$\def\{\mmlToken{mo}[stretchy="false"]‖}$
Let $X$ be a real Banach space with norm $\⋅\$.
[You may use without proof any standard results on Banach spaces, Hilbert spaces and operators between them.]
Let $W:=\{w=(x, y) ∣ x, y ∈ X\}$ and equip $W$ with the norm
\[
\(x, y)\=(\x\^2+\y\^2)^{1 / 2}
\]
so that $W$ is also a Banach space. Show that a sequence $(w_n=(x_n, y_n)) ⊂ W$ converges weakly in $W$ if and only if $(x_n)$ and $(y_n)$ converge weakly in $X$.
Let $P_1, P_2: W → X$ be the projection from $W$ onto the first and the second factor $X$ of $W$, which are clearly continuous.
Suppose $(w_n)$ converges weakly in $W$ to some $w=(x, y)$ and let $ℓ ∈ X^*$. Consider $\tilde{ℓ}(\tilde{w})=ℓ(P_1 \tilde{w})$. Clearly $\\tilde{ℓ}(\tilde{w})\ ⩽\ℓ\_*\P_1\\\tilde{w}\$ and so $\tilde{ℓ} ∈ W^*$. Hence $ℓ(x_n)=\tilde{ℓ}(w_n) → \tilde{ℓ}(w)=ℓ(x)$. This means that $x_n ⇀ x$ in $X$. Likewise $y_n ⇀ y$ in $X$.
Conversely, suppose that $x_n ⇀ x$ and $y_n ⇀ y$ in $X$, respectively. Let $\tilde{ℓ} ∈W^*$. Then $\tilde{ℓ}(w_n)=\tilde{ℓ}(x_n, 0)+\tilde{ℓ}(0, y_n)$. Since the maps $z ↦ \tilde{ℓ}(z, 0)$ and $z ↦ \tilde{ℓ}(0, z)$ belong to $X^*$, we then have by the weak convergence of $(x_n)$ and $(y_n)$ that $\tilde{ℓ}(w_n) →\tilde{ℓ}(x, 0)+\tilde{ℓ}(0, y)=\tilde{ℓ}(w)$. We thus have that $w_n ⇀ w$ in $W$.
In the rest of this question, let $Z$ be a subspace of $X$ and suppose $A: Z → X$ is a (possibly unbounded) linear map such that
\[
\text{ if }(z_n) ⊂ Z, z ∈ Z \text{ and } z_n ⇀ z \text{ in } X \text{, then } A z_n ⇀ A z \text{ in } X,
\]
and let
\[
\z\_A:=(\z\^2+\A z\^2)^{1 / 2} \text{ for } z ∈ Z .
\]
 Show that $\⋅\_A$ defines a norm on $Z$. If $(Z,\⋅\)$ is a closed subspace of $X$, must $(Z,\⋅\_A)$ be a Banach space? Justify your answer.
 Suppose in this part that $(Z,\⋅\)$ is not a closed subspace of $X$ and $(Z,\⋅\_A)$ is a Banach space. Show that $A:(Z,\⋅\) → X$ is unbounded.
 Suppose in this part that $(Z,\⋅\)$ is a closed subspace of $X$. By considering the graph of $A$, or otherwise, show that a sequence $(z_n) ⊂ Z$ converges weakly in $(Z,\⋅\_A)$ if and only if both $(z_n)$ and $(A z_n)$ converge weakly in $X$.

The positivity and homothety of $\⋅\_A$ is clear. The triangle inequality for $\⋅\_A$ follows from that of $\⋅\$, that of the Euclidean norm and the linearity of $A$. Thus $\⋅\_A$ is a norm.
Suppose that $Z$ is closed in $X$. We show that $(Z,\⋅\_A)$ is Banach. Indeed, suppose that $(z_n) ∈ Z$ is Cauchy with respect to $\⋅\_A$. Then $(z_n)$ and $(A z_n)$ are Cauchy in $X$ and hence converge, say to $z$ and $\tilde{z}$. By the closedness of $Z$, $z ∈ Z$. Also, by the given property of $A, A z_n ⇀ A z$. Since strong convergence implies weak convergence and weak limit is unique, we have $A z=\tilde{z}$. We thus have $z_n → z$ and $A z_n → A z$ in $X$, i.e. $\z_nz\_A → 0$.

Suppose by contradiction that $A$ is bounded. By assumption, there exists $(z_n) ⊂ Z$ which converges in $X$ to some $x ∈ \bar{Z} \backslash Z$. Since $A$ is bounded, $(A z_n)$ converges to $A x$. It follows that $(z_n)$ is Cauchy in $(Z,\⋅\_A)$ and hence converges with respect to $\⋅\_A$ to some $z ∈ Z$. But this implies that $(z_n)$ converges to $z$ in $X$ and so $x=z ∈ Z$, which is a contradiction.

Let $G=\{(z, A z): z ∈ Z\} ⊂ W$ be the graph of $A$. Then the map $z ↦(z, A z)$ is an isometric isomorphism from $(Z,\⋅\_A)$ to $G$. Hence $(z_n)$ converges weakly in $(Z,\⋅\_A)$ if and only if $(z_n, A z_n)$ converges weakly in $G$. Suppose that $(z_n)$ converges weakly in $(Z,\⋅\_A)$. Then $(z_n, A z_n)$ converges weakly in $G$ and hence in $W$ since the restriction of any linear functional on $W$ to $G$ is a linear functional on $G$. Apply (a), we then get that $(z_n)$ and $(A z_n)$ converges weakly in $X$.
Conversely, suppose $(z_n)$ and $(A z_n)$ converges weakly in $X$. By (a), we have $(z_n, A z_n)$ converges weakly in $W$ say to $(z, \tilde{z}) ∈ W$. We want to show that $(z, \tilde{z}) ∈ G$ and $T(z_n, A z_n) → T(z, \tilde{z})$ for every $T ∈ G^*$. Since $(z_n) ⊂ Z$ which is closed and convex and $z_n ⇀ z$ in $X$, we have by Mazur's theorem that $z ∈ Z$. By hypothesis, we have $\tilde{z}=A z$ so $(z, \tilde{z}) ∈ G$. Now, by the HahnBanach theorem, every map $T ∈ G^*$ has an extension $\tilde{T} ∈ W^*$. Then $T(z_n, A z_n)=\tilde{T}(z_n, A z_n) → \tilde{T}(z, \tilde{z})=T(z, \tilde{z})$, and so $(z_n, A z_n) ⇀(z, \tilde{z})$ in $G$.
Takeaway: weak convergence in a cloesd subspace $Z$ is equivalent to weak convergence in $X$.
Suppose that the norm $\⋅\$ on $X$ comes from an inner product $⟨⋅, ⋅⟩$ so that $X$ is a Hilbert space. Suppose also that $(Z,\⋅\_A)$ is complete. Show that a linear map $ℓ: Z → ℝ$ belongs to the dual of $(Z,\⋅\_A)$ if and only if there exists $\tilde{z} ∈ Z$ such that
\[
ℓ(z)=⟨z, \tilde{z}⟩+⟨A z, A \tilde{z}⟩ \text{ for all } z ∈ Z \text{. }
\]
Inner product is a symmetric bilinear form such that $∀x, ⟨x, x⟩ ≥ 0$ and if $x ≠ 0$, then $⟨x, x⟩ > 0$.
It is routine to checked that $Z$ equipped with the inner product
\[
⟨z_1, z_2⟩_A:=⟨z_1, z_2⟩+⟨A z_1, A z_2⟩
\]
is a Hilbert space.
(⇐) ${ℓ(z)}≤{⟨z, \tilde{z}⟩}+{⟨A z, A \tilde{z}⟩}≤\sqrt2\sqrt{{⟨z, \tilde{z}⟩}^2+{⟨A z, A \tilde{z}⟩}^2}=\z\_A$
(⇒) part follows from Riesz representation theorem.