Continuous calculus for a bounded normal operator

Original exposition and proof: OpenAI Codex (GPT-6 Astra, Ultra), October 2026. CC0-1.0.

The spectral lesson needs continuous functions of a normal operator before it constructs measurable functions. This note supplies that earlier step for every complex Hilbert space. It extends the written one-variable Bernstein construction in GP0 and the Hilbert and self-adjoint tools in BK01. The normal-calculus theorem is also treated in Jesse Peterson's freely available Notes on operator algebras, Theorems 1.2.6 and 1.3.1, pages 16–17. Here the construction and all extra approximation steps are proved directly from the two written prerequisites.

Throughout, inner products are linear in the first variable. For a bounded operator, its spectrum is the set of complex numbers for which its difference from the corresponding scalar multiple of the identity has no bounded two-sided inverse. We use no measurable calculus, spectral measure, unbounded operator theorem or representation theorem for abstract algebras.

NC1. Statement and commuting real coordinates

Theorem. Let $T\in B(H)$ be normal, with no dimension restriction on $H$. There is a unique isometric unital *-algebra isomorphism $\Psi_T:C(\sigma(T))\longrightarrow C^*(I,T),\qquad \Psi_T(z\mapsto z)=T. \tag{NC.1}$ Here $C^*(I,T)$ means the operator-norm closure of the unital *-polynomials in $T,T^*$. For every continuous $f$, $\|\Psi_T(f)\|=\max_{z\in\sigma(T)}|f(z)|, \qquad \sigma(\Psi_T(f))=f(\sigma(T)). \tag{NC.2}$ For $H=0$, the spectrum is empty and both algebras are the zero algebra; the displayed maximum over the empty set is taken to be zero. In what follows $H\ne0$.

Write $A=\frac{T+T^*}{2},\qquad B=\frac{T-T^*}{2i},\qquad T=A+iB. \tag{NC.3}$ Both $A,B$ are self-adjoint. Expansion of $TT^*-T^*T$ shows that normality is exactly $AB=BA$. Choose $C\geq\max(1,\|A\|,\|B\|)$, and identify the square $Q=[-C,C]^2$ with its image in the complex plane. The positive contractions $P=(A+CI)/(2C)$ and $R=(B+CI)/(2C)$ commute.

If two bounded positive operators $D,E$ commute, then their product is positive: the square root of $D$ in BK01 is a norm limit of polynomials in $D$, so commutes with $E$, and $DE=D^{1/2}ED^{1/2}\geq0$. This observation is the order argument needed for two-variable approximation.

NC2. A calculus on the enclosing square

For a continuous function $F$ on $[0,1]^2$, form $\mathcal B_nF(s,t)=\sum_{j,k=0}^n F(j/n,k/n)b_{n,j}(s)b_{n,k}(t), \quad b_{n,j}(s)={n\choose j}s^j(1-s)^{n-j}. \tag{NC.4}$ These polynomials converge uniformly to $F$. The scalar product weights are nonnegative and sum to one. Their mean squared distance from $(s,t)$ is $[s(1-s)+t(1-t)]/n\leq1/(2n)$, by the binomial identities proved in GP0. Thus their total weight at distance at least $\delta$ is at most $1/(2n\delta^2)$. Uniform continuity gives $\|\mathcal B_nF-F\|_\infty \leq\varepsilon+\frac{2\|F\|_\infty}{2n\delta^2} \tag{NC.5}$ when oscillations over distances below $\delta$ are at most $\varepsilon$. Compactness and uniform continuity on a square follow by applying interval bisection successively to the two coordinates; any sequence then has a convergent subsequence, and the usual pairs-with-fixed-oscillation contradiction proves uniform continuity.

The operators $b_{n,j}(P)$ and $b_{n,k}(R)$ are positive by GP0, commute, and each family sums to $I$. Their products are positive by NC1 and sum to $I$. Consequently, for real $F$, $-\|F\|_\infty I\leq\mathcal B_nF(P,R)\leq\|F\|_\infty I, \qquad \|\mathcal B_nF(P,R)\|\leq\|F\|_\infty. \tag{NC.6}$ The second conclusion uses the self-adjoint norm estimate already proved in GP0/BK01.

For a fixed real polynomial $p(s,t)$, the coefficients of $\mathcal B_np$ converge to those of $p$: for $p=s^jt^k$ the polynomial is the product of the one-variable Bernstein approximations of these two monomials, whose coefficient limits were proved in GP0. Take finite linear combinations. Substituting $P,R$ and passing to the limit in (NC.6) yields $\|p(P,R)\|\leq\|p\|_\infty$. For a complex polynomial, apply this to the real polynomial $\overline p p$ and use $p(P,R)^*p(P,R)=(\overline p p)(P,R)$ and $\|X^*X\|=\|X\|^2$.

Uniform approximation and completeness of $B(H)$, proved in GP0, therefore give a contractive unital *-homomorphism on all continuous functions on the unit square. Affine change of coordinates gives $\Phi:C(Q)\longrightarrow C^*(I,T),\quad \Phi(x)=A,\quad\Phi(y)=B,\quad\Phi(z)=T, \quad\|\Phi(F)\|\leq\|F\|_{C(Q)}. \tag{NC.7}$ The multiplication and conjugation identities hold first for polynomials and then by uniform limits. Positive continuous functions have positive images by the nonnegative Bernstein weights. Polynomial limits also show that every operator commuting with both $T$ and $T^*$ commutes with every $\Phi(F)$. No joint spectral theorem has been used.

NC3. Spectrum and localized functions in the resolvent

The geometric inverse series from BK01 shows that the resolvent is open and that $T-\lambda I$ is invertible for $|\lambda|>\|T\|$. Thus $K=\sigma(T)$ is closed and bounded, possibly empty at this stage.

In fact $K\subset Q$. For $\lambda=a+ib$, commutation of $A,B$ gives $\|(T-\lambda I)u\|^2 =\|(A-aI)u\|^2+\|(B-bI)u\|^2 =\|(T^*-\overline\lambda I)u\|^2. \tag{NC.8}$ If $|a|>C$, then $\|(A-aI)u\|\geq(|a|-C)\|u\|$; the analogous statement holds for $|b|>C$. Outside $Q$, both $T-\lambda I$ and its adjoint therefore are bounded below. Its range is closed, since a convergent sequence of images has a Cauchy sequence of inverse images. Its range is dense, since the orthogonal complement is the kernel of its adjoint by BK01. Hence it is onto with bounded inverse. The closed subset $K$ of $Q$ is compact.

Let $\lambda$ be a resolvent point, let $D_\lambda=(T-\lambda I)^{-1}$, and choose $\delta>0$ such that $q=\delta\|D_\lambda\|<1$. If $h\in C(Q)$ has support in $\{|z-\lambda|\leq\delta\}$, multiplicativity gives, for every positive integer $n$, $\Phi(h)=D_\lambda^n\Phi((z-\lambda)^nh), \qquad \|\Phi(h)\|\leq q^n\|h\|_\infty. \tag{NC.9}$ Letting $n$ tend to infinity proves $\Phi(h)=0$.

The same conclusion holds whenever the compact support $L$ of $h$ misses $K$. For every point of $L$, choose a disk as above centered at that point, and choose finitely many whose interiors cover $L$. Write their centers and radii as $\lambda_j,\delta_j$, and put $\varphi_j(z)=\max(0,\delta_j-|z-\lambda_j|),\qquad h_j(z)=\frac{h(z)\varphi_j(z)}{d(z,L)+\sum_k\varphi_k(z)}. \tag{NC.10}$ When $L\ne\varnothing$, the denominator is positive everywhere: off $L$ its first term is positive; on $L$ the covering supplies a positive summand. Each $h_j$ is continuous and supported in the closed $j$-th disk. Also $h=\sum_jh_j$, since a point where $h\ne0$ belongs to $L$, and elsewhere both sides vanish. Apply (NC.9) to each term. When $L$ is empty the conclusion is immediate. If $K$ were empty this argument would apply to $h=1$, contradicting $\Phi(1)=I\ne0$. It proves nonemptiness as well as the localized vanishing assertion.

Finally, if $h|_K=0$, put $h_\varepsilon(z)=h(z)\min\bigl(1,\max(0,2d(z,K)/\varepsilon-1)\bigr). \tag{NC.11}$ Its support is a compact set disjoint from $K$. Further, $\|h-h_\varepsilon\|_\infty\leq\sup_{d(z,K)\leq\varepsilon}|h(z)|\to0$: otherwise a sequence of such points would have a limit in $K$ with nonzero value. Contractivity now proves the full vanishing implication $h|_K=0\quad\Longrightarrow\quad\Phi(h)=0. \tag{NC.12}$

NC4. Continuous extension from the compact spectrum

We supply the scalar extension step. For any nonempty closed $K\subset Q$, every continuous complex $f$ on $K$ has an extension $F\in C(Q)$ with $\|F\|_{C(Q)}=\|f\|_{C(K)}$.

First obtain an approximate extension. Given $g\in C(K)$ and $\eta>0$, choose $r>0$ so that points of $K$ at distance below $r$ have $g$-values differing by at most $\eta$. Choose finitely many centers $k_j\in K$ whose balls of radius $r/2$ cover $K$, and set $u(z)=\frac{\sum_j\max(0,r-|z-k_j|)g(k_j)} {d(z,K)+\sum_j\max(0,r-|z-k_j|)}. \tag{NC.13}$ Its denominator is everywhere positive. It is continuous, $\|u\|_{C(Q)}\leq\|g\|_{C(K)}$, and $\|u|_K-g\|_{C(K)}\leq\eta$. Indeed, on $K$ its normalized weights sum to one, and every nonzero weight has a center within distance $r$. Off $K$, the extra nonnegative denominator term only decreases the bound.

Start with $g_0=f$. When $g_j\ne0$, take such an approximate extension $u_j$ with error at most $\|g_j\|/2$, and set $g_{j+1}=g_j-u_j|_K$. If a residual is zero, take all subsequent functions to be zero. Then $\|g_j\|\leq2^{-j}\|f\|$ and $\|u_j\|\leq2^{-j}\|f\|$. The uniformly convergent series $F_0=\sum_{j\geq0}u_j$ is continuous and restricts to $f$, by the telescoping identity and $g_j\to0$. Uniform convergence implies continuity directly by choosing one uniform tail bound and then using continuity of the finite initial sum.

If $M=\|f\|>0$, replace $F_0$ by $F(z)=\frac{M F_0(z)}{\max(M,|F_0(z)|)}. \tag{NC.14}$ This is continuous, agrees with $f$ on $K$, and has norm exactly $M$. For $M=0$, use zero. For real-valued $f$, the preceding construction stays real; for nonnegative $f$, replacing a real extension by $\max(0,F)$ preserves its boundary values and the norm bound. This proves the needed extension theorem, including the complex and positive versions, without importing a topological extension theorem.

NC5. Restriction, approximate eigenvectors, and the exact norm

For $f\in C(K)$, choose the extension in NC4 and define $\Psi_T(f)=\Phi(F)$. Equation (NC.12) makes the definition independent of the extension. Sums, products, conjugation and the constant one can therefore be computed using any extensions and give the corresponding homomorphism identities. NC4 gives $\|\Psi_T(f)\|\leq\|f\|_{C(K)}$. Its coordinate function gives $T$.

For each $\lambda\in K$, there are unit vectors $u_m$ with $\|(T-\lambda I)u_m\|\to0$. Otherwise $T-\lambda I$ would be bounded below. By (NC.8), so would its adjoint, and the closed-range/dense-range argument in NC3 would give a bounded inverse, a contradiction. Also $\|(T^*-\overline\lambda I)u_m\|\to0$, again by (NC.8).

For each *-polynomial $p$, the factorization $X^j-a^jI=\sum_{k=0}^{j-1}a^kX^{j-1-k}(X-aI)$ and the identity $T^j(T^*)^k-\lambda^j(\overline\lambda)^kI =T^j((T^*)^k-(\overline\lambda)^kI)+(\overline\lambda)^k(T^j-\lambda^jI)$ show that $\|(p(T,T^*)-p(\lambda,\overline\lambda)I)u_m\|\longrightarrow0. \tag{NC.15}$ Approximate an extension of $f$ on $Q$ uniformly by NC2 polynomials. Since $x=(z+\bar z)/2$ and $y=(z-\bar z)/(2i)$, these are *-polynomials in $z,\bar z$. Contractivity bounds both the operator error and the scalar error by the same uniform error. First choose that error, then let $m$ tend to infinity in (NC.15). This gives $\|(\Psi_T(f)-f(\lambda)I)u_m\|\longrightarrow0. \tag{NC.16}$ It follows that $\|\Psi_T(f)\|\geq|f(\lambda)|$. Taking a maximum over $K$ proves the norm equality in (NC.2).

The same extension and approximation prove that *-polynomials restricted to $K$ are uniformly dense in $C(K)$. Thus the image of $\Psi_T$ contains the *-polynomials in $T,T^*$, is contained in their norm closure by construction, and is closed because $\Psi_T$ is isometric and $C(K)$ is complete. Completeness follows by uniform limits of Cauchy sequences of continuous scalar functions. This proves the range statement in (NC.1). Uniqueness among isometric unital *-isomorphisms sending the coordinate to $T$ follows by agreement on the dense *-polynomials and continuity.

The enclosing square, three spectral points and a resolvent disk

Figure. The exact example is $T=\operatorname{diag}(-1,i,1)$ on $\mathbb C^3$, displayed in $Q=[-2,2]^2$. Its spectrum is the three marked points. At $\lambda=-i$, the inverse norm is $1/\sqrt2$. A continuous function supported in the disk of radius $1/4$ about $-i$ has zero image by (NC.9), since its image norm is at most $(1/(4\sqrt2))^n\|h\|_\infty$ for every $n$. The arrows describe the maps in (NC.7), (NC.12), and (NC.1). The colored disk is a region of vanishing, not additional spectrum. NC3 proves the same local argument for every bounded normal operator; NC4–NC5 identify the full quotient and its norm.

NC6. Spectral mapping, positivity, and commutation

If $w\notin f(K)$, then $1/(f-w)$ is continuous on $K$. Multiplicativity shows that its image is a bounded two-sided inverse of $\Psi_T(f)-wI$. Thus $\sigma(\Psi_T(f))\subset f(K)$. Conversely, if $w=f(\lambda)$ for $\lambda\in K$, (NC.16) gives unit approximate eigenvectors at $w$, precluding a bounded inverse. This proves the spectral mapping assertion in (NC.2).

For $f\geq0$, its scalar continuous square root belongs to $C(K)$, and $\Psi_T(f)=\Psi_T(\sqrt f)^*\Psi_T(\sqrt f)\geq0$. Hence order is preserved for real functions. Conversely, if $\Psi_T(f)\geq0$, then $\Psi_T(f)=\Psi_T(\bar f)$ implies $f=\bar f$ by isometry, and (NC.16) gives $f(\lambda)=\lim_m\langle\Psi_T(f)u_m,u_m\rangle\geq0$.

Every bounded operator commuting with both $T$ and $T^*$ commutes with $\Psi_T(f)$, by polynomial approximation. For a unitary $V:H\to H'$, invertibility and norms are preserved under $X\mapsto VXV^*$. The spectra agree, and polynomial approximation gives $\Psi_{VTV^*}(f)=V\Psi_T(f)V^*. \tag{NC.17}$ Finally, if a norm-closed unital *-subalgebra $\mathcal D\subseteq B(H)$ contains $T$, then every $\Psi_T(f)$ belongs to $\mathcal D$. For $\lambda\notin K$, the function $(z-\lambda)^{-1}$ proves that the inverse also belongs to $\mathcal D$. Invertibility in $\mathcal D$ therefore gives exactly the same spectrum as in $B(H)$. This is the concrete spectral-invariance conclusion; no abstract representation theorem is required.

NC7. The prerequisite supplied to the spectral lesson

NC1–NC6 prove the full continuous-calculus input for a bounded normal operator on an arbitrary Hilbert space, including exact norm, full range, *-polynomial density, commutation and spectral mapping. For a self-adjoint operator the calculus agrees with BK01: the spectrum is real, and agreement on real-coordinate polynomials plus their uniform density gives equality. For a unitary operator this supplies the circle calculus needed before the Cayley transform in SK06 and the whole-circle polynomial approximation in SK09. All these results precede the construction of any Borel calculus.

The older contract OA-MOD-OPEN-CSTAR-CFC also states a theorem for arbitrary abstract unital C*-algebras. This note does not certify that broader contract or its other consumers. It proves exactly its concrete operator case used in the spectral lesson; registration of that new proof route and the separate scalar prerequisites must be checked explicitly. Neither that registration nor this proof claims completion of the whole course.