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Let

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

I've seen people talking about the adjoint $Q:H'\to\mathcal D'$ of $\iota$. But is the notion of an adjoint even defined for $\iota$?

As far as I know the notion of an adjoint is only defined for (bounded or unbounded) linear operators between Banach spaces, but $\mathcal D$ (equipped with the usual topology) is only a locally convex topological space.

If we we forget about that, we could treat $\iota$ as a densely-defined unbounded operator and would obtain $$QF=F\iota=\left.F\right|_{\mathcal D}\;\;\;\text{for all }F\in H'\;.\tag 1$$

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Since the mapping $$ \iota\colon\mathcal{D}\to H $$ is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose $$ \iota^t \colon H' \to \mathcal{D}' $$ satisfying $$ \langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle $$ for all $\varphi\in\mathcal{D}$ and all $T\in H'$, where $$ \langle \varphi, S\rangle := S(\varphi) $$ is the duality mapping between $\mathcal{D}$ and $\mathcal{D}'$.

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  • $\begingroup$ Why is $\iota$ continuous? Is it cause $$\left\|\iota\phi\right\|_H\stackrel{\text{def}}=\left(\left\|\phi\right\|^2+ \sum_{i=1}^d\left\|\nabla\phi_i\right\|^2\right)^{\frac 12}\le\lambda(K)^{\frac 12}\left(\sup_K|\phi|^2+\sum_{i=1}^d\sup_K|\nabla\phi_i|^2\right)^{\frac 12}\stackrel{\text{def}}=\lambda(K)^{\frac 12}\left\|\phi\right\|_{C^1(K)}$$ for all compact $K\subseteq\Omega$ and $\phi\in\mathcal D$ with $\operatorname{supp}\phi\subseteq K$. $\endgroup$ – 0xbadf00d Jul 8 '16 at 14:58
  • $\begingroup$ @0xbadf00d: The continuity indeed follows from the above estimate. Note that this does not mean that $\|\iota\|\in\mathcal{D}'$ since it is not linear $\endgroup$ – Christian Jul 8 '16 at 15:09
  • $\begingroup$ Sorry, I was too hasty. $\left\|\iota\right\|_H\in\mathcal D'$ is obviously wrong, cause $\left\|\;\cdot\;\right\|_H$ is only sublinear. How does this transpose differ from the notion of an adjoint of an operator between Banach spaces? Clearly, $\mathcal D$ is not a Banach space. But I don't see anything in the linked definition of an adjoint what wouldn't make sense for an operator $\mathcal D\to H$ too. $\endgroup$ – 0xbadf00d Jul 8 '16 at 15:09
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    $\begingroup$ @0xbadf00d: I think the only real difference is the naming convention: In Banach space theory, people usually use the term "adjoint" and for topological vector spaces people use "transpose". The situation is similar to the word "bounded" which means the same as continuous for linear operators on Banach spaces. $\endgroup$ – Christian Jul 8 '16 at 15:12
  • $\begingroup$ So, $\iota^t$ is exactly the continuous linear map $Q$ from the question, right? $\endgroup$ – 0xbadf00d Jul 8 '16 at 15:25

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