Let $\Omega\subset\mathbb{R}^n$ be an open, bounded set with boundary $\partial\Omega$ of class $C^1$. $$\mathcal{A}:=\{u\in C^2(\bar\Omega):u=0\text{ on }\partial\Omega \}$$ endowed with the scalar product $$(u,v)_{\mathcal{A}}:=\int_{\Omega}(\nabla u,\nabla v)_{\mathbb{R}^n}\,dx.$$ I have to prove that $\mathcal{A}$ equipped with the induced norm is a normed space, but it is not a Banach space. I can't find a proper counterexample. Any help?

  • $\begingroup$ You know, I suppose, that the continuous functions are not complete in the $L^2$ norm? $\endgroup$ – Nate Eldredge Jun 7 '14 at 14:54

As far as I understand you only need help in showing that $\mathcal A$ is not Banach. Таке а function $f\in C^1(\bar\Omega)$ such that $f=0$ on $\partial\Omega$ and $f$ is $C^2$ everywhere in $\bar\Omega$ except at a single point $x_0\in\Omega$, where the second derivatives do not exist. Then you can approximate $f$ in the norm defined by this inner product by functions $f_n\in\mathcal A$; thus, $\mathcal A$ is not complete in this norm and hence by definition not Banach.

  • $\begingroup$ I was thinking about whether or not this would be enough, or if one should prove that such a function exists for any $\Omega$ as described, do you think this is unnecessary? $\endgroup$ – DanZimm Jun 7 '14 at 15:01
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    $\begingroup$ Without loss of generality suppose that $0\in\Omega$ and take $x_0=0$. Let $r>0$ be small enough that the ball of radius $r$ lies in $\Omega$, let $\phi(x)$ be a smooth function supported in this ball such that $\phi(0)=1$. Set $f(x)=\phi(x)x_1|x|$. $\endgroup$ – Vladimir Jun 7 '14 at 15:08
  • $\begingroup$ Beautiful, I didn't realize we could take $0 \in \Omega$ and then work from there, was trying to do something far more complicated. $\endgroup$ – DanZimm Jun 7 '14 at 15:09
  • $\begingroup$ What do you mean for $x_1$ in the definition of $f$? $\endgroup$ – avati91 Jun 7 '14 at 15:34
  • $\begingroup$ The first coordinate: $x=(x_1,x_2,\dots,x_n)$; just in case, $|x|=\sqrt{x_1^2+\dots+x_n^2}$. $\endgroup$ – Vladimir Jun 7 '14 at 15:36

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