If $H$ is a $\mathbb R$-Hilbert space, then the duality pairing $$\langle\;\cdot\;,\;\cdot\;\rangle_{H,\:H'}:H\times H'\;,\;\;\;(x,\Phi)\mapsto\Phi(x)$$ can be considered as being a mapping $H\times H\to\mathbb R$ which is identical to the inner product $\langle\;\cdot\;,\;\cdot\;\rangle_H$ in $H$ by (Riesz’ representation theorem).

If $U$ is a vector subspace of $U$, then each $x\in U$ acts on $U$ in a natural way as a bounded linear functional on $U$ via $$\langle x\rangle_{U'}:=\left.\langle\;\cdot\;,x\rangle_{H,\:H'}\right|_{U}\in U'\;.\tag 1$$

Wich assumptions on $U$ do we need, if we want that $$H\to U'\;,\;\;\;x\mapsto\langle x\rangle_{U'}\tag 2$$ is injective?

Obviously, if $x,y\in H$ with $\langle x\rangle_{U'}=\langle y\rangle_{U'}$, then $$\langle u,x-y\rangle_H=0\;\;\;\text{for all }u\in U\;.\tag 3$$ If $U$ would be dense in $H$, we would find some $(u_n)_{n\in\mathbb N}\subseteq U$ with $$0=\langle u_n,x-y\rangle_H\stackrel{n\to\infty}\to\langle x-y,x-y\rangle_H\;,\tag 4$$ i.e. $x=y$.

So, do we need density of $U$ in $H$, if we want that $(2)$ is injective? As a second question: I've seen that $\langle x\rangle_{U'}(u)$ for $x\in H$ and $u\in U$ is usually denoted by $\langle u,x\rangle_{U,\:U'}$, but isn't that a misleading notation (cause $\langle\;\cdot\;,\;\cdot\;\rangle_{U,\:U'}$ should denote the duality pairing between $U$ and $U'$)?

  • $\begingroup$ you are not capable of considering the case $H = l^2(\mathbb{Z})$ ? $\endgroup$ – reuns Jul 15 '16 at 13:13

For the map $\Phi \colon H \to U'$, $x \mapsto \langle \cdot, x \rangle$ we have $\ker \Phi = U^\perp$. Hence $\Phi$ is injective if and only if $U^\perp = 0$. Because $H = U^\perp \oplus \overline{U}$ we have $U^\perp = 0$ if and only if $\overline{U} = H$, i.e. if $U$ is dense.

  • $\begingroup$ It's clear that $\Phi$ is injective and continuous, but is $\Phi$ even an embedding, i.e. is it's inverse continuous too? $\endgroup$ – 0xbadf00d Jul 15 '16 at 17:47
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    $\begingroup$ If $U$ is dense in $H \neq 0$ then so is $U \setminus \{0\}$. Hence $$ \|T\| = \sup_{\substack{x \in H \\ x \neq 0}} \frac{\|T x\|}{\|x\|} = \sup_{\substack{y \in U \\ y \neq 0}} \frac{\|T y\|}{\|y\|} = \|\Phi(T)\|$$ for every $T \in H'$. So $\Phi$ is an isometric embedding. If I’m not mistaken then it’s also surjective and hence an isometric isomorphism. $\endgroup$ – Jendrik Stelzner Jul 16 '16 at 13:59
  • $\begingroup$ My claim $(1)$ is wrong in general, isn't it? We need that $U$ is equipped with a topology for which the inclusion $U\to H$ is continuous, right? $\endgroup$ – 0xbadf00d Jul 20 '16 at 12:35

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