Let $(H, |\cdot|)$ be a Banach space and $V$ a linear subspace of $H$. Assume that the vector space $V$ has its own norm $[ \cdot ]$ such that $(V, [\cdot])$ is a Banach space. We consider the (linear) inclusion map $i: (V, [\cdot]) \to (H, |\cdot|), v \mapsto v$. Let $(V^*, [\![ \cdot ]\!])$ be the dual space of $(V, [\cdot])$. Let $(H^*, \|\cdot\|)$ be the dual space of $(H, |\cdot|)$. I'm trying to prove a claim mentioned in this thread, i.e.,
Theorem If $i$ is continuous, then the adjoint operator of $i$ is given by $$ i^*:(H^*, \|\cdot\|) \to (V^*, [\![ \cdot ]\!]), \varphi \mapsto \varphi \restriction V. $$
Could you have a check on my below attempt?
Proof The continuity of $i$ implies that
- $i^*$ is continuous, and
- the domain of $i^*$ is the entire $H^*$, i.e., $D(i^*) = H^*$.
We have $i^*$ is characterized by $$ \langle i^* (\varphi), v \rangle_{V^*, V} = \langle \varphi, i (v) \rangle_{H^*, H} \quad \forall \varphi \in H^*, \forall v \in V. $$
Fix $\varphi \in H^*$. Notice that $i$ is the identity map, so $$ \langle i^* (\varphi), v \rangle_{V^*, V} = \langle \varphi, v \rangle_{H^*, H} \quad \forall v \in V. $$
The claim then follows.