Theorem: Bounded Linear Transformation Induces Bounded Sesquilinear Form

Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {V, \innerprod \cdot \cdot_V}$ and $\struct {U, \innerprod \cdot \cdot_U}$ be inner product spaces over $\Bbb F$. 
Let $A : V \to U$ and $B : U \to V$ be bounded linear transformations.
Let $u, v: V \times U \to \C$ be defined by:
:$\map u {h, k} := \innerprod {Ah} k_U$
:$\map v {h, k} := \innerprod h {Bk}_V$
Then $u$ and $v$ are bounded sesquilinear forms.

Proof:
We first show that $u$ and $v$ are sesquilinear, and then that they are bounded. 
Let $\alpha \in \mathbb F$ and $h_1, h_2 \in V$ and $k \in U$. 
We have: 
{{begin-eqn}}
{{eqn | l = \map u {\alpha h_1 + h_2, k}
      | r = \innerprod {\map A {\alpha h_1 + h_2} } k_U
}}
{{eqn | r = \innerprod {\alpha A h_1 + A h_2} k_U
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \alpha \innerprod {A h_1} k_U + \innerprod {A h_2} k_U
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \alpha \map u {h_1, k} + \map u {h_2, k}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map v {\alpha h_1 + h_2, k}
      | r = \innerprod {\alpha h_1 + h_2} {B k}_V
}}
{{eqn | r = \alpha \innerprod {h_1} {B k}_V + \innerprod {h_2} {B k}_V
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \alpha \map v {h_1, k} + \map v {h_2, k}
}}
{{end-eqn}}
Now, let $h \in V$ and $k_1, k_2 \in U$. 
We have: 
{{begin-eqn}}
{{eqn | l = \map u {h, \alpha k_1 + k_2}
      | r = \innerprod {A h} {\alpha k_1 + k_2}_U
}}
{{eqn | r = \overline \alpha \innerprod {A h} {k_1}_U + \innerprod {A h} {k_2}_U
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \overline \alpha \map u {h, k_1} + \map u {h, k_2}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map v {h, \alpha k_1 + k_2}
      | r = \innerprod h {\map A {\alpha k_1 + k_2} }_V
}}
{{eqn | r = \innerprod h {\alpha A k_1 + A k_2}_V
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \overline \alpha \innerprod h {A k_1}_V + \innerprod h {A k_2}_V
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \overline \alpha \map v {h, k_1} + \map v {h, k_2}
}}
{{end-eqn}}
So $u$ and $v$ are both sesquilinear. 
It remains to show that they are bounded.
Let $\norm \cdot_V$ be the inner product norm of $V$.
Let $\norm \cdot_U$ be the inner product norm of $U$.
Let $\norm A$ denote the norm on $A$.
We have that $A$ is a bounded linear transformation.
From Norm on Bounded Linear Transformation is Finite:
:$\norm A$ is finite.
Similarly, since $B$ is a bounded linear transformations, we have that:
:$\norm B$ is finite.
We then have, for all $h \in V$ and $k \in U$: 
{{begin-eqn}}
{{eqn | l = \size {\map u {h, k} }
      | r = \size {\innerprod {A h} k_U}
}}
{{eqn | o = \le
      | r = \norm {A h}_U \norm k_U
      | c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{eqn | o = \le
      | r = \norm A \norm h_V \norm k_U
      | c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
so $u$ is bounded.
Similarly:
{{begin-eqn}}
{{eqn | l = \size {\map v {h, k} }
      | r = \size {\innerprod h {B k}_V}
}}
{{eqn | o = \le
      | r = \norm h_V \size {B k}_V
      | c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{eqn | o = \le
      | r = \norm B \norm h_V \norm k_U
      | c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
so $v$ is also bounded.
{{qed}}