Existence and Uniqueness of Adjoint of Densely-Defined Linear Operator

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Theorem

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

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\Bbb F$.

Let $\struct {\map D T, T}$ be a densely defined linear operator on $\HH$.

For each $y \in \HH$, define the linear functional $f_x : \map D T \to \Bbb F$ by:

$\map {f_y} x = \innerprod {T x} y$ for each $x \in \map D T$.

Define:

$\map D {T^\ast} = \set {y \in H : f_y \text { is a bounded linear functional}}$


Then there exists a unique linear transformation $T^\ast : \map D {T^\ast} \to \HH$ with:

$\innerprod {T x} y = \innerprod x {T^\ast y}$ for all $x \in \map D T$ and $y \in \map D {T^\ast}$.


Proof

We first check that $\map D {T^\ast}$ is a linear subspace of $\HH$, so that the question whether $T^\ast$ be a linear transformation is well-posed.

Let $u, v \in \map D {T^\ast}$ and $\alpha \in \Bbb F$.

Then there exists real numbers $M_1, M_2 > 0$ such that:

$\cmod {\map {f_u} x} \le M_1 \norm x$

and:

$\cmod {\map {f_v} x} \le M_2 \norm x$

for each $x \in \map D T$.

Then, we have:

\(\ds \cmod {\map {f_{\alpha u + v} } x}\) \(=\) \(\ds \cmod {\innerprod {T x} {\alpha u + v} }\)
\(\ds \) \(=\) \(\ds \cmod {\overline \alpha \innerprod {T x} u + \innerprod {T x} v}\)
\(\ds \) \(\le\) \(\ds \cmod {\overline \alpha} \cmod {\innerprod {T x} u} + \cmod {\innerprod {T x} v}\) Triangle Inequality
\(\ds \) \(=\) \(\ds \cmod \alpha \cmod {\innerprod {T x} u} + \cmod {\innerprod {T x} v}\) Complex Modulus equals Complex Modulus of Conjugate
\(\ds \) \(\le\) \(\ds \paren {M_1 \cmod \alpha + M_2} \norm x\)

for each $x \in \map D T$.

So, $f_{\alpha u + v}$ is a bounded linear functional.

So $\alpha u + v \in \map D {T^\ast}$.

So from One-Step Vector Subspace Test, we have that $\map D {T^\ast}$ is a linear subspace of $\HH$.


Now let $y \in \map D {T^\ast}$.

Then:

$\map {f_y} x = \innerprod {T x} y$ defines a bounded linear functional $\map D T \to \Bbb F$

So, from Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain, there exists a bounded linear functional $F_y : \HH \to \Bbb F$ such that:

$\map {F_y} x = \innerprod {T x} y$ for all $x \in \map D T$.

Then from the Riesz representation theorem there exists a unique $z_y \in \HH$ such that:

$\map {F_y} x = \innerprod x {z_y}$ for all $x \in \map D T$.

For each $y \in \map D {T^\ast}$, let:

$T^\ast y = z_y$

Then:

$\innerprod {T x} y = \innerprod x {T^\ast y}$ for all $x \in \map D T$ and $y \in \map D {T^\ast}$.

It remains to verify that $T^\ast$ is linear and that it is unique.

Note that uniqueness is not immediate, since $z_y$ is the unique element of $\HH$ such that $\innerprod {T x} y = \innerprod x z$ for all $x \in \HH$, there may exist other $z$ for which this only holds for $\map D T$, and we need to verify that this is not the case.

Let $u, v \in \map D {T^\ast}$ and $\alpha \in \Bbb F$.

Then for all $x \in \map D T$ we have:

\(\ds \innerprod x {\map {T^\ast} {\alpha u + v} }\) \(=\) \(\ds \innerprod {T x} {\alpha u + v}\)
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod {T x} u + \innerprod {T x} v\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \overline \alpha \innerprod x {T^\ast u} + \innerprod x {T^\ast v}\)
\(\ds \) \(=\) \(\ds \innerprod x {\alpha T^\ast u + T^\ast v}\) Inner Product is Sesquilinear

Since $\map D T$ is dense in $\HH$, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ converging to $\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v}$.

We have:

$\innerprod {x_n} {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } = 0$ for each $n \in \N$.

So, from Inner Product is Continuous, we have:

$\innerprod {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } = 0$

So, from the definition of the inner product, we have:

$\map {T^\ast} {\alpha u + v} = \alpha T^\ast u + T^\ast v$


Now, we look to uniqueness.

Suppose for some $y \in \map D {T^\ast}$ that $z_y^{(1)}, z_y^{(2)}$ are such that:

$\innerprod {T x} y = \innerprod x {z_y^{(1)} } = \innerprod x {z_y^{(2)} }$ for $y \in \map D {T^\ast}$ and all $x \in \map D T$.

Then, we would have:

$\innerprod x {z_y^{(1)} - z_y^{(2)} } = 0$ for each $x \in \map D T$.

Since $\map D T$ is dense, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ converging to $z_y^{(1)} - z_y^{(2)}$.

Then:

$\innerprod {x_n} {z_y^{(1)} - z_y^{(2)} } = 0$ for each $n \in \N$.

Then, taking $n \to \infty$, we have:

$\innerprod {z_y^{(1)} - z_y^{(2)} } {z_y^{(1)} - z_y^{(2)} } = 0$

from Inner Product is Continuous.

So, from the definition of the inner product, we have:

$z_y^{(1)} = z_y^{(2)}$

justifying that $T^\ast$ is unique.

$\blacksquare$


Sources

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