Delta u_\epsilon+\kappa u_\epsilon,v)_{L^2} = \langle \iota (-\Delta u_\epsilon+ \kappa u_\epsilon), v\rangle. $$ Thus $$ \langle F - \iota(-\Delta u_\epsilon + \kappa u_\epsilon),v\rangle = \int_{\Lambda}\left[ (\nabla u -\nabla u_\epsilon) \cdot \nabla v + \kappa (u-u_\epsilon)v\right] $$ and we can estimate $$ \Vert F - \iota(-\Delta u_\epsilon+ \kappa u_\epsilon)\Vert_{H^{-1}} = \sup_{\Vert v \Vert_{H^1_0} \le 1} \langle F - \iota(-\Delta u_\epsilon+ \kappa u_\epsilon),v\rangle \le \Vert u - u_\epsilon \Vert_{H^1_0} < \epsilon. $$ Since $\epsilon >0$ was arbitrary we deduce that $\iota L^2$ is dense in $H^{-1}$.

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    There is a mistake in your answer: We don't have $$\Vert u - u_\epsilon \Vert_{H^1_0} = \left( \int_\Lambda |\nabla u - \nabla u_\epsilon|^2 \right)^{1/2}\;,$$ but the right-hand side is not greater than the left-hand side. Maybe you had in mind that $$H_0^1(\Lambda)\ni u\mapsto\left\|\nabla u\right\|_{L^2(\Lambda,\:\mathbb R^d)}$$ is an equivalent norm on $H_0^1(\Lambda)$, **if $\Lambda$ is bounded** (which is not the case here). – 2017-01-22
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    You've came up with a direct proof. I'm interested in how the desired statement follows from the fact that $\iota L^2(\Lambda)$ separates the points of $H_0^1(\Lambda)$. How can we show that fact and from which theorem does the density follow? – 2017-01-22
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    Ah, I didn't notice that $\Lambda$ wasn't bounded. I can fix it easily. I'm going to make an edit. – 2017-01-22
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    Typically $H^1_0(\Lambda)$ is defined as the closure of $C^\infty_c(\Lambda)$ w.r.t. the norm $\Vert \nabla u \Vert_{L^2}$ when $\Lambda$ is unbounded. Nevertheless I've edited to given you a proof that doesn't depend directly on Poincare's inequality. – 2017-01-22
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    I guess you mean that "typically $H_0^1(\Lambda)$ is defined to be the closure of $C_c^\infty(\Lambda)$ in $H^1(\Lambda)$". Or you mean "when $\Lambda$ is bounded", instead of "unbounded". If $\Lambda$ is unbounded (and hence the *Poincaré inequality* doesn't hold), then $H_0^1(\Lambda)\ni u\mapsto\left\|\nabla u\right\|_{L^2(\Lambda)}$ is not positive definite. – 2017-01-22
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    As for your second question, I am not sure what you mean. You asked for density, and I gave you a proof of density. Nowhere did you mention you wanted the proof done in any particular way. – 2017-01-22
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52255/discussion-between-glitch-and-0xbadf00d). – 2017-01-22
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    It's not clear that $\kappa uv$ is inside the integral in the definition of your new inner product on $H_0^1(\Lambda)$. – 2017-01-22
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    Why do you need the $\kappa$ at all? From your formulation, we can choose any $\kappa>0$. So, why shouldn't we choose $\kappa=1$, which would yield the usual inner product on $H_0^1(\Lambda)$? – 2017-01-22