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# Unbounded Operators

## Definitions

## Hilbert Spaces

### The Adjoint

Notes on basic properties of unbounded operators.

Definition 1. 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, \(D\).

Definition 2. An unbounded operator \(T\) is said to be *densely defined* if \(D(T)\)
is dense in \(X\).

Definition 3. 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 \(y=Tx\).

Example 4. 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.

Example 5. 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 \(g = Tf\).

Definition 6. 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 \(\Gamma(S)\subseteq\Gamma(T)\).

Definition 7. 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 \(a_1 \overline{T}(x_1) + a_2 \overline{T}(x_2) = \overline{T}(a_1 x_1 + a_2 x_2)\).

Definition 8. 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.

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 \(y\in D(T^*)\), \[ \langle Tx, y\rangle = \phi_y(x) = \langle x, T^*y\rangle \]

Lemma 9. 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 \(x_0 = T^*y_0\).

Remark 10. 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 \(T^*\).

Definition 11. An unbounded operator from the Hilbert space \(\H\) to itself is *symmetric*
if \(T^*\) is an extension of \(T\). That is, if \(\Gamma(T)\subseteq\Gamma(T^*)\).

Definition 12. 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 \[ \int_0^1 f'(x) \overline{g(x)}\,dx = f(1)g(1) - f(0)g(0) - \int_0^1 f(x) \overline{g'(x)}\,dx \]

Example 13. 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.

Example 14. 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.

Example 15. 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

Lemma 16. Every symmetric operator is closeable.

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