In an exercise I'm having problems with the following:
Let $E=\mathbb{R}^{d} \times L^{2}((0,+\infty), \mathbb{R}^d)$ Hilbert space with norm: $$ \|\alpha\|^{2}:=|x|^{2}+\|z\|_{L^{2}}^{2}, \forall \alpha=(x, z) \in E $$ and the corresponding inner product $\langle., .\rangle$. Let $T$ be the linear unbounded operator on $E$ with $D(T)=\mathbb{R}^{d} \times H^{1}((0,+\infty), \mathbb{R}^d)$ and defined by $$ T(y, w)=(y-w(0),-\dot{w}), \forall(y, w) \in D(T) $$ How do you show that its adjoint $T^{*}$ has domain $D\left(T^{*}\right)=E_{0}=\left\{(x, z) \in E: z \in H^{1}, z(0)=x\right\}$ and is given by $$ T^{*}(x, z)=(z(0), \dot{z})=(x, \dot{z}), \forall(x, z) \in D\left(T^{*}\right)=E_{0} $$ I can see by Riesz theorem and by integration by parts that $\dot{z}$ is the second component of the operator $T^*$ and then the domain for this part must be $H^1$ but why the domain is $z(0)=x$?
The domain of $T^*$ is given by $D(T^*) = \left\{(x, z) \, \middle\vert \, (y,w) \mapsto \langle T(y, w), (x, z)\rangle \, \text{is a bounded linear functional}\right\}$.
If $x \neq z(0)$, the boundary terms in the partial integration you mentioned don't cancel the term $- \langle w(0), x \rangle $ in $\langle T(y, w), (x, z)\rangle$. Therefore the corresponding linear functional cannot be bounded as $w(0)$ can be arbitrarily large while the $L^2$-Norm of $w$ stays small.