# Existence and Uniqueness of Adjoint

## Theorem

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

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$.

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

Then there exists a unique bounded linear transformation $B: \KK \to \HH$ such that:

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

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

That is:

each bounded linear transformation between Hilbert spaces has a unique adjoint.

## Proof

We first show that such a unique mapping $B$ exists, without first insisting on a bounded linear transformation.

## 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 $y \in \KK$. </onlyinclude>

## 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.

$\norm A$ 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.

$\blacksquare$ $\Box$

We now show that $B$ is a linear transformation.

### Lemma 2

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.

$\Box$

Finally, we show that $B$ is bounded.

### Lemma 3

$B$ is a bounded linear transformation.

$\Box$

So $B$ is the unique bounded linear transformation such that:

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

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

$\blacksquare$