Section 8.2 Adjoint Operators
In this section we prove the spectral theorem for symmetric operators. For this we need the concept of an adjoint operator.Definition 8.2.1.
Let \(A:X\to Y\) be linear with dense domain \(D(A)\text{.}\) We define the adjoint operator on the dual spaces \(A^*: Y^*\to X^*\) by
\begin{equation*}
A^*(y)=w \qquad\mbox{if and only if} \qquad y(Au) = w(u), \qquad y\in Y^*, w\in X^*, u\in X,
\end{equation*}
i.e., \(A^* y\) has the action of \(y\circ A\) on \(u\in X\text{.}\) We call \((A^*, D(A^*))\) the adjoint operator, where \(D(A^*)\) denotes the natural domain of definition of \(A^*\text{.}\)Theorem 8.2.2.
Let \(A\in \mathcal{L}(X, Y)\text{.}\) Then \(D(A^*) = Y^* \) and \(A^*\) is bounded with
\begin{equation*}
\|A^*\|=\|A\|.
\end{equation*}
Proof.
[cross-reference to target(s) "HHH" missing or not unique] of the Hahn-Banach Theorem.Definition 8.2.3.
Let \(A:H_1\to H_2\) be a linear map between Hilbert spaces. Then \(A^*:H_2\to H_1\) such that
\begin{equation*}
(v, Au)_{H_2} = (A^* v, u)_{H_1} .
\end{equation*}
\((A^*, D(A^*))\) is called the Hilbert adjoint, or simply the adjoint of \(A\text{.}\)Theorem 8.2.4.
Assume \((A, D(A))\) is given on a Hilbert space \(H\) with a dense domain \(D(A)\text{.}\) Then
\begin{equation*}
(\mbox{Range}(A))^\perp = N(A^*).
\end{equation*}
If in addition \(\mbox{Range}(A)\) is closed, then the reverse is also true:
\begin{equation*}
\mbox{Range}(A) = N(A^*)^\perp.
\end{equation*}
Proof.
\begin{equation*}
\mbox{Range}(A) = \mbox{Range(A)}^{\perp\perp} = N(A^*)^\perp.
\end{equation*}
Definition 8.2.5.
A is symmetric if \((Au, v) = (u, Av)\) for all \(u,v\in D(A)\text{.}\) \(A\) is self-adjoint if it is symmetric and \(D(A) = D(A^*)\text{.}\)Example 8.2.6.
\begin{equation*}
D(A) = \{ u\in H_0^1(\Omega)\cap H^2(\Omega): u\Big|_{\partial \Omega}=0\}.
\end{equation*}
Then we have
\begin{equation}
(u, Av) = -\int_\Omega u \Delta v \, dx = \int_\Omega \nabla u \nabla v dx = -\int_\Omega \Delta u \; v\, dx = (Au, v). \tag{8.2.1}
\end{equation}
Hence \(A^*=A\) and we can chose \(D(A^*)=D(A)\) to obtain a self-adjoint operator.Example 8.2.7.
\begin{equation*}
(u, Av) = \int u v' dx = -\int u' v dx = - (Au, v).
\end{equation*}
Hence \(A^* = -A\) and \(D(A^*)=D(A)\) and we call \(A\) to be skew-adjoint. Proposition 8.2.8.
Assume \(A:X\to Y\) is a linear and surjective operator, which is bounded below \(\|Ax\|\geq \delta\|x\| >0\) for all \(x\in X\text{.}\) Then \(A^{-1}\) exists and is bounded.Proof.
\begin{equation}
\|A^{-1}\| = \sup_{y\in Y} \frac{\|A^{-1} y\|}{\|y \|}
= \sup_{x\in X} \frac{\|A^{-1}(Ax)\|}{\|Ax\|}
= \sup_{x\in X}\frac{\|x\|}{\|Ax\|}\leq \frac{1}{\inf_{x} \frac{\|Ax\|}{\|x\|}} \leq \frac{1}{\delta}. \tag{8.2.2}
\end{equation}
The inverse is well defined, since when there are two \(x_1, x_2\) with \(Ax_1 = Ax_2 =y\text{,}\) then
\begin{equation*}
\|x_1-x_2\| = \|A^{-1}y - A^{-1} y\| = 0,
\end{equation*}
since \(A^{-1}\) is bounded.Corollary 8.2.9.
Assume \(A-\lambda I\) is invertible and bounded below away from \(0\text{,}\) then
\begin{equation*}
\|R_\lambda(A) \| \leq \frac{1}{\| A-\lambda I\|}.
\end{equation*}
Theorem 8.2.10.
Let \((A, D(A))\) be a densely defined symmetric operator on a Hilbert space \(H\text{.}\) Then- \((Ax, x)\) is real for all \(x\in D(A)\text{.}\)
- All eigenvalues of \(A\) are real.
- Eigenvectors to distinct eigenvalues are orthogonal.
- The continuous spectrum is real: \(\sigma_c(A) \subset \RR\text{.}\)
Proof.
- On a complex Hilbert space we have \((a,b)=\overline{(b,a)}\text{.}\) Then\begin{equation*} \overline{(Ax, x)} = (x, Ax) = (Ax, x) \end{equation*}and \((Ax, x)\in \RR\text{.}\)
- Let \((\lambda, \varphi)\) be an eigenpair of \(A\) (i.e. \(\lambda\) is an eigenvalue and \(\varphi\) is a corresponding eigenvector). Then\begin{equation*} (A\varphi, \varphi) = \lambda \|\varphi\|^2 \end{equation*}and \(\lambda\) must be real.
- Let \((\lambda_1,\varphi_1), (\lambda_2,\varphi_2)\) be two eigenpairs. Then\begin{equation*} \lambda_1(\varphi_1, \varphi_2) = (A\varphi_1, \varphi_2) = (\varphi_1, A\varphi_2) = \lambda_2(\varphi_1, \varphi_2) \end{equation*}and either \(\lambda_1=\lambda_2\) or \(\varphi_1 \perp\varphi_2\text{.}\)
- Let \(\lambda\in \sigma_c(A)\) and \(\lambda = \gamma+i\mu\text{,}\) \(\gamma, \mu \in \RR\text{.}\) We want to show that \(\mu=0\text{.}\) We compute\begin{align*} \|(A-\lambda I) x\|^2 \amp=\amp (Ax-\gamma x-i\mu x, Ax-\gamma x-i\mu x)\\ \amp=\amp (Ax -\gamma x, Ax-\gamma x) + (i\mu x, \gamma x) + (i\mu x, i\mu x)\\ \amp\amp - (i\mu x, Ax) - (Ax, i\mu x) + (\gamma x, i \mu x)\\ \amp=\amp \|Ax-\gamma x\|^2 -i(\mu x, \gamma x) +i(\gamma x,\mu x)\\ \amp\amp + i \mu(x, Ax) - i \mu (Ax, x) + \mu^2 \|x\|^2\\ \amp=\amp\|Ax-\gamma x\|^2 +\mu^2\|x\|^2. \end{align*}If \(\mu>0\text{,}\) then \(\|(A-\lambda I)x\|\) is bounded below, away from \(0\text{.}\) Then, by Corollary Corollary 8.2.9, the inverse \(R_\lambda(A)\) exists and is bounded. But in this case \(\lambda\not\in \sigma_c(A)\text{,}\) which is a contradiction. Hence we must have \(\mu=0\) and \(\lambda\) is real.
