Spectral measures with the original operator domain retained

A sharp spectral cutoff can be constructed without an eigenbasis. This companion builds the scalar measures, the bounded positive-operator calculus, and the exact passage back to a lower-bounded self-adjoint operator. Every product keeps its actual operator domain. The construction applies to the unchanged weighted Hilbert space in the elliptic course, including noncompact domains.

The first two parts supply the measure and Hilbert foundations used by the operator proof. The third constructs the full calculus from positive Bernstein polynomial weights. The fourth recovers the original lower-bounded operator, its recursive powers, and both endpoints of a sharp cutoff. The fifth gives continuous and atomic models with complete domain calculations. The sixth proves the Hilbert geometry, square roots, closed-operator polar decomposition tensor completion, complete direct sums and geometric inverses needed by the trace, operator and divergence arguments. Inner products are linear in the first variable throughout.

1. Constructing the scalar measure from a positive functional

The space throughout is $K=[0,1]$, with its usual metric and relative topology. All Borel sets and all supports below are relative to $K$. A positive complex-linear functional means a complex-linear map $L:C(K;\mathbb C)\to\mathbb C$ such that $L(f)$ is a nonnegative real number whenever $f$ is real and nonnegative. Inner products and all sesquilinear forms are linear in their first argument and conjugate-linear in their second argument.

This proof constructs the measure, proves countable additivity and regularity, develops the integration facts used, and then proves the exact complex-measure statements for operator pairings. Its measure-theoretic input is set theory and elementary compactness of the interval; it does not use a representation theorem or a spectral theorem. Completeness of a Hilbert space is used only in the final operator-representation step, whose proof is also included.

1.1 Positivity and the mass constant

Let $L$ be positive, and put $M=L(1)\geq0$. For real continuous $u$, the functions $u_+=\max(u,0)$ and $u_-=\max(-u,0)$ are continuous and nonnegative, so $L(u)=L(u_+)-L(u_-)$ is real. Consequently $L(\overline f)=\overline{L(f)}$ for complex continuous $f$. If $u\leq v$ are real, positivity gives $L(u)\leq L(v)$.

For complex $f$, if $L(f)\ne0$, choose a scalar $\zeta$ of modulus one with $\zeta L(f)=|L(f)|$. Then $\operatorname{Re}(\zeta f)\leq|f|$ pointwise, and complex linearity and the preceding real-valued property give

$$|L(f)|=L(\operatorname{Re}(\zeta f))\leq L(|f|)\leq M\|f\|_\infty. \tag{PR1}$$

The same inequality holds when $L(f)=0$. Thus positivity implies boundedness, $\|L\|\leq M$, and evaluation on $1$ gives $\|L\|=M$. In particular $M=0$ implies $L=0$. This case is included in every construction below.

1.2 Mass assigned to open sets

For each open $U\subset K$, define

$$c(U)=\sup\{L(f):f\in C(K;\mathbb R),\ 0\leq f\leq1,\ \operatorname{supp}f\subset U\}. \tag{PR2}$$

The zero function is allowed, and the empty support causes no exception. We have $0\leq c(U)\leq M$, $c(\varnothing)=0$, $c(K)=M$, and $c$ is monotone under inclusion.

We first prove the finite decomposition needed for countable subadditivity. Let $S\subset\bigcup_{i=1}^rU_i$ be a nonempty compact set, with each $U_i$ open. Put $d_i(x)=\operatorname{dist}(x,K\setminus U_i)$ when $U_i\ne K$, and put $d_i=1$ when $U_i=K$. These functions are continuous, nonnegative, and positive at exactly the points of $U_i$. The continuous function $\max_i d_i$ is positive on $S$, so its minimum there is positive. Choose $\delta>0$ smaller than that minimum and also smaller than one. Set

$$h_i=(d_i-\delta/2)_+,\qquad h_0(x)=\operatorname{dist}(x,S),\qquad \psi_i=\frac{h_i}{h_0+\sum_{j=1}^rh_j}. \tag{PR3}$$

The denominator is positive on $S$ by the choice of $\delta$, and positive off $S$ because $h_0>0$ there. Thus each $\psi_i$ is continuous and between zero and one. Its support is contained in the compact set $\{d_i\geq\delta/2\}\subset U_i$. On $S$, the sum of the $\psi_i$ is one.

Suppose that $U\subset\bigcup_{j=1}^\infty U_j$, with all these sets open, and take an admissible $f$ in (PR2) for $U$. If $f=0$, there is nothing to prove. Otherwise its compact support $S$ has a finite subcover from the $U_j$. Apply (PR3) to that subcover and set $f_i=f\psi_i$. Each $f_i$ is admissible for its corresponding $U_i$, and $\sum_i f_i=f$: on $S$ the partition sums to one, and off $S$ the function $f$ is zero. Therefore

$$L(f)=\sum_iL(f_i)\leq\sum_{j=1}^\infty c(U_j). \quad\text{Hence}\quad c(U)\leq\sum_{j=1}^\infty c(U_j). \tag{PR4}$$

For finitely many pairwise disjoint open sets $U_1,\ldots,U_r$, admissible functions $f_i$ have disjoint supports, so $0\leq\sum_i f_i\leq1$ and its support is contained in $\bigcup_iU_i$. Approximating each supremum in (PR2) within an arbitrarily small positive error gives $c(\bigcup_iU_i)\geq\sum_i c(U_i)$. Combined with (PR4), this proves equality. For countably many disjoint open sets, monotonicity gives the lower bound by every finite partial sum, while (PR4) gives the upper bound. Thus $c$ is countably additive on disjoint open unions.

1.3 Outer measure and separation by positive distance

For every subset $E\subset K$, define

$$\mu^*(E)=\inf\{c(U):U\subset K\text{ open},\ E\subset U\}. \tag{PR5}$$

The family is never empty, since $K$ is allowed. The function $\mu^*$ is monotone, vanishes on the empty set, and is at most $M$. For subsets $E_j$ and $\epsilon>0$, choose open $U_j\supset E_j$ with $c(U_j)\leq\mu^*(E_j)+\epsilon2^{-j}$, where indices begin at one. By (PR4),

$$\mu^*\Bigl(\bigcup_jE_j\Bigr) \leq c\Bigl(\bigcup_jU_j\Bigr) \leq\sum_j\mu^*(E_j)+\epsilon. \tag{PR6}$$

Letting $\epsilon\downarrow0$ proves countable subadditivity. Thus $\mu^*$ is an outer measure. For open $U$, (PR5) and monotonicity of $c$ show $\mu^*(U)=c(U)$; in particular $\mu^*(K)=M$.

If nonempty subsets $A,B$ have distance $d>0$, take any open $U\supset A\cup B$ and form $U_A=U\cap\{x:\operatorname{dist}(x,A)<d/3\}$ and $U_B=U\cap\{x:\operatorname{dist}(x,B)<d/3\}$. They are disjoint open supersets of $A,B$. To see disjointness, a point in both would admit points of $A,B$ whose distance is less than $2d/3$ plus arbitrarily small errors, contradicting the definition of $d$. Open-set additivity and monotonicity give $c(U)\geq c(U_A)+c(U_B)\geq\mu^*(A)+\mu^*(B)$. Take the infimum over $U$ and use outer subadditivity for the reverse inequality. The result, also immediate if one set is empty, is

$$\operatorname{dist}(A,B)>0 \quad\Longrightarrow\quad \mu^*(A\cup B)=\mu^*(A)+\mu^*(B). \tag{PR7}$$

1.4 Borel measurability and countable additivity

Call a set $A\subset K$ measurable for $\mu^*$ when, for every subset $E\subset K$,

$$\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A). \tag{PR8}$$

First every closed set $F$ satisfies (PR8). The cases $F=\varnothing,K$ are immediate. Otherwise the function $d_F(x)=\operatorname{dist}(x,F)$ is continuous and satisfies $|d_F(x)-d_F(y)|\leq|x-y|$, by the triangle inequality and taking infima. Fix any subset $E$, and put

$$E_i=\{x\in E:d_F(x)\geq1/i\},\qquad A_j=\{x\in E:1/(j+1)\leq d_F(x)<1/j\}. \tag{PR9}$$

The set $E_i$ has distance at least $1/i$ from $E\cap F$. Hence (PR7) and monotonicity give $\mu^*(E)\geq\mu^*(E\cap F)+\mu^*(E_i)$. The sets $A_j$ with indices of one fixed parity are separated from one another by positive distance in each finite collection: for indices $j,j+2$, the gap between the indicated distance intervals is at least $1/(j+1)-1/(j+2)>0$, and the distance function is 1-Lipschitz. Iterating (PR7) over a finite collection of even indices, and then over odd indices, shows that each of the two series of their outer masses has sum at most $M$. Therefore $\sum_{j\geq1}\mu^*(A_j)$ converges and its tails tend to zero.

The set $E\setminus F$ is contained in $E_i\cup\bigcup_{j\geq i}A_j$. Outer subadditivity now gives

$$\mu^*(E\setminus F) \leq\mu^*(E_i)+\sum_{j\geq i}\mu^*(A_j). \tag{PR10}$$

Combining the preceding two inequalities and letting $i\to\infty$ proves $\mu^*(E)\geq\mu^*(E\cap F)+\mu^*(E\setminus F)$. Outer subadditivity gives the opposite inequality. This proves (PR8) for every closed $F$, without assuming continuity from below for an outer measure.

Here is the complete algebra argument for the measurable sets. Complementation preserves (PR8). If $A,B$ satisfy (PR8), split $E$ first by $A$, then split $E\setminus A$ by $B$. This gives $\mu^*(E)=\mu^*(E\cap A)+\mu^*((E\setminus A)\cap B)+\mu^*(E\setminus(A\cup B))$. The first two terms are at least $\mu^*(E\cap(A\cup B))$ by subadditivity. Together with the reverse subadditivity bound, this proves (PR8) for $A\cup B$. Thus the measurable sets form an algebra.

For pairwise disjoint measurable $A_j$, repeated splitting gives, with $A=\bigcup_jA_j$,

$$\mu^*(E)\geq\sum_{j=1}^r\mu^*(E\cap A_j)+\mu^*(E\setminus A) \qquad(r\geq1). \tag{PR11}$$

Let $r\to\infty$. The sum is at least $\mu^*(E\cap A)$ by countable subadditivity. Again the reverse inequality is automatic, so $A$ is measurable. Any countable union can be made disjoint by replacing its terms by their differences from the finite preceding union; those differences belong to the algebra. Consequently the measurable sets form a sigma-algebra. Since they contain every closed set, they contain every Borel set.

For disjoint measurable $A_j$, take $E=\bigcup_jA_j$ in (PR11) and compare with countable subadditivity. The result is exact countable additivity. Restriction to Borel sets therefore gives a finite positive Borel measure

$$\mu(E)=\mu^*(E)\quad(E\text{ Borel}),\qquad \mu(K)=M,\qquad \mu(U)=c(U)\quad(U\text{ open}). \tag{PR12}$$

No representation or integration theorem has been used to obtain this measure.

1.5 Outer and inner regularity

Equation (PR5), together with (PR12), gives for every Borel set $E$

$$\mu(E)=\inf_{U\supset E,\ U\text{ open}}\mu(U). \tag{PR13}$$

To prove inner regularity, choose open $V\supset K\setminus E$ with $\mu(V)<\mu(K\setminus E)+\epsilon$, and put $F=K\setminus V$. Then $F$ is compact and contained in $E$. Finite additivity on the disjoint parts of $V$ gives $\mu(E\setminus F)=\mu(E\cap V)=\mu(V)-\mu(K\setminus E)<\epsilon$. It follows that

$$\mu(E)=\sup_{F\subset E,\ F\text{ compact}}\mu(F). \tag{PR14}$$

Both inequalities also hold for empty sets and for zero mass. This is the claimed regularity on every Borel set, rather than only on open sets.

1.6 Integration and the convergence statements actually used

The following construction and proofs apply to any finite positive measure $\rho$ on a sigma-algebra, including $\mu$ on the Borel sets. A nonnegative simple function can be written as $s=\sum_{i=1}^ra_i1_{E_i}$ with disjoint measurable $E_i$ and $a_i\geq0$. Define $\int s\,d\rho=\sum_i a_i\rho(E_i)$. Refining two finite partitions by their intersections proves independence of the representation, additivity for simple functions, and monotonicity. For a nonnegative measurable function $f$, define

$$\int f\,d\rho=\sup\{\int s\,d\rho:0\leq s\leq f,\ s\text{ simple}\}. \tag{PR15}$$

Countable additivity first implies continuity from below for sets. If $E_k\uparrow E$, the disjoint sets $E_1,E_2\setminus E_1,\ldots$ partition $E$, so $\rho(E_k)\uparrow\rho(E)$. Taking complements inside a set of finite mass also proves continuity from above.

If $0\leq f_k\uparrow f$, monotonicity gives $\lim_k\int f_k\leq\int f$. For the reverse inequality, take a nonnegative simple $s\leq f$ and $0<c<1$. On every point where $s>0$, the increasing sequence $f_k$ eventually exceeds $cs$. Thus the sets $E_k=\{f_k\geq cs\}$ increase to the entire underlying space (points where $s=0$ already belong to every $E_k$). Continuity from below for each set in the finite simple partition gives $\int s1_{E_k}\to\int s$, while $f_k\geq cs1_{E_k}$. Hence $\lim_k\int f_k\geq c\int s$. Take the supremum over $s$ and then let $c\uparrow1$. This proves monotone convergence, including infinite integrals:

$$0\leq f_k\uparrow f\quad\Longrightarrow\quad \int f_k\,d\rho\uparrow\int f\,d\rho. \tag{PR16}$$

Every nonnegative measurable $f$ has the increasing simple approximants $s_k=2^{-k}\lfloor2^k\min(f,k)\rfloor$. They are simple because the cap is finite, are measurable because their level sets are measurable, are increasing because the caps increase and each grid refines its predecessor, and tend pointwise to $f$. Applying (PR16) to these approximants for $f$, $g$, and their sums proves $\int(f+g)=\int f+\int g$. Positive homogeneity follows in the same way or directly from (PR15).

For real $f$ with $\int|f|<\infty$, define $\int f=\int f_+-\int f_-$. To prove additivity for real integrable $f,g$, use the pointwise identity $f_++g_++(f+g)_-=f_-+g_-+(f+g)_+$. All terms have finite integrals because $|f+g|\leq|f|+|g|$; additivity for nonnegative functions and rearrangement therefore give $\int(f+g)=\int f+\int g$. Positive homogeneity is already proved, and $(-f)_+=f_-$, $(-f)_-=f_+$ proves the negative-scalar case. For complex integrable $f$, define its integral by its real and imaginary parts; real linearity and multiplication by $i$ prove complex linearity. If its integral is nonzero, multiply by a constant phase making that integral positive real and use $\operatorname{Re}(\zeta f)\leq|f|$, as in (PR1). This proves

$$\left|\int f\,d\rho\right|\leq\int|f|\,d\rho. \tag{PR17}$$

Fatou’s inequality follows directly from (PR16): for nonnegative measurable $f_k$, set $h_j=\inf_{k\geq j}f_k$. These measurable functions increase to $\liminf f_k$, and $\int h_j\leq\inf_{k\geq j}\int f_k$. Therefore

$$\int\liminf_k f_k\,d\rho\leq\liminf_k\int f_k\,d\rho. \tag{PR18}$$

If complex measurable $f_k\to f$ pointwise and $|f_k|\leq g$ with $\int g<\infty$, then $|f|\leq g$. Apply (PR18) to the nonnegative functions $2g-|f_k-f|$, whose limit is $2g$. Their integrals are $2\int g-\int|f_k-f|$, so the resulting inequality forces $\limsup_k\int|f_k-f|\leq0$. Thus

$$\int|f_k-f|\,d\rho\longrightarrow0, \qquad \int f_k\,d\rho\longrightarrow\int f\,d\rho. \tag{PR19}$$

The same proof applies to convergence outside a measurable null set by changing the functions to zero there. In particular, on a finite measure space, pointwise convergence of a uniformly bounded sequence of bounded Borel functions implies convergence of their integrals. All monotone and dominated convergence uses below refer to the proofs (PR16)–(PR19), not to a spectral-theoretic convergence assertion.

1.7 Recovery of the original functional

Let $f\in C(K;\mathbb R)$ be nonnegative, and let $B=\|f\|_\infty$. If $B=0$, the assertion is immediate. Otherwise fix $\delta>0$, let $r=\lceil B/\delta\rceil$, and put $U_i=\{f>i\delta\}$, $1\leq i\leq r$. For each $i$, choose any function $g_i$ admissible in (PR2) for $U_i$. At a point $x$, the number of indices with $g_i(x)>0$ is at most the number of positive integers $i$ for which $i\delta<f(x)$. Therefore $0\leq\delta\sum_i g_i(x)\leq f(x)$. Positivity and approximation of the finitely many suprema give

$$L(f)\geq\delta\sum_{i=1}^r c(U_i) =\int s_\delta\,d\mu, \qquad s_\delta=\delta\sum_{i=1}^r1_{\{f>i\delta\}}. \tag{PR20}$$

The approximation step is finite: choose each $L(g_i)$ within $\epsilon/(r\delta)$ of $c(U_i)$, and then let $\epsilon\downarrow0$. Pointwise $0\leq f-s_\delta\leq\delta$, including values of $f$ that are exact multiples of $\delta$. By the integration facts already proved, $\int s_\delta\geq\int f-\delta M$. Let $\delta\downarrow0$; then $L(f)\geq\int f$.

Apply this same inequality to the nonnegative continuous function $B-f$. Since $\mu(K)=M=L(1)$, it gives $BM-L(f)\geq BM-\int f$, hence $L(f)\leq\int f$. Equality follows. Real continuous functions are differences of their continuous positive and negative parts, and complex continuous functions have real and imaginary parts. Linearity therefore proves

$$L(f)=\int_K f\,d\mu\quad(f\in C(K;\mathbb C)), \qquad \mu(K)=L(1). \tag{PR21}$$

This proves existence of the representing finite positive regular Borel measure for every positive bounded complex-linear functional.

1.8 Uniqueness and lower semicontinuous approximation

For an open proper nonempty $U\subset K$, the continuous functions $g_k(x)=\min(1,(k\operatorname{dist}(x,K\setminus U)-1)_+)$ increase to $1_U$. They are between zero and one, and each support lies in $\{x:\operatorname{dist}(x,K\setminus U)\geq1/k\}\subset U$. For $U=K$ use the constant sequence one, and for $U=\varnothing$ use zero. If another finite positive Borel measure $\nu$ has the integrals (PR21), monotone convergence for both measures gives $\nu(U)=\lim_k\int g_k\,d\nu=\lim_kL(g_k)=\mu(U)$.

We give the elementary set argument extending this equality to all Borel sets. A family $\mathcal D$ containing $K$, closed under complements in $K$, and closed under disjoint countable unions is called a lambda-system. It is also closed under differences $A\setminus B$ when $B\subset A$ and both sets belong to it: take the complement of the disjoint union $(K\setminus A)\cup B$. A family $\mathcal P$ closed under finite intersections is a pi-system. If a lambda-system contains a pi-system $\mathcal P$ that contains $K$, it contains the sigma-algebra generated by $\mathcal P$. Here is a proof. Let $\mathcal D_0$ be the smallest lambda-system containing $\mathcal P$. For fixed $A\in\mathcal P$, the sets $B$ satisfying $A\cap B\in\mathcal D_0$ form a lambda-system: the complement step uses the nested-difference property inside $A$. This lambda-system contains $\mathcal P$, so it contains $\mathcal D_0$. Next fix $B\in\mathcal D_0$ and repeat the argument with the sets $A$ satisfying $A\cap B\in\mathcal D_0$. The first step shows this new lambda-system contains $\mathcal P$, hence $\mathcal D_0$. Thus $\mathcal D_0$ is closed under finite intersections. Complements and finite intersections give finite unions and differences, and disjointizing any countable union then proves it is a sigma-algebra. This proves the assertion.

The Borel sets $E$ with $\mu(E)=\nu(E)$ form a lambda-system: total masses agree, complementation subtracts from that common finite mass, and disjoint countable unions use countable additivity. The open sets are a pi-system containing $K$ and generating the Borel sigma-algebra. The preceding argument proves $\mu=\nu$ on every Borel set. No regularity assumption on the competing measure was needed.

For later use, let $f:K\to[0,B]$ be lower semicontinuous. For every positive integer $k$, set

$$f_k(x)=\inf_{y\in K}\bigl(f(y)+k|x-y|\bigr). \tag{PR22}$$

The infimum is attained: a minimizing sequence has a convergent subsequence in compact $K$, and lower semicontinuity gives a minimizing limit. The inequality $\big||x-y|-|x'-y|\big|\leq|x-x'|$ implies $|f_k(x)-f_k(x')|\leq k|x-x'|$ by comparing infima in both directions. Thus $f_k$ is continuous. It satisfies $0\leq f_k\leq f$ and increases with $k$. If $y_k$ is a minimizer for a fixed $x$, then $k|x-y_k|\leq f_k(x)\leq f(x)\leq B$, so $y_k\to x$. Lower semicontinuity yields $\liminf_k f_k(x)\geq\liminf_k f(y_k)\geq f(x)$. Together with $f_k\leq f$, this proves $f_k\uparrow f$. Lower semicontinuity makes $f$ Borel because its strict upper level sets are open. Consequently, by (PR16),

$$\int f\,d\mu =\sup\{L(g):g\in C(K;\mathbb R),\ 0\leq g\leq f\}. \tag{PR23}$$

The inequality from right to left is monotonicity; the reverse inequality follows by choosing the functions (PR22).

1.9 Finite complex measures, variation, and integration

The complex measures needed here will be finite linear combinations $\nu=\sum_{r=1}^sc_r\rho_r$ of finite positive Borel measures. They are countably additive because every summand is. Define the positive measure $\lambda=\sum_r|c_r|\rho_r$; then $|\nu(E)|\leq\lambda(E)$. Define total variation by finite measurable partitions:

$$|\nu|(E)=\sup\Bigl\{\sum_{j=1}^r|\nu(E_j)|: E=\bigsqcup_{j=1}^rE_j,\ E_j\text{ Borel}\Bigr\}. \tag{PR24}$$

It satisfies $|\nu|(E)\leq\lambda(E)$. It is a measure, as follows directly. For disjoint Borel sets $E_i$ and $E=\bigcup_iE_i$, combine near-maximizing finite partitions of the first finitely many $E_i$, together with the remaining subset of $E$, to obtain $|\nu|(E)\geq\sum_{i=1}^r|\nu|(E_i)$ for every $r$. In the other direction, for any finite partition $E=\bigsqcup_jF_j$, countable additivity and triangle inequality give $\sum_j|\nu(F_j)|\leq\sum_i\sum_j|\nu(F_j\cap E_i)|\leq\sum_i|\nu|(E_i)$. Take the supremum over the partition and then let $r\to\infty$ in the first inequality. This proves countable additivity of $|\nu|$.

For complex simple $s=\sum_jz_j1_{E_j}$ on a finite disjoint partition, define $\int s\,d\nu=\sum_jz_j\nu(E_j)$. Refinement proves independence of the partition, and (PR24) gives $|\int s\,d\nu|\leq\int|s|\,d|\nu|$. Every bounded complex Borel function has uniformly convergent simple approximants, obtained by rounding its real and imaginary parts to successively finer finite grids on their bounded ranges. The preceding inequality makes their integrals Cauchy, with a limit independent of the approximating sequence. This defines $\int f\,d\nu$, proves linearity, and gives

$$\left|\int f\,d\nu\right|\leq\int|f|\,d|\nu|, \qquad \int f\,d\nu=\sum_{r=1}^sc_r\int f\,d\rho_r \quad(f\text{ bounded Borel}). \tag{PR25}$$

Both assertions follow first for simple functions, then by uniform approximation; $\big||s|-|f|\big|\leq|s-f|$ handles the absolute-value integral. If $f_k\to f$ pointwise and all functions are uniformly bounded, (PR19) for $|\nu|$ and (PR25) show convergence of the complex integrals. More generally, if $|f_k|\leq g$ with $g$ integrable for $|\nu|$, define integrals of such functions by truncation; the bound (PR25) makes the truncations Cauchy because the absolute-value tail integrals tend to zero by (PR19). The same bound and (PR19) prove dominated convergence in that case as well.

If the positive measures $\rho_r$ are regular, so is $\lambda$: for a Borel $E$, choose compact subsets $F_r\subset E$ and open supersets $U_r\supset E$ making each $|c_r|\rho_r(U_r\setminus F_r)$ as small as prescribed, ignoring zero coefficients. The compact union $F=\bigcup_rF_r$ is contained in $E$, and the open intersection $U=\bigcap_rU_r$ contains $E$. Since $U\setminus F\subset U_r\setminus F_r$ for every $r$, their weighted sum can be made arbitrarily small. As $|\nu|\leq\lambda$, the same $F,U$ prove regularity of $|\nu|$, and $|\nu(E)-\nu(F)|\leq|\nu|(E\setminus F)$, $|\nu(U)-\nu(E)|\leq|\nu|(U\setminus E)$ give the corresponding approximation for $\nu$.

The uniqueness argument extends to any finite complex Borel measure of finite total variation. Indeed, once finiteness of (PR24) is given, the proof that variation is a measure and the integration construction above apply unchanged; the finite positive-measure decomposition was used to establish finiteness and regularity, not in those subsequent proofs. If $\int f\,d\nu=0$ for every continuous $f$, the continuous open-set approximants in Section 1.8 and dominated convergence for $|\nu|$ give $\nu(U)=0$ for every open $U$. The family of Borel sets on which $\nu$ vanishes is a lambda-system, since $\nu(K)=0$, and countable additivity handles disjoint unions. The pi-system argument already proved therefore gives $\nu=0$. Applying this to a difference proves uniqueness whenever two finite complex measures have equal continuous integrals.

2. Complex measure pairings and Hilbert representation

2.1 Polarization with the linear-first convention

Let $H$ be a complex Hilbert space. Suppose $F_f(x,y)$, for continuous complex $f$, is complex-linear in $f$, linear in $x$, conjugate-linear in $y$, and satisfies

$$F_f(x,x)\geq0\quad\text{when }f\geq0, \qquad 0\leq F_1(x,x)\leq C\|x\|^2 \quad(x\in H), \tag{PR26}$$

for a fixed finite $C\geq0$. These are precisely the measure-stage hypotheses, for example when $F_f(x,y)=\langle\Phi(f)x,y\rangle$ and $\Phi$ is a positive linear map with $\|\Phi(1)\|\leq C$. A unital continuous $*$-homomorphism has $C=1$: for real $f\geq0$, its continuous square root gives $\Phi(f)=\Phi(\sqrt f)^*\Phi(\sqrt f)\geq0$, and $\Phi(1)=I$.

For each $x$, Section 1.1 and Section 1.7 give a unique finite positive regular measure $\mu_x$ such that

$$F_f(x,x)=\int f\,d\mu_x,\qquad \mu_x(K)=F_1(x,x)\leq C\|x\|^2. \tag{PR27}$$

For any sesquilinear form $B$ with the linear-first convention, writing $q(z)=B(z,z)$ gives the algebraic identity

$$B(x,y)=\frac14\sum_{k=0}^3i^kq(x+i^ky) =\frac14\bigl(q(x+y)-q(x-y)+i q(x+iy)-i q(x-iy)\bigr). \tag{PR28}$$

This identity does not require $B$ to be Hermitian. Indeed, for $t=i^k$, $q(x+ty)=q(x)+q(y)+\overline t B(x,y)+tB(y,x)$. After multiplication by $t$ and summation, the constant terms vanish because $\sum t=0$, the $B(y,x)$ term vanishes because $\sum t^2=0$, and the $B(x,y)$ term is multiplied by $\sum|t|^2=4$. This proves the formula and in particular fixes the plus sign before $i q(x+iy)$.

Define the finite regular complex measure

$$\mu_{x,y}=\frac14\bigl(\mu_{x+y}-\mu_{x-y}+i\mu_{x+iy}-i\mu_{x-iy}\bigr). \tag{PR29}$$

Equation (PR25), followed by (PR28) applied to $F_f$, proves

$$\int f\,d\mu_{x,y}=F_f(x,y)\qquad(f\in C(K;\mathbb C)). \tag{PR30}$$

The complex-measure uniqueness proved in Section 1.9 now shows that $(x,y)\mapsto\mu_{x,y}$ is sesquilinear as a measure-valued map. For example, the continuous integrals of $\mu_{ax+bz,y}-a\mu_{x,y}-b\mu_{z,y}$ vanish for all $f$, so this measure is zero; the second variable is treated with conjugated scalars. The same uniqueness gives $\mu_{x,x}=\mu_x$. This step proves sesquilinearity of the actual measures; it is not inferred just from the appearance of the polarization expression.

Fix a Borel set $E$. Then $B_E(x,y)=\mu_{x,y}(E)$ is a sesquilinear form with nonnegative real diagonal $B_E(x,x)=\mu_x(E)$. Every such form is Hermitian and satisfies Cauchy–Schwarz. For completeness, reality of $B_E(x+t y,x+t y)$ for real $t$ gives that $B_E(x,y)+B_E(y,x)$ is real; reality for $it$ gives that $i(B_E(y,x)-B_E(x,y))$ is real. Together these two equalities imply $B_E(y,x)=\overline{B_E(x,y)}$. If $B_E(y,y)>0$, substitute $t=-B_E(x,y)/B_E(y,y)$ into the nonnegative quadratic expression for $B_E(x+t y,x+t y)$. Its value is $B_E(x,x)-|B_E(x,y)|^2/B_E(y,y)$, proving the inequality. If $B_E(y,y)=0$ and $B_E(x,y)\ne0$, substituting $t=-sB_E(x,y)$ with large positive real $s$ makes that quadratic expression negative, a contradiction. Thus

$$\mu_{y,x}(E)=\overline{\mu_{x,y}(E)},\qquad |\mu_{x,y}(E)|^2\leq\mu_x(E)\mu_y(E). \tag{PR31}$$

For a finite partition $E=\bigsqcup_jE_j$, use (PR31), then the finite scalar Cauchy–Schwarz inequality, to obtain $\sum_j|\mu_{x,y}(E_j)|\leq\sum_j\sqrt{\mu_x(E_j)\mu_y(E_j)}\leq\sqrt{\mu_x(E)\mu_y(E)}$. The scalar inequality itself follows from nonnegativity of $\sum_j|u_j+t v_j|^2$ by the same quadratic minimization just used. Taking the supremum over partitions yields the sharper local variation estimate

$$|\mu_{x,y}|(E)\leq\sqrt{\mu_x(E)\mu_y(E)}, \qquad |\mu_{x,y}|(K)\leq C\|x\|\|y\|. \tag{PR32}$$

In particular this estimate handles zero vectors and zero diagonal mass exactly, without division by their norms or masses.

2.2 Bounded Borel pairings and their convergence

For bounded complex Borel $g$, define

$$B_g(x,y)=\int g\,d\mu_{x,y}. \tag{PR33}$$

Measure-valued sesquilinearity and integration linearity show that this is a sesquilinear form. By (PR25) and (PR32),

$$|B_g(x,y)|\leq C\|g\|_\infty\|x\|\|y\|. \tag{PR34}$$

It agrees with $F_g$ for continuous $g$, and $B_g(x,x)=\int g\,d\mu_x$. It is positive when $g\geq0$, and (PR31) gives $B_{\overline g}(x,y)=\overline{B_g(y,x)}$.

One additional estimate is useful for strong convergence. For a simple function $g=\sum_jz_j1_{E_j}$ on a finite disjoint partition of $K$, (PR31) and finite Cauchy–Schwarz give $|B_g(x,y)|\leq\sum_j|z_j|\sqrt{\mu_x(E_j)\mu_y(E_j)}\leq(\sum_j|z_j|^2\mu_x(E_j))^{1/2}\mu_y(K)^{1/2}$. Uniform simple approximation extends this to every bounded Borel $g$: uniform convergence of $g_j$ implies uniform convergence of $|g_j|^2$ because the sequence is uniformly bounded. Therefore

$$|B_g(x,y)|^2\leq\left(\int|g|^2\,d\mu_x\right)\mu_y(K) \leq C\|y\|^2\int|g|^2\,d\mu_x. \tag{PR35}$$

If $g_k$ are uniformly bounded Borel functions and converge pointwise to $g$, dominated convergence for $|\mu_{x,y}|$ proves $B_{g_k}(x,y)\to B_g(x,y)$ for every $x,y$. In fact (PR35) applied to $g_k-g$, with (PR19) for $\mu_x$, proves

$$\sup_{\|y\|\leq1}|B_{g_k-g}(x,y)|\longrightarrow0 \qquad(x\in H). \tag{PR36}$$

For $0\leq g_k\uparrow g$ with a common finite bound, (PR16) also gives the increasing convergence $B_{g_k}(x,x)\uparrow B_g(x,x)$. For bounded nonnegative lower semicontinuous $g$, the explicit continuous functions (PR22) give $B_g(x,x)=\sup\{F_f(x,x):f\in C(K;\mathbb R),0\leq f\leq g\}$ by (PR23). These conclusions specify the convergence and lower semicontinuous supremum used in the operator application.

2.3 The corresponding bounded operators, with the Hilbert-space step proved

We first prove the Hilbert representation needed to turn (PR33) into an operator. Let $\ell$ be a bounded conjugate-linear functional on a complex Hilbert space with the linear-first convention. If it is zero, its representing vector is zero. Otherwise $F(y)=\overline{\ell(y)}$ is a nonzero bounded linear functional. The closed affine set $A=\{y:F(y)=1\}$ is nonempty and has distance $d>0$ from the origin, since $1\leq\|F\|\|y\|$ on it. Choose $v_k\in A$ with $\|v_k\|\to d$. Its midpoints remain in $A$, and the parallelogram identity yields

$$\|v_k-v_l\|^2 =2\|v_k\|^2+2\|v_l\|^2-4\|(v_k+v_l)/2\|^2 \leq2\|v_k\|^2+2\|v_l\|^2-4d^2\longrightarrow0. \tag{PR37}$$

Completeness gives a limit $v\in A$ of norm $d$. For every $w\in\ker F$, the vector $v+t w$ lies in $A$ for every complex $t$. Expanding its squared norm and using the minimality of $\|v\|$ for real $t$ and for purely imaginary $t$ proves $\langle w,v\rangle=0$. Since $y-F(y)v\in\ker F$, it follows that $\langle y,v\rangle=F(y)\|v\|^2$. Thus $\ell(y)=\langle v/\|v\|^2,y\rangle$. Uniqueness follows by taking $y$ equal to the difference of two representing vectors. Its norm equals $\|\ell\|$: Cauchy–Schwarz gives one inequality, and evaluation at the unit vector in the representing vector’s direction gives the other. The Cauchy–Schwarz inequality for the Hilbert inner product follows from the positive-form quadratic proof in Section 2.1, so no separate representation result is being imported here.

For fixed $g$ and $x$, apply this result to $\ell(y)=B_g(x,y)$. There is a unique vector $\Psi(g)x$ with

$$\langle\Psi(g)x,y\rangle=B_g(x,y)=\int g\,d\mu_{x,y}. \tag{PR38}$$

Linearity in $x$ follows from sesquilinearity and uniqueness of the representing vector. The bound (PR34) gives $\|\Psi(g)\|\leq C\|g\|_\infty$. Linearity in $g$, positivity for $g\geq0$, and $\Psi(\overline g)=\Psi(g)^*$ follow by testing the corresponding vector equalities against every $y$, using (PR33)–(PR34) and their conjugation identity. If the original pairings came from $\Phi$, then $\Psi(f)=\Phi(f)$ for continuous $f$, by (PR30) and uniqueness. If $F_1(x,y)=\langle x,y\rangle$, then $\Psi(1)=I$.

Equation (PR35), followed by the equality between the norm of a vector and the norm of its represented functional, gives

$$\|\Psi(g)x\|^2\leq C\int|g|^2\,d\mu_x. \tag{PR39}$$

Consequently uniformly bounded pointwise convergence of a sequence of Borel functions implies strong operator convergence on each vector, by (PR19): $\|\Psi(g_k)x-\Psi(g)x\|^2\leq C\int|g_k-g|^2\,d\mu_x\to0$. Weak convergence follows as well. All convergence assertions here concern sequences.

The constructed map is the unique linear extension of its continuous-function values that preserves uniformly bounded pointwise sequential convergence in every operator pairing. To prove this, let $\widetilde\Psi$ be another extension with that convergence property, and fix $x,y$. The continuous approximants to $1_U$ in Section 1.8 show equality of the two pairings on indicators of open sets. The Borel sets on whose indicators the two pairings agree form a lambda-system: they include $K$, complementation uses the common value on $1$, and a disjoint countable union follows from linearity for partial sums of indicators followed by the assumed sequential convergence. The proved pi-system argument extends equality to all Borel indicators. Linearity gives equality on simple functions, and uniform simple approximation, which is in particular uniformly bounded pointwise sequential approximation, gives equality for every bounded Borel function. This is true for every $x,y$, so the operators agree.

Multiplicativity is a further property when the original continuous map is a homomorphism; no multiplicativity follows from positivity alone, and none is assumed in the construction above. All representation, measure, polarization, variation, bounded-Borel pairing, regularity, uniqueness, and convergence assertions used in (PR38)–(PR39) have been proved here.

3. The full calculus of a bounded positive contraction

3.1 Foundations and exact statement

Let $H$ be a complex Hilbert space and let $S\in\mathcal B(H)$ satisfy $$S=S^*,\qquad (Sx,x)\geq0,\qquad\|S\|\leq1, \qquad\ker S=\{0\}. \tag{BS1}$$ Set $K=[0,1]$. We prove that there is a unique strongly countably additive orthogonal projection-valued measure $E$ on the Borel subsets of $K$, with $E(K)=I$, such that $$S=\int_K t\,dE(t). \tag{BS2}$$ For every bounded Borel function $f:K\to\mathbb C$ the integral is a bounded operator. For every Borel function finite at each point, including unbounded functions, its closed densely defined integral has exactly the domain $$D(f(S))=\left\{x\in H:\int_K|f|^2\,d\mu_x<\infty\right\}, \quad\mu_x(B)=(E(B)x,x), \quad\|f(S)x\|^2=\int_K|f|^2\,d\mu_x. \tag{BS3}$$ We prove all adjoint and product domains and $E(\{0\})=0$. No claim that $S\geq cI$ for a positive $c$ is made.

The scalar interval measure theorem, integration, convergence and finite complex-measure uniqueness are proved in Section 1. The representation of bounded Hilbert functionals is proved in Section 2. The remaining foundations are the Hilbert-space axioms, completeness of bounded operators in operator norm, and the elementary algebra and compactness used below.

Every finite positive Borel measure $\nu$ on the interval is regular: apply the already proved representation theorem to $f\mapsto\int f\,d\nu$; its regular representing measure agrees with $\nu$ by Section 1.8, whose uniqueness proof does not assume regularity of the competing measure. If $\mu$ is a finite complex measure of finite variation and $f$ is bounded Borel, the set function $f\mu:B\mapsto\int_B f\,d\mu$ is countably additive. Indeed the finite partial-union indicators converge boundedly pointwise, and dominated convergence for $|\mu|$ passes the integrals to the union. Finite partitions give $$|f\mu|(B)\leq\int_B|f|\,d|\mu| \leq\|f\|_\infty|\mu|(B).$$ The finite positive measure $|\mu|$ is regular by the preceding argument. Given a Borel $B$, choose compact $F\subset B\subset U$ with $U$ open and $|\mu|(U\setminus F)$ arbitrarily small. The displayed bound then gives arbitrarily small $|f\mu|(U\setminus F)$. Thus $f\mu$ is regular as well. This proves the weighted-measure fact used in the multiplicativity argument.

3.2 Elementary positivity facts and Bernstein operator weights

If $B$ is a bounded positive self-adjoint operator, the scalar nonnegativity of $(B(x+zy),x+zy)$, for all $z\in\mathbb C$, proves $$|(Bx,y)|^2\leq(Bx,x)(By,y). \tag{BS4}$$ Indeed, if $(By,y)>0$, choose the phase and modulus of $z$ to minimize that quadratic expression. If $(By,y)=0$, any nonzero cross term would make the expression negative for a suitable phase and sufficiently large modulus; thus the cross term is zero. This proves (BS4) in both cases.

Write $\beta=\sup_{\|x\|=1}(Bx,x)$. Taking the supremum over unit $y$ in (BS4) gives $\|Bx\|^2\leq\beta(Bx,x)$. Therefore $\|B\|\leq\beta$, and the opposite inequality follows from Cauchy–Schwarz. We have proved $$\|B\|=\beta,\qquad \|Bx\|^2\leq\|B\|(Bx,x). \tag{BS5}$$ Editorial zero-space case. The unit-sphere supremum immediately above is used when $H\ne\{0\}$. If $H=\{0\}$, its unit sphere is empty and that real supremum does not define $\beta$. In this case take $\beta=0$ separately: the only operator is zero, so both equalities in (BS5) hold with both sides zero. Every later vector, projection and integral on this space is zero. In particular the essential operator norm is zero, using the infimum over nonnegative bounds, and the unit-vector arguments for a nonzero projection apply only when such a projection exists. No assumption of a nonzero Hilbert space is added to the theorem.

Apply this to $S$. The assumptions imply that $I-S$ is positive. Moreover $$((S-S^2)x,x)=(Sx,x)-\|Sx\|^2\geq0. \tag{BS6}$$ Consequently all four operators $I,S,I-S,S(I-S)$ are positive.

For integers $n\geq1$ and $0\leq k\leq n$, define $$W_{n,k}=\binom nk S^k(I-S)^{n-k}. \tag{BS7}$$ These operators are positive, as can be proved without taking an operator square root. Write $k=2a+\varepsilon$ and $n-k=2b+\delta$, where $\varepsilon,\delta\in\{0,1\}$, and put $Q=S^a(I-S)^b$. All factors are polynomials in the self-adjoint $S$, so they commute, and $Q=Q^*$. Then $$W_{n,k}=\binom nk Q\,S^\varepsilon(I-S)^\delta Q.$$ The middle factor is one of the four positive operators just proved positive. For each $x$, the quadratic form is its nonnegative value at $Qx$, multiplied by $\binom nk$. This proves positivity. The binomial formula for the commuting operators $S,I-S$ gives $$\sum_{k=0}^nW_{n,k}=I. \tag{BS8}$$