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Section 4.6 Closed operators in Hilbert spaces

When \(X\) is a Hilbert space, there is a further actor in the closedness business.

As we will see in Section 8.2, in a Hilbert space \((H,\langle\cdot,\cdot\rangle)\) to every densely defined operator \(A:D(A)\subset H\to H\) one can associate another linear operator \(A^*:D(A^*)\subset H\to H\) so that
\begin{equation*} \langle Ax,y\rangle = \langle x,A^*y\rangle\text{ for all } x\in D(A), y\in D(A^*). \end{equation*}
In case of finite-dimensional vector spaces, \(A^*\) is precisely the transposed matrix of \(A\) (Hermition conjugate in case of complex vector spaces).
(1) is a direct consequence of the fact that \(\Gamma_{A^*}\) is the orthogonal space, in \(H\times H\text{,}\) of the set \(V=\{x\in H\,:\,(-Ax,x)\}\text{.}\)

Indeed, let \((u,v)\in V^\perp\subset H\times H\text{.}\) Then
\begin{equation*} \langle(u,v),(-Ax,x)\rangle_{H\times H} = \langle u,-Ax\rangle_H + \langle v,x\rangle_H = 0\text{ for all }x\in X. \end{equation*}
By definition of \(A^*\text{,}\) this means precisely that \(u\in D(A^*)\) and \(v=A^*u\text{,}\) namely that \((u,v)\in\Gamma_{A^*}\text{.}\)

Since orthogonal spaces are always closed, \(A^*\) is a closed operator.

(2) Assume first that \(A\) is closable. Then \(\overline{\Gamma(A)}\) is the graph of a closed operator \(\bar A\text{,}\) the closure of \(A\text{.}\) From (1), we also know that \(\Gamma_{A^*} = V^\perp\text{.}\) Now, suppose that \(A^*\) is not dense. Then there is a \(w\in D(A^*)^\perp\text{,}\) with \(w\neq0\text{.}\) Then
\begin{equation*} \langle(w,0),(x,A^*x)\rangle_{H\times H} = \langle w,x\rangle_H = 0\text{ for all }x\in X, \end{equation*}
namely
\begin{equation*} (w,0)\in\Gamma_{A^*}^\perp = (V^\perp)^\perp=\overline{V}. \end{equation*}
Hence, there is a sequence \(x_n\in D(A)\) such that \(x_n\to 0\) and \(Ax_n\to-w\text{.}\) Since \(A\) is closable, this would mean that \(w=0\text{,}\) against the assumption.

Assume now that \(D(A^*)\) is dense in \(H\text{.}\) We need to prove that \((0,y)\in \overline{\Gamma(A)}\) implies \(y=0\text{.}\) Now, if \((0,y)\in \overline{\Gamma(A)}\text{,}\) then there is a sequence \(x_n\in D(A)\) with \(x_n\to0\) and \(Ax_n\to y\text{.}\) Recall that \(w\in D(A^*)\) if
\begin{equation*} \langle w,Ax\rangle_{H} = \langle A^*w,x\rangle_H \text{ for all }x\in X. \end{equation*}
Hence, in particular,
\begin{equation*} \langle w,Ax_n\rangle_{H} = \langle A^*w,x_n\rangle_H \text{ for all }n\geq1 \end{equation*}
so that, in the limit for \(n\to\infty\text{,}\)
\begin{equation*} \langle w,y\rangle_{H} = \langle A^*w,0\rangle_H = 0. \end{equation*}
Since this is true for every \(w\in D(A^*)\text{,}\) this means that \(y\in D(A^*)^\perp\text{,}\) against the hypothesis that \(D(A^*)\) is dense in \(H\text{.}\)

(3) TBA
Consider, for \(k=1,2,\dots\text{,}\) the operators
\begin{equation*} T_k:L^2([0,1])\to L^2([0,1]), \end{equation*}
\begin{equation*} D(T_k)=\{f\in C^k([0,1])\,:\,f(0)=f(1)=0\} \end{equation*}
defined by
\begin{equation*} T_k(f) = \frac{d}{dx}f. \end{equation*}
Then
\begin{equation*} T_1\supset T_2\supset T_3\supset\dots. \end{equation*}
Moreover, as already discussed in the case \(k=1\text{,}\) each \(T_k\) is closable but none of them is closed. On the other side,
\begin{equation*} \overline{T_1}=\overline{T_2}=\overline{T_3}=\dots. \end{equation*}
This operator, that we denote by \(\overline{T}\text{,}\) is \(\frac{d}{dx}\) with domain \(D(\overline{T})=H^1_0([0,1])\text{.}\)

On the other side,
\begin{equation*} T_1^*=T_2^*=T_3^*=\dots. \end{equation*}
This operator, that we denote by \(T^*\text{,}\) is \(\frac{d}{dx}\) with domain \(D(T^*)=H^1([0,1])\text{.}\) Notice that \(T^{**}=\overline T\text{.}\)
Definition 4.6.3.
Let \(A:D(A)\longrightarrow H\) on a Hilbert space \(H\text{.}\) Assume that \(D(A)\) is dense in \(H\text{,}\) so that \(A^*\) is uniquely defined. \(A\) is said:
  • symmetric, if \(A\subset A^*\text{,}\) namely if \(D(A)\subset D(A^*)\) and
    \begin{equation*} \langle Ax,y\rangle = \langle x,Ay\rangle\text{ for all }x,y\in D(A). \end{equation*}
  • self-adjoint, if \(A=A^*\text{,}\) i.e. if and only if \(A\) is symmetric and \(D(A)=D(A^*)\text{;}\)
  • essentially self-adjoint, if \(\bar A\) is self-adjoint.

Below we summarize the relations between closure and symmetry of an operator:
  • \(A\) symmetric: \(A\subset\bar A=A^{**} \subset A^*\text{;}\)
  • \(A\) symmetric and closed: \(A=\bar A=A^{**} \subset A^*\text{;}\)
  • \(A\) essentially self-adjoint: \(A\subset\bar A = A^{**} = A^*\text{;}\)
  • \(A\) self-adjoint: \(A=\bar A = A^{**} = A^*\text{;}\)

In particular:
  1. If \(A\) is closable, \(A^{**}=\overline{A}\text{.}\)
  2. A symmetric operator is always closable.
  3. A self-adjoint operator is always closed.
If \(k(x,y)=k(y,x)\text{,}\) then \(K\) is symmetric on \(L^2(\Omega)\)

\begin{align*} (Kf,g)\amp=\amp\int \int k(x,y)f(y)g(x)dydx\\ \amp=\amp\int\int f(y)k(y,x)g(x)dxdy\\ \amp=\amp(f,Kg). \end{align*}

Consider
\begin{equation*} A=-\frac{d^2}{dx^2}, \quad D(A)=\left\{f\in L^2[0,1], \text{ } Af\in L^2[0,1],\text{ } \frac{df}{dx}(0)=0,\text{ } \frac{df}{dx}(1)=0\right\} \end{equation*}
Then
\begin{align*} (Au,v)\amp=\amp\int_0^1-\frac{d^2}{dx^2}u(x)v(x)dx\\ \amp=\amp-\frac{d}{dx}u v\Big|_0^1+\int_0^1\frac{d}{dx}u\frac{d}{dx}v dx\\ \amp=\amp u\frac{d}{dx}v\Big|_0^1-\int_0^1u\frac{d^2}{dx^2}vdx\\ \amp=\amp(u,Av), \end{align*}
hence \((A, D(A))\) is symmetric.

(1) Let \(x_n\in D(A)\) with \(x_n\to 0\) and \(A x_n\to w\text{.}\) Then every \(x_n\) is a finite linear combinations of \(e_i\) so that
\begin{equation*} \lim_{n\to\infty}\langle x_n,e_i\rangle = 0, \; \lim_{n\to\infty}\langle Ax_n,e_i\rangle = \lambda_i\langle x_n,e_i\rangle = \langle w,e_i \rangle. \end{equation*}
Hence \(\langle w,e_i \rangle = 0\) for all \(i=1,2,\dots\text{,}\) so that \(w=0\text{.}\)

(2) Let \(x\in D(\overline{A})\text{.}\) Then there is a sequence \(x_n\in D(A)\) such that \(x_n\to x\) and \(A x_n\to y\text{.}\) Then
\begin{equation*} \lim_{n\to\infty}\langle x_n,e_i\rangle = \langle x,e_i\rangle,\;\lim_{n\to\infty}\lambda_i\langle x_n,e_i\rangle=\lim_{n\to\infty}\langle A x_n,e_i\rangle = \langle y,e_i\rangle. \end{equation*}
Since \(x,y\in H\text{,}\) we have that
\begin{equation*} \sum_{i=1}^\infty |\langle x,e_i\rangle|^2\infty,\;\sum_{i=1}^\infty |\lambda_i|\cdot|\langle x,e_i\rangle|^2\infty. \end{equation*}
We now prove that this condition is also sufficient. So, let \(x=\sum_{i=1}^\infty c_i e_i\in H\) such that
\begin{equation*} \sum_{i=1}^\infty |c_i|^2\infty,\;\sum_{i=1}^\infty |\lambda_i|\cdot|c_i|^2\infty \end{equation*}
and set
\begin{equation*} x_n = \sum_{i=1}^n c_i e_i,\;y_n = A x_n = \sum_{i=1}^n \lambda_i c_i e_i,\;y=\sum_{i=1}^\infty \lambda_i c_i e_i. \end{equation*}
Then
\begin{equation*} x_n \to x\text{ and }Ax_n\to y. \end{equation*}
Hence, \(x\in D(\overline{A})\) and \(\overline{A}(x)=y\text{.}\)

Definition 4.6.7.
An operator \(A\) on a Hilbert space is positive if there is a constant \(k>0\) such that
\begin{equation*} (Au,u)\geq k\|u\|^2, \quad u\in D(A) \end{equation*}
In this case, all its eigenvalues are positive (see Spectral Theory in Chapter Chapter 8).

If \(k(x,y)\geq \delta >0\text{,}\) then \(K\) is positive on \(L^2_+\text{.}\)
\begin{align*} (Ku,u) \amp=\amp \int \int k(x,y)u(x)u(y)dxdy\\ \amp\geq\amp\delta \int\int u(x)u(y) dx dy\\ \amp\geq\amp\delta\int\int_{\{y=x\}}u(x)u(y)dydx\\ \amp=\amp\delta \|u\|_2^2. \end{align*}

\(A=-\frac{d^2}{dx^2}, \quad D(A)\) as above is positive.
\begin{align*} (Au,u)\amp=\amp\int_0^1 -\frac{d^2}{dx^2}u(x)u(x)dx\\ \amp=\amp\int_0^1 \frac{d}{dx}u\frac{d}{dx}u dx\\ \amp=\amp\int_0^1\Big|\frac{d}{dx}u\Big|^2dx\\ \text{(Poincar\'e Inequality)} \amp\geq\amp \int_0^1 | u|^2dx\\ \amp=\amp\|u\|^2. \end{align*}

Consider \(x_n\longrightarrow x\) with \(x_n\in D(A)\) and \(Ax_n\longrightarrow y\) in \(Y\text{.}\) We define the extension as follows
\begin{equation*} D(\hat{A}):=\left\{x\in X: \mbox{ there exists a sequence } \{x_n\}\subset D(A): x_n\longrightarrow x \right\}. \end{equation*}
Define \(\hat{A}\) by \(y=\hat{A}x\text{.}\) \(\hat{A}\) is well-defined from the above limits. If there are two sequences \(x_n\rightarrow x\) and \(x_n^*\rightarrow x\) with \(Ax_n\rightarrow y\text{,}\) \(A x_n^*\rightarrow y^*\) then for each \(u\in D(A)\) we have
\begin{align*} (y^*-y,u)\amp=\amp \lim_{n\rightarrow \infty} (Ax_n^*-Ax_n,u)\\ \amp=\amp\lim_{n\rightarrow \infty} (x_n^*-x_n,Au)\\ \amp=\amp0 \quad \mbox{for all }\quad u\in D(A). \end{align*}
Since \(D(A)\) is dense it follows that \(y^*=y\) and the extension \(\hat{A}\) is well defined and \(x\in D(\hat{A})\) and \(\hat{A}\) is closed.

Subsection 4.6.1 An important example: a symmetric operator with uncountably many self-adjoint extensions

In this example we use notations and results from Subsection 4.5.4. Denote by \(A\) the first-derivative operator
\begin{equation*} i\frac{d}{dx}:C_c^\infty((0,1))\subset L^2([0,1])\to L^2([0,1]). \end{equation*}
One can check directly with an integration by parts that \(A\) is symmetric. However, \(A\) is not self-adjoint. Indeed, \(A^*\) satisfies
\begin{equation*} \langle Af,g\rangle = \langle f,A^*g\rangle\text{ for all }f\in D(A),\,g\in D(A^*). \end{equation*}
Integrating by parts, we also see that
\begin{equation*} \langle Af,g\rangle - \langle f,A^*g\rangle = \int_{[0,1]}(f'(s) g(s)-f(s) g'(s))ds = \overline{f(1)}g(1)-\overline{f(0)}g(0). \end{equation*}
Since \(D(A)=C_c^\infty((0,1))\text{,}\) this quantity is zero no matter what the values of \(g(1)\) and \(g(0)\) are. This is why, indeed, \(D(A^*)=H^1([0,1])\text{,}\) namely \(A=W\) (see Subsection 4.5.4). The closure of \(A\) is equal to the operator \(A^{**}\text{,}\) that is not self-adjoint since \((A^{**})^*=A^*\text{.}\)

Among the uncountably many closed extensions \(A_\lambda\) of \(A\text{,}\) instead, there are self-adjoint operators. These are precisely those with \(|\lambda|=1\text{.}\) Indeed, recall that \(H^1_0([0,1])\subsetneq D_\lambda\subsetneq H^1([0,1])\) for every \(\lambda\in\bC\text{.}\) Then, for \(f\in D_\lambda\setminus H^1_0([0,1])\text{,}\) we get that
\begin{equation*} 0=\langle A_\lambda f,f\rangle - \langle f,A_\lambda f\rangle = |f(1)|^2-|f(0)|^2. \end{equation*}
Hence, \(|f(1)|^2=|f(0)|^2\text{,}\) so that \(|\lambda|^2=1\text{.}\) Now, let \(f\in D_\lambda\) and \(g\in D(\overline{A_\lambda})\text{.}\) Then
\begin{equation*} 0=\langle A_\lambda f,g\rangle - \langle f,A^*_\lambda g\rangle = \bar\alpha\overline{f(0)}g(1) - \overline{f(0)}g(0), \end{equation*}
so that \(g\) must satisfy \(g(1)=\lambda g(0)\) as well. Hence, \(A^*_\lambda=A_\lambda\text{.}\)