Bounded Linear Transformation Induces Bounded Sesquilinear Form

Theorem
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:

and:

Now, let $h \in V$ and $k_1, k_2 \in U$.

We have:

and:

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 $k \in U$:

so $u$ is bounded.

Similarly:

so $v$ is also bounded.

Also see

 * Classification of Bounded Sesquilinear Forms, which states that all sesquilinear forms are of this type.