THE MEASURABLE FUNCTIONAL CALCULUS

In this note, we will outline a proof of the measurable functional calculus for bounded self-adjoint operators. The result we want (Theorem 2 below) is stated as Theorem VII.2 of Reed & Simon's Methods of Modern Mathematical Physics. We will assume the continuous functional calculus (Theorem VII.1):

Theorem 1 (Continuous Functional Calculus) Let A be a self-adjoint operator on a Hilbert space ${\cal H}$. Then there is a unique map $\phi : C(\sigma(A)) \to {\cal L}({\cal H})$ with the following properties:

(a)

$\phi$ is an algebraic $\ast$-homomorphism; that is

(b)

Let f be the function f(t) = t; then $\phi(f) = A$.

Moreover, $\phi$ has the following properties:

(c)

$\Vert \phi(f)\Vert = \Vert f\Vert_\infty$.

(d)

If $A x = \lambda x$, then $\phi(f) x = f(\lambda) x$.

(e)

If $f \geq 0$, then $\phi(f) \geq 0$.

(f)

$\sigma(\phi(f)) = \{ f(\lambda) : \lambda \in \sigma(A) \}$ (Spectral Mapping Theorem.)

(The operator $\phi(f) \in {\cal L}({\cal H})$ is usually denoted by f(A).) Our aim is to extend the map $\phi$ to a map $\hat{\phi}: {\cal B}(\sigma(A)) \to {\cal L}({\cal H})$ with essentially the same properties, and thus to define f(A) for any Borel function f on $\sigma(A)$. We begin with a proposition about Borel functions which is of some independant interest. Let us say that a sequence of functions $\{ f_n \}$converges strongly to f on a set E, if $f_n \to f$pointwise on E and $\Vert f_n\Vert_\infty$ is bounded.

Lemma 1 Let $C( {{\rm I}\kern-.18em{\rm R}})$ denote the algebra of bounded, continuous functions on ${{\rm I}\kern-.18em{\rm R}}$, and ${\cal B}( {{\rm I}\kern-.18em{\rm R}})$ the algebra of bounded, Borel functions.

(a)

${\cal B}( {{\rm I}\kern-.18em{\rm R}})$ is the smallest algebra of functions which contains $C( {{\rm I}\kern-.18em{\rm R}})$ and is closed under strong limits.

(b)

${\cal B}( {{\rm I}\kern-.18em{\rm R}})$ is the smallest vector space of functions which contains $C( {{\rm I}\kern-.18em{\rm R}})$ and is closed under strong limits.

Proof (a) Let ${\cal A}$ denote the smallest strongly-closed algebra containing $C( {{\rm I}\kern-.18em{\rm R}})$. Certainly ${\cal A} \subseteq {\cal B}( {{\rm I}\kern-.18em{\rm R}})$. It is easy to see that ${\cal A}$contains all characteristic functions of open intervals. Now let ${\cal C}$ denote the collection of subsets of ${{\rm I}\kern-.18em{\rm R}}$ whose characteristic functions are in ${\cal A}$. Since ${\cal A}$ is an algebra, ${\cal C}$ is an algebra of sets (that is, closed under complements and finite intersections); since ${\cal A}$is closed under strong limits, it follows that ${\cal C}$ is in fact a $\sigma$-algebra, so that ${\cal C}$ contains all Borel sets. It follows easily that ${\cal A} \supseteq {\cal B}( {{\rm I}\kern-.18em{\rm R}})$.

(b) Let ${\cal V}$ denote the smallest strongly-closed vector space containing $C( {{\rm I}\kern-.18em{\rm R}})$. Certainly ${\cal V} \subseteq {\cal B}( {{\rm I}\kern-.18em{\rm R}})$. Now define

$${\cal V}_1 := \{ f \in {\cal V} : fg \in {\cal V} \quad \forall g \in C( {{\rm I}\kern-.18em{\rm R}})\} ,$$

and observe that ${\cal V}_1$ is a vector space which contains $C( {{\rm I}\kern-.18em{\rm R}})$ and is closed under strong limits. Thus ${\cal V}_1 = {\cal V}$. Similarly, define

$${\cal V}_2 := \{ f \in {\cal V} : fg \in {\cal V} \quad \forall g \in {\cal V} \} ;$$

then ${\cal V}_2$ is a vector space which contains $C( {{\rm I}\kern-.18em{\rm R}})$ (since ${\cal V}_1 = {\cal V}$) and is closed under strong limits. Thus ${\cal V}_2 = {\cal V}$. Hence ${\cal V}$ is an algebra, and the result follows from part (a)

We will actually apply the above lemma with ${{\rm I}\kern-.18em{\rm R}}$ replaced by $\sigma(A)$; since $\sigma(A)$ is closed, it can easily be checked that the lemma, and its proof, are valid in this case.

We now show how to construct the required extension $\hat{\phi}$of $\phi$. Theorem 1 shows that, for fixed $x, y \in {\cal H}$, the map $f \mapsto \langle x ,\phi(f)y\rangle$ is a bounded linear functional on $C(\sigma(A))$. It follows from the Riesz representation theorem that there is a (complex) measure $\mu_{x,y}$ on $\sigma(A)$ such that  

$$\langle x ,\phi(f)y\rangle = \int_{\sigma(A)} f(\lambda)\,d\mu_{x,y} (\lambda) ,$$ (1)

for all $f \in C(\sigma(A))$. The integral in (1) is defined for any Borel function f, and so we use it as the definition of $\langle x ,\hat{\phi}(f)y\rangle$in this case. We will write $\mu_x$ for $\mu_{x,x}$; note that $\mu_x$ is actually a positive measure. (The above construction is usually done by observing that $f \mapsto \langle x ,\phi(f)x\rangle$ is a positive linear functional and deducing the existence of $\mu_x$ first, but a rather subtle polarisation argument is then required to find $\mu_{x,y}$.) Since $\phi(f)$ is known to belong to ${\cal L}({\cal H})$ when f is continuous, it is easy to see that the map $(x,y) \mapsto \mu_{x,y}$ is sesquilinear; this in turn shows that $\hat{\phi}(f) \in {\cal L}({\cal H})$ for all $f \in {\cal B}(\sigma(A))$.In summary: $\hat{\phi}$ is a well-defined map from ${\cal B}(\sigma(A))$ into ${\cal L}({\cal H})$.)

Our next lemma will be strengthened by part (c) of Theorem 2, but it seems to be a necessary preliminary step in the proof.

Lemma 2 Suppose $\{ f_n \}$ is a sequence of Borel functions with $f_n \to f$ strongly on $\sigma(A)$. Then $\hat{\phi}(f_n) \to \hat{\phi}(f)$ in the weak operator topology in ${\cal L}({\cal H})$.

Proof For fixed x and y in ${\cal H}$, we have

We can now state and prove our main theorem.

Theorem 2 (Measurable Functional Calculus) Let A be a bounded self-adjoint operator on ${\cal H}$. There is a unique map $\hat{\phi}: {\cal B}( {{\rm I}\kern-.18em{\rm R}})\to {\cal L}({\cal H})$ with the following properties:

(a)

$\hat{\phi}$ is an algebraic $\ast$-homomorphism.

(b)

Let f be the function f(t) = t; then $\hat{\phi}(f) = A$.

(c)

If $f_n \to f$ strongly on $\sigma(A)$, then $\hat{\phi}(f_n) \to \hat{\phi}(f)$ in the strong operator topology in ${\cal L}({\cal H})$.

Moreover, $\hat{\phi}$ has the following properties:

(d)

$\hat{\phi}$ is norm continuous: $\Vert \hat{\phi}(f)\Vert \leq \Vert f\Vert_\infty$.

(e)

If $A x = \lambda x$, then $\hat{\phi}(f) x = f(\lambda) x$.

(f)

If $f \geq 0$, then $\hat{\phi}(f) \geq 0$.

(g)

If AB = BA, then $\hat{\phi}(f)B = B\hat{\phi}(f)$.

Proof Uniqueness is (almost) obvious, so the proof comes down to showing that the map $\hat{\phi}$ we have described above has the required properties. We first observe that part (a) is obvious from the definition of $\hat{\phi}$, except perhaps for the multiplicative property $\hat{\phi}(fg) = \hat{\phi}(f)\hat{\phi}(g)$. To see this, first define

$${\cal B}_1 := \{ f \in {\cal B}(\sigma(A)) : \hat{\phi}(fg)... ...phi}(f)\hat{\phi}(g) \ \quad \forall g \in C(\sigma(A)) \} .$$

From Theorem 1, we know that ${\cal B}_1 \supseteq C(\sigma(A))$; it is clear that ${\cal B}_1$ is a vector space; and ${\cal B}_1$ is closed under strong limits, because if $g \in C(\sigma(A))$, $f_n \in {\cal B}_1$ and $f_n \to f$ strongly on $\sigma(A)$, then $\hat{\phi}(f_n g) = \hat{\phi}(f_n) \hat{\phi}(g)$, and we can use Lemma 2 to let $n \to \infty$ in this equation, obtaining $\hat{\phi}(fg) = \hat{\phi}(f)\hat{\phi}(g)$. It follows from Lemma 1 that ${\cal B}_1 = {\cal B}(\sigma(A))$. Now define

$${\cal B}_2 := \{ f \in {\cal B}(\sigma(A)) : \hat{\phi}(fg)... ...)\hat{\phi}(g) \quad \forall g \in {\cal B}(\sigma(A)) \} .$$

${\cal B}_2 \supseteq C(\sigma(A))$ since ${\cal B}_1 = {\cal B}(\sigma(A))$; as with ${\cal B}_1$, we see that ${\cal B}_2$ is a vector space closed under strong limits. Hence ${\cal B}_2 = {\cal B}(\sigma(A))$, which completes the proof of part (a) of the theorem.

We turn now to part (d). We must show that if $f_n \to f$ strongly on $\sigma(A)$, then $\hat{\phi}(f_n) x \to \hat{\phi}(f) x$ (in ${\cal H}$-norm) for each $x \in {\cal H}$. But

which goes to by dominated convergence. This proves (d); with (a) and (d) proved, the remaining parts of the theorem follow easily from Theorem 1

We conclude with a few words about the spectral mapping theorem in the Borel case. It is easy to see that it cannot be strictly true, because $\hat{\phi}(f) = \hat{\phi}(g)$ whenever f and g agree almost everywhere with respect to each of the measures $\mu_x$. Let V be the largest open subset of ${\rm C}\kern-.47em {\vrule height1.4ex width.08em depth-.04ex} \;\;$ with the property that $\mu_x (f^{-1}(V)) = 0$ for every $x \in {\cal H}$, and define the essential range of f to be the complement of V. (Note that with this definition, the essential range of f depends on A as well as f.) One can then show that $\sigma(f(A))$ is the essential range of f, for any $f \in {\cal B}(\sigma(A))$.We can also define the essential supremum of f, $\Vert f\Vert_\infty'$, as the supremum (in fact, maximum) of $\vert \lambda\vert$ as $\lambda$ runs through the essential range of f. Part (d) of the Theorem can then be strengthened to the equality $\Vert \hat{\phi}(f)\Vert = \Vert f\Vert_\infty'$.