Bounded Linear Transformation Induces Bounded Sesquilinear Form

Theorem
Let $\mathbb F \in \set {\R, \C}$.

Let $\HH$ be a Hilbert space over $\mathbb F$ with inner product $\innerprod \cdot \cdot_\HH$.

Let $\KK$ be a Hilbert space over $\mathbb F$ with inner product $\innerprod \cdot \cdot_\KK$.

Let $A : \HH \to \KK$ and $B : \KK \to \HH$ be bounded linear transformations.

Let $u, v: \HH \times \KK \to \Bbb F$ be defined by:


 * $\map u {h, k} := \innerprod {Ah} k_\KK$
 * $\map v {h, k} := \innerprod h {Bk}_\HH$

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 $h_1, h_2 \in \HH$ and $k \in \KK$.

We have:

and:

Now, let $h \in \HH$ and $k_1, k_2 \in \KK$.

We have:

and:

So $u$ and $v$ are both sesquilinear.

It remains to show that they are bounded.

Let $\norm \cdot_\HH$ be the inner product norm of $\HH$.

Let $\norm \cdot_\KK$ be the inner product norm of $\KK$.

Since $A$ is a bounded linear transformations, we have that:


 * the operator norm, $\norm A$ of $A$ is finite

from Operator Norm is Finite.

Similarly, since $B$ is a bounded linear transformations, we have that:


 * $\norm B$ is finite.

We then have, for all $h \in \HH$ and $k \in \KK$:

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.