Existence and Uniqueness of Adjoint/Lemma 2

From ProofWiki
Jump to navigation Jump to search

Lemma

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 $A : \HH \to \KK$ be a bounded linear transformation.

Let $B : \KK \to \HH$ be the unique mapping satisfying:

$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$

for each $x \in \HH$ and $y \in \KK$.


Then we have:

$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$

for all $\alpha, \beta \in \mathbb F$ and $x, y \in \KK$.


That is, $B$ is a linear transformation.


Proof

Let $\alpha, \beta \in \mathbb F$.

Let $x, y \in \KK$.

We have:

\(\ds \innerprod x {\map B {\alpha x + \beta y} }_\KK\) \(=\) \(\ds \innerprod {A x} {\alpha x + \beta y}_\HH\) Definition of $B$
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod {A x} x_\HH + \overline \beta \innerprod {A x} y_\HH\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod x {B x}_\KK + \overline \beta \innerprod x {B y}_\KK\) Definition of $B$
\(\ds \) \(=\) \(\ds \innerprod x {\alpha B x}_\KK + \innerprod x {\beta B y}_\KK\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \innerprod x {\alpha B x + \beta B y}_\KK\) Inner Product is Sesquilinear

Recall that there exists exactly one $z \in \HH$ such that:

$\innerprod {A x} {\alpha x + \beta y}_\HH = \innerprod x z_\KK$

So we must have:

$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$

So:

$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$

for each $\alpha, \beta \in \mathbb F$ and $x, y \in \KK$.

So:

$B$ is a linear transformation.

$\blacksquare$