Existence and Uniqueness of Adjoint/Lemma 1

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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.


Then:

There exists a unique mapping $B : \KK \to \HH$ such that:
$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$

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


Proof

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

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


For each $y \in \KK$, define the linear functional $f_y : \HH \to \mathbb F$ by:

$\map {f_y} x = \innerprod {\map A x} y_\KK$


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.


We therefore have:

\(\ds \size {\map {f_y} x}\) \(=\) \(\ds \size {\innerprod {A x} y_\KK}\)
\(\ds \) \(\le\) \(\ds \norm {A x}_\KK \norm y_\KK\) Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
\(\ds \) \(=\) \(\ds \norm A \norm x_\HH \norm y_\KK\) Fundamental Property of Norm on Bounded Linear Transformation

Taking $M = \norm A \norm y_\KK$, we have:

$\size {\map {f_y} x} \le M \norm x_\HH$

for each $x \in \HH$, with $M$ independent of $x$.

So, $f_y$ is bounded.

So, by the Riesz Representation Theorem (Hilbert Spaces), there exists a unique $\map z y \in \HH$ such that:

$\map {f_y} x = \innerprod x {\map z y}_\HH$

for each $x \in \HH$.

That is, for each $y \in \KK$ there exists precisely one $\map z y \in \HH$ such that:

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

for all $x \in \HH$.


Define the mapping $B : \KK \to \HH$ by:

$B y = \map z y$

for each $y \in \KK$.

This map has:

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

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

Since the choice of $\map z y$ was unique, the map $B$ must also be unique, so $B$ is the unique map with the required properties.

$\blacksquare$