# Joan Lindsay Orr

$$\newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\H}{\mathcal{H}} \newcommand{\e}{\epsilon} \newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}}$$

# Unbounded Operators

Notes on basic properties of unbounded operators.

## Definitions

Let $$X$$ and $$Y$$ be normed vector spaces. Formally, an unbounded operator is a 2-tuple $$(T, D)$$ where $$D\subseteq X$$ is a linear subspace of $$X$$ and $$T:D\rightarrow Y$$ is a linear map. In practice we refer to $$T$$ as the unbounded operator, and write $$D(T)$$ for its domain, $$D$$.

An unbounded operator $$T$$ is said to be densely defined if $$D(T)$$ is dense in $$X$$.

The graph, $$\Gamma(T)$$ of $$T$$, is defined to be $\Gamma(T) = \{(x, Tx) : x \in D(T)\}$ This is a linear subspace of $$X\oplus Y$$. If $$\Gamma(T)$$ is a closed subset of $$X\oplus Y$$, we say that $$T$$ is closed.

It doesn't matter which of the equivalent norms we put on $$X\oplus Y$$. If $$X$$ and $$Y$$ are Hilbert spaces we can use the Euclidean norm $$\sqrt{\|x\|^2 + \|y\|^2}$$; if $$X$$ and $$Y$$ are function spaces with the uniform norm, we can use $$\max\{\|x\|, \|y\|\}$$, and so forth.

An equiavlent condition is that $$T$$ is closed if, whenever there is a sequence $$x_n$$ in $$D(T)$$ such that $$x_n\rightarrow x$$ in $$X$$ and $$Tx_n\rightarrow y$$ in $$Y$$, then $$x\in D(T)$$ and $$y=Tx$$.

Let $$X = Y = C([0, 1])$$ and let $$D(T) = C^1([0, 1])$$ and $$(Tf)(x) = f'(x)$$. $$T$$ is densely defined by, for example, the Weierstrass Approximation Theorem, from which it follows that $$C^\infty([0, 1])$$ is dense in $$C([0, 1])$$. Moreover, $$T$$ is closed.

To see this, suppose $$f_n$$ is a sequence of continuously differentiable functions such that $$f_n$$ converges uniformly to $$f$$ and $$f_n'$$ converges uniformly to $$g$$. Define $G(x) = f(0) + \int_0^x g(t)\, dt$ By the Fundamental Theorem of Calculus $f_n(x) = f_n(0) + \int_0^x f_n'(t)\,dt$ and clearly for all $$x\in[0,1]$$ $|f_n(x) - G(x)| \le |f_n(0) - f(0)| + \int_0^1 |f'(t) - g(t)|\, dt \le \|f_n - f\| + \|f'_n - g\|$ Thus $$f_n$$ converges uniformly to $$G$$, hence $$f=G$$. Moreover, $$G$$ is differentiable and $$G'=g$$ (proof). Thus $$f\in C^1([0,1]) = D(T)$$ and $$Tf = f' = G' = g$$, as desired.

Let $$X = Y = L^2([0,1])$$ and let $$D(T)$$ be the set of (a.e.-equivalence classes of) absolutely continuous functions with derivative in $$L^2$$. (Absolutely continuous functions are differentiable a.e. with integrable derivatives, but we need the additional requirement that the derivatives are in $$L^2$$.) Define $$Tf = f'$$ on $$D(T)$$. To see that $$T$$ is closed, suppose that $$f_n\in D(T)$$ where $$\|f_n-f\|_2\rightarrow 0$$ and $$\|f_n'-g\|_2\rightarrow 0$$. Note that there is a subsequence $$f_{n_k}$$ which converges to $$f$$ pointwise almost everywhere. Thus for almost every $$a$$ and $$x$$ in $$[0,1]$$, \begin{align} f(x) - f(a) &= \lim_{k\rightarrow \infty} f_{n_k}(x) - f_{n_k}(a) \\ &= \lim_{k\rightarrow \infty} \int_a^x f'_{n_k} \\ &= \int_a^x g \\ \end{align} Fix $$a$$ such that this holds for almost every $$x$$ and define $\tilde{f}(x) = f(a) + \int_a^x g$ for all $$x\in[0,1]$$. Then $$\tilde{f}$$ is absolutely continuous and $$\tilde{f}' = g$$ almost everywhere. Since $$f = \tilde{f}$$ almost everywhere, it follows that $$f\in D(T)$$ and $$g = Tf$$.

Let $$S$$ and $$T$$ be unbounded operators from $$X$$ to $$Y$$. Say that $$T$$ is an extension of $$S$$ if $$D(S)\subseteq D(T)$$ and $$Tx = Sx$$ for all $$x\in D(S)$$. More concisely, $$T$$ is an extension of $$S$$ if $$\Gamma(S)\subseteq\Gamma(T)$$.

An unbounded operator is closeable if it has a closed extension. Equivalently, it is closeable if $$\Gamma(T)$$ is the graph of an unbounded operator, which we write as $$\overline{T}$$.

To show that an unbounded operator $$T$$ is closeable, it is enough to check that for each $$x\in X$$, there is at most one $$y\in Y$$ with $$(x,y)\in\overline{\Gamma(T)}$$. For $$\Gamma(T)$$ is a linear space, therefore $$\overline{\Gamma(T)}$$ is a linear space. Let $$D(\overline{T}) = P_X(\overline{\Gamma(T)})$$, which is thus a linear space. For each $$x\in D(\overline{T})$$ there is a unique $$y\in Y$$ with $$(x,y)\in\overline{\Gamma(T)}$$, by assumption; define $$\overline{T}(x)$$ to be this $$y$$. If $$x_1, x_2$$ are in $$D(\overline{T})$$ then by definition $$(x_i, \overline{T}(x_i))\in\overline{\Gamma(T)}$$ for $$i=1,2$$. Thus $$(a_1 x_1 + a_2 x_2, a_1 \overline{T}(x_1) + a_2 \overline{T}(x_2) \in \overline{\Gamma(T)}$$ which implies by definition that $$a_1 \overline{T}(x_1) + a_2 \overline{T}(x_2) = \overline{T}(a_1 x_1 + a_2 x_2)$$.

Let $$S$$ be an unbounded operator from $$X$$ to $$Y$$ and $$T$$ an unbounded operator from $$Y$$ to $$Z$$. Define $$D(TS)$$ to be the preimage $$S^{-1}(D(T))$$ and (of course) for $$x\in D(TS)$$, let $$TSx = T(S(x))$$.The composition is well-defined as a function because $$D(TS)$$ cannot be empty since it must contain $$0$$, but it need not necessarily contain anything else.

## Hilbert Spaces

Let $$\H_1$$ and $$\H_2$$ be Hilbert spaces and $$T$$ a densely defined unbounded operator from $$\H_1$$ to $$\H_2$$. For each $$y\in\H$$, define the linear map $$\phi_y:D(T)\rightarrow \CC$$ by $\phi_y : x\in D(T) \mapsto \langle Tx, y\rangle$ Let $$D^*$$ be the set of all $$y\in\H_2$$ for which $$\phi_y$$ is bounded. Clearly $$D^*$$ is a linear space. Since $$T$$ is assumed to be densely defined, for each $$y\in D^*$$ we can extend $$\phi_y$$ uniquely to a bounded linear functional on $$\H_1$$. By the Riesz Representation Theorem there is a unique $$z\in\H_1$$ such that $$\phi_y(x) = \langle x, z\rangle$$ for all $$x\in D(T)$$. Define $$D(T^*) = D^*$$ and for each $$y\in D(T^*)$$ define $$T^*y$$ to be the corresponding $$z$$ (which was uniquely derived from $$y$$). Thus when $$x\in D(T)$$ and $$y\in D(T^*)$$, $\langle Tx, y\rangle = \phi_y(x) = \langle x, T^*y\rangle$

Let $$T$$ be a densely defined unbounded operator from $$\H_1$$ to $$\H_2$$. Then $$T^*$$ is a closed unbounded operator from $$\H_2$$ to $$\H_1$$.

Proof. Let $$y_n\in D(T^*)$$ be such that $$y_n\rightarrow y_0$$ and $$T^*y_n\rightarrow x_0$$. We must show that $$y_0\in D(T^*)$$ and that $$T^*y_0 = x_0$$.

For $$x\in D(T)$$, \begin{align} \phi_{y_0}(x) &= \langle Tx, y_0 \rangle \\ &= \lim_{n\rightarrow\infty}\langle Tx, y_n \rangle \\ &= \lim_{n\rightarrow\infty}\langle x, T^*y_n \rangle \\ &= \langle x, x_0 \rangle \\ \end{align}

Thus $$\phi_{y_0}$$ is bounded on $$D(T)$$, and so $$y_0\in D(T^*)$$. Also, $$x_0$$ satisfies $$\phi_{y_0}(x) = \langle x, x_0 \rangle$$ for all $$x\in D(T)$$, and so $$x_0 = T^*y_0$$.

Clearly if $$T$$ extends $$S$$ then the condition for membership in $$D(T^*)$$ is more stringent than for membership in $$D(S^*)$$, and so $$D(T^*)\subseteq D(S^*)$$. Thus $$S^*$$ extends $$T^*$$.

An unbounded operator from the Hilbert space $$\H$$ to itself is symmetric if $$T^*$$ is an extension of $$T$$. That is, if $$\Gamma(T)\subseteq\Gamma(T^*)$$.

An unbounded operator from $$\H$$ to itself is self-adjoint if $$\Gamma(T)=\Gamma(T^*)$$.

In the following series of examples $$T$$ will be the normalized quantum momentum operator $$-i d/dx$$ (setting $$\hbar = 1$$). What changes between the examples is the domain defined for $$T$$. In all cases, $$D(T^*)$$ is the set of $$g\in L^2([0,1])$$ such that $f \mapsto \langle Tf, g \rangle = \int_0^1 -if'(x) \overline{g(x)}\,dx$ is bounded on $$D(T)$$. If $$f$$ and $$g$$ are both smooth then $\int_0^1 f'(x) \overline{g(x)}\,dx = f(1)g(1) - f(0)g(0) - \int_0^1 f(x) \overline{g'(x)}\,dx$

Let $$D(T)$$ be $$C^\infty([0,1])$$, the smooth functions on $$[0,1]$$. Taking $$g(x) \equiv 1$$, we see $\langle Tf, g \rangle = -i(f(1) - f(0))$ and this is clearly not bounded as $$f$$ ranges over the $$L^2$$-norm unit vectors in $$D(T)$$. Thus $$D(T)\not\subseteq D(T^*)$$, and $$T$$ is not symmetric.

Let $$D(T)$$ be the smooth functions $$f$$ on $$[0,1]$$ with $$f(0)=f(1)=0$$. If $$g$$ is any smooth function, $\langle Tf, g \rangle = i \int_0^1 f(x) \overline{g'(x)}\,dx = \int_0^1 f(x) (\overline{-ig'(x)})\, = \langle f, Tg \rangle$ This is clearly $$L^2$$-bounded as $$f$$ ranges over $$D(T)$$ and so we can conclude $$g\in D(T^*)$$. Thus $$D(T) \subsetneq C^\infty([0,1]) \subseteq D(T^*)$$ and so $$T$$ is symmetric but not self-adjoint.

Let $$X = Y = L^2([0,1])$$ and let $$D(T)$$ be the set of (a.e.-equivalence classes of) absolutely continuous functions with derivative in $$L^2$$ and satisfying $$f(0)=f(1)$$. If $$f$$ and $$g$$ are both in this set then the integrated term $$f(1)g(1)-f(0)g(0)$$ in the integration by parts vanishes, and $$\langle Tf, g \rangle = \langle f, Tg \rangle$$ as before. Thus $$D(T)\subseteq D(T^*)$$. To see the reverse inclusion, let $$g\in D(T^*)$$. Then there is an $$h$$ in $$L^2([0,1])$$ such that $\int_0^1 f'(x)\overline{g(x)}\, dx = - \int_0^1 f(x) \overline{h(x)}\, dx$ for all $$f\in D(T)$$. Fix $$x_0$$ in $$[0, 1)$$ and let $$f_n$$ be a sequence of smooth functions satisfying $$f_n(x)\ge 0$$ on $$[0,1]$$, $$f_n(x)=0$$ for $$x\not\in[x_0, x_0+\frac{1}{n}]$$, and $$\int_0^1 f_n(x)\, dx = 1$$. Let $$F_n(x) = \int_0^x f_n(t)\, dt$$. $\overline{g(x_0)} = \lim_{n\rightarrow\infty}\int_0^1 f_n(x)\overline{g(x)}\, dx = - \lim_{n\rightarrow\infty}\int_0^1 F_n(x) \overline{h(x)}\, dx = - \int_{x_0}^1 \overline{h(x)} \, dx$ (The RHS comverges by the DCT, and the LHS by a routine continuity argument.) Letting $$x_0$$ vary, $$g$$ is absolutely continuous and $$h$$ is equal (almost everywhere) to $$g'$$. But then, putting $$g$$ into the familiar integration by parts identity, $f(1)g(1) - f(0)g(0) = 0$ for any $$f$$ in $$D(T)$$, and so $$g(0) = g(1)$$ and $$g\in D(T)$$. It follows that $$D(T) = D(T^*)$$ and $$T$$ is self-adjoint

Every symmetric operator is closeable.

Proof. If $$T$$ is symmetric then $$\Gamma(T)\subseteq\Gamma(T^*)$$ and so $$T^*$$ provides a closed extension of $$T$$.