Bounded Linear Transformation Induces Bounded Sesquilinear Form

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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:

\(\ds \map u {\alpha h_1 + h_2, k}\) \(=\) \(\ds \innerprod {\map A {\alpha h_1 + h_2} } k_U\)
\(\ds \) \(=\) \(\ds \innerprod {\alpha A h_1 + A h_2} k_U\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \alpha \innerprod {A h_1} k_U + \innerprod {A h_2} k_U\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \alpha \map u {h_1, k} + \map u {h_2, k}\)

and:

\(\ds \map v {\alpha h_1 + h_2, k}\) \(=\) \(\ds \innerprod {\alpha h_1 + h_2} {B k}_V\)
\(\ds \) \(=\) \(\ds \alpha \innerprod {h_1} {B k}_V + \innerprod {h_2} {B k}_V\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \alpha \map v {h_1, k} + \map v {h_2, k}\)

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

We have:

\(\ds \map u {h, \alpha k_1 + k_2}\) \(=\) \(\ds \innerprod {A h} {\alpha k_1 + k_2}_U\)
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod {A h} {k_1}_U + \innerprod {A h} {k_2}_U\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \overline \alpha \map u {h, k_1} + \map u {h, k_2}\)

and:

\(\ds \map v {h, \alpha k_1 + k_2}\) \(=\) \(\ds \innerprod h {\map A {\alpha k_1 + k_2} }_V\)
\(\ds \) \(=\) \(\ds \innerprod h {\alpha A k_1 + A k_2}_V\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod h {A k_1}_V + \innerprod h {A k_2}_V\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \overline \alpha \map v {h, k_1} + \map v {h, k_2}\)

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 $k \in U$:

\(\ds \size {\map u {h, k} }\) \(=\) \(\ds \size {\innerprod {A h} k_U}\)
\(\ds \) \(\le\) \(\ds \norm {A h}_U \norm k_U\) Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces
\(\ds \) \(\le\) \(\ds \norm A \norm h_V \norm k_U\) Fundamental Property of Norm on Bounded Linear Transformation

so $u$ is bounded.

Similarly:

\(\ds \size {\map v {h, k} }\) \(=\) \(\ds \size {\innerprod h {B k}_V}\)
\(\ds \) \(\le\) \(\ds \norm h_V \size {B k}_V\) Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces
\(\ds \) \(\le\) \(\ds \norm B \norm h_V \norm k_U\) Fundamental Property of Norm on Bounded Linear Transformation

so $v$ is also bounded.

$\blacksquare$


Also see


Sources