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:
\(\ds \map u {\alpha h_1 + h_2, k}\) | \(=\) | \(\ds \innerprod {\map A {\alpha h_1 + h_2} } k_U\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod {\alpha A h_1 + A h_2} k_U\) | 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 for 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 for 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
- Classification of Bounded Sesquilinear Forms, which states that all sesquilinear forms are of this type.
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\S \text {II}.2$