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Let

  • $d\in\mathbb N$
  • $\Omega\subseteq\mathbb R^d$ be open
  • $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the inner product on $L^2(\Omega,\mathbb R^d)$
  • $\mathcal D:=C_c^\infty(\Omega,\mathbb R^d)$ and $$H:=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle+\sum_{i=1}^d\langle\nabla\phi_i,\nabla\psi_i\rangle\;\;\;\text{for }\phi,\psi\in\mathcal D$$
  • $F\in H'$ and $f:=\left.F\right|_{\mathcal D}$

Can we prove the following claim? If so, how?

Let $K\subseteq\Omega$ be compact and $\phi\in\mathcal D$ with $\operatorname{supp}\phi\subseteq K$ $\Rightarrow$ $\exists c\ge 0,k\in\mathbb N_0$ with $$\left|f(\phi)\right|\le c\sum_{\left|\alpha\right|\le k}\sup_K\left|\partial^\alpha\phi\right|\;.\tag 1$$

By the bounded linear transformation theorem, $F$ is the continuous linear extension of $f$ and hence $$\left\|F\right\|_{H'}=\left\|f\right\|_{\mathcal D'}\;.\tag 2$$ I'm new to distribution theory, but I could imagine that $\left\|f\right\|_{\mathcal D'}<\infty$ iff $(1)$ is satisfied.

EDIT: I was too hasty. $\mathcal D$ (equipped with the usual locally convex topology) is not a normed space and hence the mentioned theorem cannot be applied.

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1 Answer 1

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Let me handle the case $d = 1$ for simplicity of notation. Since $H$ is a Hilbert space, $F$ is represented by an element $g \in H$. Thus, for all $\phi \in \mathcal{D} \subset H$ with $\operatorname{supp} \phi = K$ we have

$$ |F(\phi)| = \left| \left<\phi, g \right>_H \right| \leq ||g||_H ||\phi||_H = ||g||_H \left( \int_{\Omega} \phi \cdot \phi + \nabla \phi \cdot \nabla \phi \right)^{\frac{1}{2}} \leq ||g||_H \left( \int_{K} ||\phi||_{\infty,K}^2 + ||\nabla \phi||_{\infty,K}^2\right)^{\frac{1}{2}} \leq ||g||_H \cdot \mu(K)^{\frac{1}{2}} \cdot \max (||\phi||_{\infty,K}, ||\nabla \phi||_{\infty,K}) $$

and so $f = F|_{\mathcal{D}}$ is a distribution of order (at most) o

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    $\begingroup$ What is $||f||_{\mathcal{D}'}$? The space of distributions $\mathcal{D}'$ with the usual topology placed on it is not a normed vector space. $\endgroup$
    – levap
    Commented Jul 4, 2016 at 16:34
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    $\begingroup$ @0xbadf00d: The operator norm is relevant for operators between two normed spaces but the space $\mathcal{D}$ (again, with the relevant topology for the theory of distributions) is not normed... $\endgroup$
    – levap
    Commented Jul 4, 2016 at 16:50
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    $\begingroup$ @user1952009: By definition of the topology on $\mathcal{D}$, if $\phi_n \rightarrow 0$ then there exists a compact $K \subset \Omega$ with $\operatorname{supp} \phi_n \subseteq K$ for all $n \in \mathbb{N}$. Then we can choose an open bounded set $\Omega'$ such that $K \subset \Omega' \subseteq \Omega$ and then $||\phi_n||_{H^k(\Omega)} = ||\phi_n||_{H^k(\Omega')}$ so if you believe the result for bounded domains, it follows for general domains. $\endgroup$
    – levap
    Commented Jul 4, 2016 at 17:02
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    $\begingroup$ @user1952009: By the definition of the topology on $\mathcal{D}$ - see en.wikipedia.org/wiki/Distribution_(mathematics). $\endgroup$
    – levap
    Commented Jul 4, 2016 at 17:23
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    $\begingroup$ @levap Are you sure that your edit is correct? Since $\mathcal D$ is dense in $H$, the Hahn-Banach theorem should yield $$f=0\Leftrightarrow F=0\;.$$ EDIT: Didn't read your whole edit ;) Please note that $H$ is defined to be $H_0^1(\Omega,\mathbb R^d)$. $\endgroup$
    – 0xbadf00d
    Commented Jul 4, 2016 at 19:02

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