# Existence and Uniqueness of Adjoint

## Theorem

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

This article, or a section of it, needs explaining.In particular: How sure are we that this does not hold for ALL subfields of $\C$, not just these ones? At least, the Hilbert assumptions forces $\mathbb F$ to be complete. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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 1

## 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.

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.

$\blacksquare$ $\Box$

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 $y \in \KK$.

$\blacksquare$

## Sources

- 2020: James C. Robinson:
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