Theorem 8.3.1.
Let \(A \in \L(V, W)\) where \(V, W\) are Hilbert spaces. There is a unique operator \(A\ad \in \L(W, V)\) such that
\begin{equation*}
\ip{Ax}{y}_W = \ip{x}{A\ad y}_V
\end{equation*}
for all \(x \in V, y \in W\text{.}\)
Section 8.3 Adjoint operators
Restricting to our attention to \(\C^n\) for the moment, let \(A\) be an \(m \times n\) matrix (where we think of \(A: \C^n \to \C^m\)). Let \(x \in \C^n\) and \(y \in \C^n\text{.}\) Denote by \(A\ad\) the Hermitian transpose (or conjugate transpose) of \(A\) (which if course acts in the other direction \(A: \C^m \to \C^n\)). Then in the inner product, we get the relationship
\begin{align*}
\ip{Ax}{y} \amp= y\ad (Ax)\\
\amp = (y\ad A) x \\
\amp= (A\ad y)\ad x\\
\amp=\ip{x}{A\ad y}.
\end{align*}
We seek to define a similar notion of the adjoint operator in the infinite dimensional case by way of the Riesz representation theorem.
Proof.
For any \(y \in W\text{,}\) the map
\begin{equation*}
x \mapsto \ip{Ax}{y}_W
\end{equation*}
is a continuous linear functional on \(V\text{.}\) Then Theorem 7.2.1 gives the existence of a unique vector \(z \in V\) so that
\begin{equation*}
\ip{Ax}{y}_W = \ip{x}{z}_V
\end{equation*}
for all \(x \in V\text{.}\) Then we define the map \(A\ad: W \to V\) by
\begin{equation*}
A\ad y = z.
\end{equation*}
It remains to show that \(A\ad\) is a continuous linear functional.
To see that \(A\ad\) is linear, choose \(\alpha, \beta \in \C\) and \(u, v \in W\text{.}\) Then for all \(x \in V\text{,}\) we have
\begin{align*}
\ip{x} {A\ad(\alpha u + \beta v)} \amp = \ip{Ax}{\alpha u + \beta v}\\
\amp= \cc\alpha \ip{Ax}{u} + \cc\beta\ip{Ax}{v}\\
\amp = \cc\alpha \ip{x}{A\ad u} + \cc\beta\ip{x}{A\ad v}\\
\amp= \ip{x}{ \alpha A\ad u + \beta A\ad v}.
\end{align*}
Since this holds for all \(x \in V\text{,}\) we must have
\begin{equation*}
A\ad(\alpha u + \beta v) = \alpha A\ad u + \beta A\ad v.
\end{equation*}
To see that \(A\ad\) is bounded, notice that for all \(y \in W,\)
\begin{align*}
\norm{A\ad y}^2 \amp= \ip{A\ad y}{A\ad y} \\
\amp=\ip{A A\ad y}{y}\\
\amp\leq \norm{A A\ad y}\norm{y}\\
\amp\leq \norm{A} \norm{A\ad y}\norm{y}.
\end{align*}
Dividing through, we get
\begin{equation*}
\norm{A\ad y} \leq \norm{A} \norm{y}
\end{equation*}
which gives that \(\norm{A\ad} \leq \norm{A}\text{,}\) and so \(A\ad\) is bounded.
Uniqueness follows from the observation that if \(A\) had two adjoints, then for all \(x \in V\text{,}\) we’d have
\begin{equation*}
\ip{x}{A_1\ad y} = \ip{x}{A_2\ad y},
\end{equation*}
and so \(A_1\ad = A_2\ad\text{.}\)
In matrices, \((A\ad)\ad = A\text{.}\) The same is true of operators on Hilbert spaces.
Theorem 8.3.2.
Suppose that \(V, W\) are Hilbert spaces and \(A \in \L(V, W)\text{.}\) Then \(A^{**} = A\) and \(\norm{A\ad} = \norm{A}\text{.}\)
Proof.
Suppose that \(A \in \L(V, W), B = A\ad \in \L(W, V)\text{,}\) and \(B\ad \in \L(V, W)\text{.}\) We claim that \(B\ad = A\text{.}\) For all \(x \in V, y \in W\text{,}\) we have
\begin{align*}
\ip{y}{B\ad x} \amp= \ip{By}{x}\\
\amp= \ip{A\ad y}{x}\\
\amp= \cc{\ip{x}{A\ad y}}\\
\amp= \cc{\ip{Ax}{y}}\\
\amp= \ip{y}{Ax}.
\end{align*}
Then \(B\ad x = Ax\) for all \(x\in V\) and so \(B\ad = A\text{.}\)
To see that \(\norm{A\ad} = \norm{A}\text{,}\) recall that the proof of the previous theorem established that \(\norm{A\ad} \leq \norm{A}\text{.}\) Applying this to \(A^{**}\) gives that \(\norm{A^{**}}\leq \norm{A\ad}\) and hence the result.
