Spectral Theorem III: general bounded operators via functional calculus

The finite-dimensional spectral theorem rests on two facts: every operator has an eigenvalue, because the characteristic polynomial has a root, and Gram–Schmidt induction terminates after finitely many steps. Neither survives in infinite dimensions. An operator on a Hilbert space can have no eigenvectors at all, and there is no dimension to induct on. This post rebuilds the spectral theorem for any bounded self-adjoint operator on a Hilbert space, replacing the eigenbasis wholesale with continuous functional calculus and a projection-valued measure. The series so far: Spectral Theorem I built the finite-dimensional theory; Spectral Theorem II handled compact operators, where eigenvectors survive but a discrete-sum eigendecomposition replaces the finite basis. This Part III handles the rest, and the Operator SVD capstone collects all three into the polar-form construction of the SVD.

The goal, and what breaks in infinite dimensions

The target is a spectral theorem for bounded self-adjoint operators: a decomposition of any self-adjoint $A \in \mathcal{B}(H)$ analogous to the orthonormal eigenbasis of Part I. The SVD application is collected separately in the Operator SVD capstone (given such a theorem applied to $L^* L$, the polar form $L = U|L|$ follows and the SVD comes from spectralizing $|L|$). This post is about the spectral theorem itself.

What breaks in infinite dimensions is the finite-dimensional proof. The argument there picked an eigenvalue (a root of the characteristic polynomial), peeled off its eigenspace, and inducted on dimension. In infinite dimensions:

• There is no characteristic polynomial.
• An operator can have no eigenvectors at all. The multiplication operator $(M_t f)(x) = x f(x)$ on $L^2[0, 1]$ is bounded and self-adjoint with spectrum $\sigma(M_t) = [0, 1]$, but $M_t f = \lambda f$ would force $f$ to be supported on the single point $\{x = \lambda\}$, a set of measure zero, so $f = 0$. Its point spectrum is empty.
• There is no finite dimension to induct on.

The eigenbasis that drove the finite-dimensional proof has to be replaced wholesale. The replacement, following Lax (2002, §31.2–31.3), is to build the functional calculus first and then extract a projection-valued measure from it. The rest of this post is that construction.

The adjoint: Banach-space definition, Hilbert specialization

The self-adjoint operators that the spectral theorem decomposes are defined by an inner-product identity, but the right level of generality for “adjoint” is one rung lower, on Banach spaces.

For a bounded operator $T: X \to Y$ between Banach spaces, the adjoint is the bounded operator $T^*: Y^* \to X^*$ defined by

$(T^* y^*)(x) := y^*(T x), \qquad x \in X,\ y^* \in Y^*.$

This is what “adjoint” means without extra structure: the dual map, the canonical morphism between dual spaces induced by $T$. It satisfies $\|T^*\| = \|T\|$ (Conway, 2007, Ch. VI).

When $X = Y = H$ is a Hilbert space, the Riesz representation theorem gives an antilinear isometry $H \to H^*$, $y \mapsto \langle \cdot,\, y\rangle$. Pulling the Banach adjoint $T^*: H^* \to H^*$ back through this isometry produces the familiar Hilbert-space adjoint $T^*: H \to H$ characterized by

$\langle T x,\, y\rangle = \langle x,\, T^* y\rangle, \qquad x, y \in H.$

The two are the same map under Riesz. The Hilbert form is what we use to write $L^* L$ and to talk about self-adjointness. The Banach form is what carries the compactness arguments in Part II, via Schauder’s theorem on $T^*(\overline{B_{Y^*}})$.

Functional calculus: from polynomials to continuous functions

For a bounded self-adjoint operator $A$ and a polynomial $p(t) = \sum_k c_k t^k$, the operator $p(A) = \sum_k c_k A^k$ is defined by ordinary algebra. The content is an estimate that controls it.

Three standard results combine into one estimate. First, $p(A)$ is normal: as a polynomial in the self-adjoint $A$ it commutes with its adjoint $\bar p(A)$. Second, for a normal operator the operator norm equals the spectral radius, $\|p(A)\| = \sup_{\mu \in \sigma(p(A))} |\mu|$ (Lax, 2002). Third, the spectral mapping theorem identifies that spectrum, $\sigma(p(A)) = p(\sigma(A))$. Together,

$\|p(A)\| = \max_{\lambda \in \sigma(A)} |p(\lambda)|.$

So $p \mapsto p(A)$ is an isometry from the algebra of polynomials, normed by the supremum over $\sigma(A)$, into the bounded operators $\mathcal{B}(H)$. The spectrum $\sigma(A)$ is compact, being closed and contained in $[-\|A\|, \|A\|]$, so by the Stone–Weierstrass theorem the polynomials are dense in $C(\sigma(A))$ under the supremum norm.

The extension to all of $C(\sigma(A))$ is the bounded linear extension theorem: a bounded linear map from a dense subspace of a normed space into a Banach space extends uniquely to the whole space, with the same operator norm. Here the dense subspace is the polynomials inside $C(\sigma(A))$, the Banach space is $\mathcal{B}(H)$, and the map is the isometry $p \mapsto p(A)$. So for $f \in C(\sigma(A))$, choose polynomials $p_n \to f$ uniformly on $\sigma(A)$ and set

$f(A) := \lim_{n \to \infty} p_n(A).$

The limit exists because $\{p_n(A)\}$ is Cauchy in the complete space $\mathcal{B}(H)$, and the extension theorem makes it independent of the approximating sequence. This is the continuous functional calculus: an isometric $*$-homomorphism $C(\sigma(A)) \to \mathcal{B}(H)$, $f \mapsto f(A)$, sending the identity function to $A$ and constants to scalar multiples of $I$.

The construction never mentions an eigenvector. It needs only the spectrum and the norm estimate, and the uniform approximation of continuous functions by polynomials stands in for the eigenbasis.

Resolution of identity: from functional calculus to a projection-valued measure

The functional calculus is a map from functions to operators. Riesz representation turns it into a measure.

Fix $x, y \in H$. The map $f \mapsto \langle f(A) x, y\rangle$ is a bounded linear functional on $C(\sigma(A))$, since $|\langle f(A) x, y\rangle| \le \|f(A)\| \, \|x\| \, \|y\| = \|f\|_\infty \|x\| \, \|y\|$. By the Riesz–Markov representation theorem there is a unique complex Borel measure $m_{x, y}$ on $\sigma(A)$ with

$\langle f(A) x, y\rangle = \int_{\sigma(A)} f \, dm_{x, y}.$

For each Borel set $S \subseteq \sigma(A)$, the number $m_{x, y}(S)$ is sesquilinear and bounded in $(x, y)$, so by the Riesz representation theorem for bounded sesquilinear forms there is a unique bounded operator $E(S)$ with $m_{x, y}(S) = \langle E(S) x, y\rangle$. The multiplicativity $f(A) g(A) = (fg)(A)$ of the functional calculus then forces the family $\{E(S)\}$ to be a projection-valued measure: each $E(S)$ is an orthogonal projection, $E(\emptyset) = 0$, $E(\sigma(A)) = I$, $E(S \cap T) = E(S) E(T)$, and $E$ is countably additive in the strong operator topology (Lax, 2002).

The strong operator topology

Convergence $T_n \to T$ in the strong operator topology means $T_n x \to T x$ for each fixed $x \in H$, which is weaker than norm convergence $\|T_n - T\| \to 0$. Concretely, for disjoint $S_1, S_2, \ldots$ the partial sums $\sum_{k=1}^N E(S_k)$ converge to $E(\bigcup_k S_k)$ vector by vector, but not in operator norm.

Setting $f \equiv 1$ recovers the resolution of identity $I = \int dE$; setting $f(\lambda) = \lambda$ gives the spectral resolution

$A = \int_{\sigma(A)} \lambda \, dE(\lambda).$

This is the spectral theorem for bounded self-adjoint operators. It requires no eigenvalues: the measure $E$ may have no atoms, in which case $A$ has no eigenvectors and the “eigenbasis” is genuinely a measure rather than a sum. The multiplication operator $M_t$ on $L^2[0, 1]$ is exactly this case. Its projection-valued measure is $E(S) =$ multiplication by the indicator $\mathbf{1}_S$, purely continuous with no atoms, and $M_t = \int_0^1 \lambda \, dE(\lambda)$.

Gelfand–Naimark: the structural summary

The construction above is one instance of a single structural fact.

The norm closure of the polynomials in $A$ is a commutative unital $C^*$-algebra $\mathcal{A}(A) \subseteq \mathcal{B}(H)$: commutative because $A$ commutes with itself, and a $C^*$-algebra because the operator norm satisfies the $C^*$-identity $\|x^* x\| = \|x\|^2$. The Gelfand–Naimark theorem (commutative version) states that every unital commutative $C^*$-algebra is isometrically $*$-isomorphic to $C(X)$, where $X$ is its character space, a compact Hausdorff space (Conway, 2007). For $\mathcal{A}(A)$ the character space is $\sigma(A)$, and the isomorphism $C(\sigma(A)) \to \mathcal{A}(A)$ is precisely the functional calculus $f \mapsto f(A)$.

So the functional calculus and the projection-valued measure are not ad hoc devices: they are the concrete face of the isomorphism $\mathcal{A}(A) \cong C(\sigma(A))$. The reason to state Gelfand–Naimark after the construction rather than before is that the construction is where the work actually happens, in the norm estimate and Weierstrass approximation of the functional calculus and the Riesz representation behind the projection-valued measure. The abstract isomorphism is the summary, not the engine.

Recovering the compact case: atomic spectral measures

The spectral theorem just proved holds for any bounded self-adjoint operator, but it produces an integral, not a sum. The discrete decomposition of Part II is the special case where the integral collapses.

For $A$ compact self-adjoint, Part II gives the discrete eigendecomposition

$A = \sum_n \lambda_n P_n, \qquad |\lambda_n| \to 0,$

with each $P_n$ a finite-rank orthogonal projection onto the $\lambda_n$-eigenspace. From the projection-valued-measure perspective above, this is the statement that the spectral measure $E$ of $A$ is purely atomic: each eigenvalue $\lambda_n$ carries a point mass $E(\{\lambda_n\}) = P_n$, and the integral $A = \int \lambda \, dE(\lambda)$ collapses to the sum. Without compactness $E$ may carry a continuous component, and the integral does not reduce; the multiplication operator $M_t$ on $L^2[0,1]$ above is exactly that case. The Operator SVD capstone walks the consequence for the SVD: compactness is exactly the condition that makes a discrete-sum SVD exist.

What it amounts to

The two proofs are parallel but mechanically different. The finite-dimensional route picks an eigenvalue, peels off its eigenspace, inducts on dimension, and assembles the eigenbasis. The operator route here builds the polynomial calculus, extends it to continuous functions by Weierstrass approximation, extracts a projection-valued measure by Riesz representation, and reads off the spectral theorem as an integral; the compact intermediate (Part II) is where eigenvalues come back and the integral collapses to a sum.

The substance of the infinite-dimensional case is concentrated in that contrast. Self-adjoint operators always admit a spectral theorem; eigenvalues are a separate question that the compactness of the operator decides. For the Koopman operator of a stochastic process on $L^2(\pi)$, whether it is compact, merely bounded, or unbounded is a substantive question about the dynamics, taken up in the companion Koopman post.

References

1. Lax, P. D. (2002). Functional Analysis. Wiley-Interscience.
2. Conway, J. B. (2007). A Course in Functional Analysis (2nd ed.). Springer.