5 The Spectral Theorem

Functional Analysis by Walter Rudin 1991, extract from Chapter 12

It should perhaps be stated explicitly that the spectrum \(\sigma (T)\) of an operator \(T \in \mathcal{B}(H)\) will always refer to the full algebra \(\mathcal{B}(H)\). In other words, \(\lambda \in \sigma (T)\) if and only if \(T - \lambda I\) has no inverse in \(\mathcal{B}(H)\). Sometimes we shall also be concerned with closed subalgebras \(A\) of \(\mathcal{B}(H)\) which have the additional property that \(I \in A\) and \(T^* \in A\) whenever \(T \in A\). (Such algebras are sometimes called \(*\)-algebras.)

Let \(A\) be such an algebra, and suppose that \(T \in A\) and \(T^{-1} \in \mathcal{B}(H)\). Since \(TT^*\) is self-adjoint, \(\sigma (TT^*)\) is a compact subset of the real line (Theorem 12.15), hence does not separate \(\mathbb {C}\), and therefore \(\sigma _A(TT^*) = \sigma (TT^*)\), by the corollary to Theorem 10.18. Since \(TT^*\) is invertible in \(\mathcal{B}(H)\), this equality shows that \((TT^*)^{-1} \in A\), and therefore \(T^{-1} = T^(TT^*)^{-1}\) is also in \(A\).

Thus \(T\) has the same spectrum relative to all closed *-algebras in \(\mathcal{B}(H)\) that contain \(T\).

Theorem 12.23 will be obtained as a special case of the following result, which deals with normal algebras of operators rather than with individual ones.

Theorem 25 12.22

If \(A\) is a closed normal subalgebra of \(\mathcal{B}(H)\) which contains the identity operator \(I\) and if \(\Delta \) is the maximal ideal space of \(A\), then the following assertions are true:

There exists a unique resolution \(E\) of the identity on the Borel subsets of \(\Delta \) which satisfies

\begin{equation} \label{eq:1} T = \int _\Delta \widehat{T} \ dE \end{equation}
1

for every \(T \in A\), where \(\widehat{T}\) is the Gelfand transform of \(T\).
The inverse of the Gelfand transform (i.e., the map that takes \(\widehat{T}\) back to \(T\)) extends to an isometric *-isomorphism of the algebra \(L^\infty (E)\) onto a closed subalgebra \(B\) of \(\mathcal{B}(H)\), \(B\supset A\), given by

\begin{equation} \label{eq:2} \Phi f = \int _\Delta f \ dE \quad (f \in L^\infty (E)). \end{equation}
2

Explicitly, \(\Phi \) is linear and multiplicative and satisfies

\[ \Phi (\bar{f}) = (\Phi f)^*, \| \Phi f \| = \| f \| _{\infty } \quad (f \in L^\infty (E)). \]
\(B\) is the closure [in the norm topology of \(\mathcal{B}(H)\)] of the set of all finite linear combinations of the projections \(E(\omega )\).
If \(\omega \subset \Delta \) is open and nonempty, then \(E(\omega ) \neq 0\).
An operator \(S \in \mathcal{B}(H)\) commutes with every \(T \in A\) if and only if \(S\) com mutes with every projection \(E(\omega )\).

Proof ▶

Recall that 1 is an abbreviation for

\begin{equation} \label{eq:4} (Tx , y) = \int _\Delta \widehat{T} \ dE_{x,y} \quad (x,y \in H, T \in A). \end{equation}

Since \(\mathcal{B}(H)\) is a \(B^*\)-algebra (Section 12.9), our given algebra \(A\) is a commutative \(B^*\)-algebra. The Gelfand-Naimark theorem 11.18 asserts therefore that \(T \to \widehat{T}\) is an isometric *-isomorphism of \(A\) onto \(C(\Delta )\).

This leads to an easy proof of the uniqueness of \(E\). Suppose \(E\) satisfies 3. Since \(\widehat{T}\) ranges over all of \(C(\Delta )\), the assumed regularity of the complex Borel measures \(E_{x,y}\) shows that each \(E_{x,y}\) is uniquely determined by 3; this follows from the uniqueness assertion that is part of the Riesz representation theorem ([23], Th. 6.19) 11. Since, by definition, \((E(\omega )x, y) = E_{x,y}(\omega )\), each projection \(E(\omega ))\) is also uniquely determined by 3.

This uniqueness proof motivates the following proof of the existence of \(E\). If \(x \in H\) and \(y \in H\), Theorem 11.18 shows that \( \widehat{T} \mapsto (Tx, y)\) is a bounded linear functional on \(C(\Delta )\), of norm \(\leq \| x\| | \| y\| \), since \(\| \widehat{T}\| _{\infty } = \| T\| \). The Riesz representation theorem supplies us therefore with unique regular complex Borel measures \(\mu _{x,y}\) on \(\Delta \) such that

\begin{equation} \label{eq:5} (Tx , y) = \int _\Delta \widehat{T} \ d\mu _{x,y} \quad (x,y \in H, T \in A). \end{equation}

For fixed \( T \), the left side of 4 is a bounded sesquilinear functional on \( H \), hence so is the right side, and it remains so if the continuous function \( \widehat{T} \) is replaced by an arbitrary bounded Borel function \( f \). To each such \( f \) corresponds therefore an operator \( \Phi f \in \mathcal{B}(H) \) (see Theorem 12.8) such that

\begin{equation} \label{eq:6} ((\Phi f)x, y) = \int _{\Delta } f \ d\mu _{x,y} \quad (x, y \in H). \end{equation}

Comparison of 4 and 5 shows that \( \Phi \hat{T} = T \). Thus \( \Phi \) is an extension of the inverse of the Gelfand transform.

It is clear that \( \Phi \) is linear.

Part of the Gelfand-Naimark theorem states that \( T \) is self-adjoint if and only if \( \hat{T} \) is real-valued. For such \( T \),

\[ \int _{\Delta } \widehat{T} \ d\mu _{x,y} = (Tx, y) = (x, Ty) = \overline{(Ty, x)} = \overline{\int _{\Delta } \hat{T} d\mu _{y,x}}, \]

and this implies that \( \mu _{y,x} = \overline{\mu _{x,y}} \). Hence,

\[ ((\Phi \overline{f})x, y) = \int _{\Delta } \bar{f} \ d\mu _{x,y} = \overline{\int _{\Delta } f \, d\mu _{y,x}} = \overline{((\Phi f)y, x)} = (x, (\Phi f)y) \]

for all \( x, y \in H \), so that

\begin{equation} \label{eq:7} \Phi \bar{f} = (\Phi f)^*. \end{equation}

Our next objective is the equality

\begin{equation} \label{eq:8} \Phi (fg) = (\Phi f)(\Phi g) \end{equation}

for bounded Borel functions \( f, g \) on \( \Delta \). If \( S \in A \) and \( T \in A \), then \( (ST)^{\wedge } = \widehat{S} \widehat{T} \); hence

\[ \int _{\Delta } \hat{S} \hat{T} \ d\mu _{x,y} = (STx, y) = \int _{\Delta } \widehat{S} \ d\mu _{Tx,y}. \]

This holds for every \( \widehat{S} \in C(\Delta ) \); hence the two integrals are equal if \( \widehat{S} \) is replaced by any bounded Borel function \( f \). Thus

\[ \int _{\Delta } f \widehat{T} d\mu _{x,y} = \int _{\Delta } f \ d\mu _{Tx,y} = ((\Phi f)Tx, y) = (Tx, z) = \int _{\Delta } \hat{T} d\mu _{x,z}, \]

where we put \( z = (\Phi f)^* y \). Again, the first and last integrals remain equal if \( \widehat{T} \) is replaced by \( g \). This gives

\[ \begin{aligned} (\Phi (fg)x, y) & = \int _{\Delta } fg \ d\mu _{x,y} = \int _{\Delta } g \ d\mu _{x,z} \\ & = ((\Phi g)x, z) = ((\Phi g)x, (\Phi f)^* y) = (\Phi (f) \Phi (g)x, y), \end{aligned} \]

and 7 is proved.

We are finally ready to define \( E \): If \( \omega \) is a Borel subset of \( \Delta \), let \( \chi _{\omega } \) be its characteristic function, and put

\[ E(\omega ) = \Phi (\chi _{\omega }). \]

By 7, \( E(\omega \cap \omega ') = E(\omega )E(\omega ') \). With \( \omega ' = \omega \), this shows that each \( E(\omega ) \) is a projection. Since \( \Phi f \) is self-adjoint when \( f \) is real, by 6, each \( E(\omega ) \) is self-adjoint. It is clear that \( E(\emptyset ) = \Phi (0) = 0 \). That \( E(\Delta ) = I \) follows from 4 and 5. The finite additivity of \( E \) is a consequence of 5, and, for all \( x, y \in H \),

\[ E_{x,y}(\omega ) = (E(\omega )x, y) = \int _{\Delta } \chi _{\omega } \ d\mu _{x,y} = \mu _{x,y}(\omega ). \]

Thus 5 becomes 2. That \( \| \Phi f\| = \| f\| _{\infty } \) follows now from Theorem 12.21.

This completes the proof of (1) and (2).

Part (3) is now clear because every \( f \in L^{\infty }(E) \) is a uniform limit of simple functions (i.e., of functions with only finitely many values).

Suppose next that \( \omega \) is open and \( E(\omega ) = 0 \). If \( T \in A \) and \( \widehat{T} \) has its support in \( \omega \), 1 implies that \( T = 0 \); hence \( \widehat{T} = 0 \). Since \( \widehat{A} = C(\Delta ) \), Urysohn’s lemma implies now that \( \omega = \emptyset \). This proves (4).

To prove (5), choose \( S \in \mathcal{B}(H) \), \( x \in H \), \( y \in H \), and put \( z = S^* y \). For any \( T \in A \) and any Borel set \( \omega \subset \Delta \) we then have

\begin{equation} \label{eq:10} (STx, y) = (Tx, z) = \int _{\Delta } \widehat{T} \ dE_{x,z}, \end{equation}

\begin{equation} \label{eq:11} (TSx, y) = \int _{\Delta } \hat{T} \ dE_{Sx,y}, \end{equation}

\[ (SE(\omega )x, y) = (E(\omega )x, z) = E_{x,z}(\omega ), \]

\[ (E(\omega )Sx, y) = E_{Sx,y}(\omega ). \]

If \( ST = TS \) for every \( T \in A \), the measures in 8 and 9 are equal, so that \( SE(\omega ) = E(\omega )S \). The same argument establishes the converse.