Joan Lindsay Orr

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Unbounded Operators

Notes on basic properties of unbounded operators.


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

The Adjoint

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 \(T\).