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The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

For a normal operator with disconnected spectrum $K=K_1\bigsqcup K_2$, the indicator function of a nonempty proper clopen component union is continuous on $K$. The continuous functional calculus makes it a nonzero proper self-adjoint linear projection commuting with $T,T^{\ast }$. Its range is therefore a nontrivial closed reducing subspace. No eigenvector is needed.

Here strictly positive means positive and invertible. Set $k=a^{-1/2}b{a}^{-1/2}\ge 1$ using congruence preserves positivity in a C-star algebra. The continuous functional calculus gives $k^{-1}\le 1$, since $1-t^{-1}\ge 0$ for $t\ge 1$. Then $a^{-1}-b^{-1}=a^{-1/2}(1-k^{-1})a^{-1/2}\ge 0$. The proof works without commutativity. If $1\le a$, functional calculus also gives $a^{-1}\le 1$.

For a unital star-subalgebra $A\subseteq B(H)$, its unit ball is dense in the unit ball of $A^{\prime \prime }$ in the strong operator topology. The same holds for self-adjoint unit balls and for positive unit balls, and the full unit balls may be approximated strongly together with their adjoints. Passing first to the norm closure allows continuous functional calculus; resolvent identities prove strong continuity of continuous functions vanishing at infinity, even for unbounded approximating nets of self-adjoint operators. A cutoff function supplies the norm bound, and a two-by-two self-adjoint matrix removes the self-adjoint restriction.

For a normal C-star element $x$ in a nonunital algebra, take its spectrum of an element in the C-star unitization. Zero belongs to that spectrum. Continuous functions on this compact spectrum that vanish at zero form an ideal canonically identified with the displayed function algebra. Their values under the unital continuous functional calculus belong to the original algebra because the scalar quotient evaluates them at zero. Star-polynomial approximation with the constant term removed gives exactly the image $C^{\ast }(x)$.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. The inequality $\Vert a{\Vert }^2\le \Vert a^{\ast }\Vert \Vert a\Vert$, and then the same inequality for $a^{\ast }$, show that the involution is isometric. In a nonzero unital C-star algebra, the identity has norm one.

For a Hermitian algebra element $h$, the exponential $e^{ith}$ is unitary for real $t$ and has norm one. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (e^{ith})$: factoring $e^{itz}-e^{it\lambda }=(z-\lambda )q(z)$ with entire $q$ proves this, since the evaluated factors commute. Hence $|e^{it\lambda }|\le 1$ for all positive and negative $t$, forcing $\mathrm{Im}\lambda =0$. Thus Hermitian algebra elements have real Banach algebra spectrum.

For a normal C-star element $a$, commutation of $a,a^{\ast }$ givesThe middle equality of norms uses the C-star identity on the Hermitian algebra element $a^{\ast }a$. Iterating along powers $2^n$ and applying the spectral radius formula proves $r(a)=\Vert a\Vert$.

Now suppose $A$ is commutative and unital. Every maximal ideal is closed: if it were dense, it would contain an element within distance less than one of the identity, hence an invertible element by the Neumann series, which is impossible for a proper ideal. Its quotient is a complex Banach division algebra and is therefore $\mathbb{C}$ by the Gelfand-Mazur theorem. Maximal ideals thus give characters. Conversely, every character $\phi$ satisfies $\phi (a)\in {\sigma }_A(a)$, because applying it to an inverse of $a-\phi (a)1$ would give a contradiction. Characters are consequently contractive. The converse spectral inclusion follows by placing the proper ideal generated by $a-\lambda 1$ in a maximal ideal. We have provedThe character space ${\Phi }_A$ is a weak-star closed subset of the dual unit ball, characterized by $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$. The Banach-Alaoglu theorem makes it compact Hausdorff. Define the Gelfand transform by $\hat{a}(\phi )=\phi (a)$. For Hermitian $h$, its values are real; writing $a=h+ik$ with both terms Hermitian shows $\widehat{a^{\ast }}=\overline{\hat{a}}$. In the commutative algebra every element is normal, soThe transform is therefore an isometric unital star-homomorphism and has closed range in $C({\Phi }_A)$. Its range separates characters by their definition and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence equal to $C({\Phi }_A)$. This proves the Commutative Gelfand--Naimark theorem:For a nonunital C-star algebra, its C-star unitization can be constructed without assuming an operator representation on a Hilbert space. Represent $A$ by left multiplication on itself, which is isometric because testing $L_a$ on $a^{\ast }/\Vert a\Vert$ gives $\Vert L_a\Vert =\Vert a\Vert$. Adjoin the identity operator and set $\Vert z\Vert =\Vert L_z\Vert$. For $z=a+\lambda 1$ and $b\in A$, the C-star identity in $A$ gives $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Consequently $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Applying this also to $z^{\ast }$ proves that the new involution is isometric and that its C-star identity holds. The extension by the identity is complete, since the isometric image of $A$ is closed and one adjoins just one dimension. Apply the unital result to a commutative $A^{+}$; the character $\infty$ given by $a+\lambda 1\mapsto \lambda$ has kernel $A$. Under the isomorphism, this kernel consists of the continuous functions vanishing at $\infty$. Thus $A\cong C_0({\Phi }_A)$ in the nonunital case, with locally compact character space ${\Phi }_A={\Phi }_{A^{+}}\setminus \{\infty \}$.

For a Hermitian $h$ in a possibly noncommutative algebra, apply this result to the commutative unital C-star subalgebra $C^{\ast }(1,h)$ in the unitization. Its Banach algebra spectrum at $h$ equals the ambient Banach algebra spectrum: both Banach algebra spectra are real, the complement of the ambient compact real Banach algebra spectrum is connected, and the resolvent-component result of Question 1 retains that whole complement in the smaller algebra. Characters of this subalgebra are determined by their values at $h$, so its character space is homeomorphic to $\sigma (h)$. We obtain the continuous functional calculus $C(\sigma (h))\to C^{\ast }(1,h)$.

Interpret ${\mathbb{R}}^{+}$ here as $[0,\infty )$. If $\sigma (h)\subseteq [0,\infty )$, apply this calculus to $t\mapsto \sqrt{t}$, obtaining a Hermitian $k$ with $k^2=h$. In a nonunital algebra this element remains in $A$ because the defining scalar function vanishes at zero. Conversely, if $h=k^2$ with $k=k^{\ast }$, spectral mapping theorem and the real Banach algebra spectrum of $k$ give $\sigma (h)=\{t^2:t\in \sigma (k)\}\subseteq [0,\infty )$. Hence nonnegative Banach algebra spectrum is equivalent to being the square of a Hermitian algebra element. A strict-positive interpretation would exclude $h=0$, and would not be equivalent to the assertion about an arbitrary Hermitian square.

Finally, use the two continuous real functions $t\mapsto \sqrt{\max (t,0)}$ and $t\mapsto \sqrt{\max (-t,0)}$ on $\sigma (x)$. Their calculus values $u,v$ are Hermitian, belong to $A$ also in the nonunital case, and satisfy

Put $K=\sigma (T)$ and use the continuous functional calculus $\pi :C(K)\to \mathcal{B}(H)$. It is a unital star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^{\ast }(I,T,T^{\ast })$, together with spectral permanence for C-star subalgebras, which is among the permitted C-star results.

Take the Hilbert space inner product linear in its first variable. For a vector $\xi$, let $H_{\xi }=\overline{\{\pi (f)\xi :f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi (f)$ and its adjoint, so it is reducing. The functional $f\mapsto ⟨\pi (f)\xi ,\xi ⟩$ is positive because $f\ge 0$ implies $\pi (f)=\pi (\sqrt{f}{)}^{\ast }\pi (\sqrt{f})$. It has, by the Riesz-Markov-Kakutani representation theorem, a finite regular positive Borel measure ${\mu }_{\xi }$ on $K$. The identitymakes $f\mapsto \pi (f)\xi$ an isometry on its quotient by the functions of zero $L^2$ norm. Density of continuous functions in $L^2({\mu }_{\xi })$ extends it to a unitary $U_{\xi }:L^2({\mu }_{\xi })\to H_{\xi }$. Multiplication by a continuous function intertwines with $\pi$ on this subspace.

Choose a maximal family of mutually orthogonal nonzero cyclic reducing subspaces. Their orthogonal sum is all of $H$: otherwise its reducing orthogonal complement contains a nonzero vector generating another such subspace. This argument does not require separability. For a bounded Borel measurable function $g$ on $K$, multiplication $M_g$ on every $L^2({\mu }_{\xi })$ is bounded with norm at most $\Vert g{\Vert }_{\infty }$. Define on that orthogonal sumMultiplication operators satisfy $M_{fg}=M_f{M}_g$, $M_{\overline{g}}=M_g^{\ast }$ and $M_1=I$. The same identities hold for their direct sums, and the norm bound is uniform in $\xi$. Thus ${\beta }_T$ is a norm-decreasing unital star-homomorphism on the algebra of actual bounded Borel measurable functions, and ${\beta }_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.

Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ for $z\ne 0$, taking $\mathrm{Arg}\in [0,2\pi )$. Let $S={\beta }_T(s)$. ThenThus every bounded normal operator has a normal square root. The Borel branch is essential in general: on a circle winding about zero a continuous square-root branch need not exist.

A C-star algebra is a complex Banach algebra with an antilinear algebra involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. First take a nonzero unital algebra. Its identity has norm one, since $\Vert 1{\Vert }^2=\Vert 1\Vert$. Alsoand exchanging $a,a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the algebra involution is continuous. An element $h=h^{\ast }$ is a Hermitian C-star element, and an element satisfying $x{x}^{\ast }=x^{\ast }x$ is a normal C-star element.

We establish the spectral facts needed for the representation, to avoid assuming the conclusion. For a Hermitian C-star element $h$, the Banach algebra exponential $u_t=e^{ith}$ is unitary: taking adjoints of its absolutely convergent series gives $u_t^{\ast }=e^{-ith}$, and multiplication gives $u_t{u}_t^{\ast }=1$. Hence $\Vert u_t\Vert =1$. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (u_t)$. Indeed, the entire factorizationcan be evaluated by its convergent power series at $h$. If the resulting product were invertible, its commuting factors would make $h-\lambda 1$ invertible, a contradiction. The spectral bound therefore gives $|e^{it\lambda }|\le 1$ for every real $t$. Both signs of $t$ force $\mathrm{Im}\lambda =0$. Thus Hermitian elements have real spectrum.

For completeness, the spectral radius formula needed below follows from the same resolvent theory. Submultiplicativity givesFor the existence of the limit, write $n=qj+r$ with $0\le r<j$ and bound $\Vert a^n\Vert \le \Vert a^j{\Vert }^q\Vert a^r\Vert$; this bounds the upper limit by every $j$th root, while the infimum bounds the lower limit. If a power vanishes, the conclusion follows directly. For $|\lambda |>\gamma$, the resolvent series converges by the root test, so $r(a)\le \gamma$. Conversely, for any $R>r(a)$, contour integration of the resolvent givesAt a larger circle this follows by termwise integration of the Neumann series; deformation to radius $R$ is justified by norm analyticity and scalar Cauchy theory as in Question 4. Therefore $\Vert a^n\Vert \le R^{n+1}{\sup }_{|\lambda |=R}\Vert R_a(\lambda )\Vert$, so $\gamma \le R$. Letting $R\downarrow r(a)$ proves $r(a)=\gamma$.

For a normal C-star element $a$, its adjoint commutes with it. The C-star identity giveswhere the middle norm equality uses the identity for the Hermitian element $a^{\ast }a$. Every power is again normal, so $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral-radius formula yieldsThis derives the spectral radius norm equality for normal elements from the axioms.

Now let $A$ be commutative and unital. Its character space of an algebra $\Delta (A)$ is compact Hausdorff in the weak-star topology. Every algebra character has norm at most one by the argument in Question 4, and this space is the weak-star closed subset of the dual unit ball defined by $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Compactness follows from the Banach-Alaoglu theorem; weak-star separation makes it Hausdorff.

We also need the full character-spectrum correspondence, not just one inclusion. Every proper ideal is contained in a maximal ideal. Such a maximal ideal $M$ is closed: if its closure contained one, it would contain an element within norm distance less than one of the identity, which is invertible by the Neumann series, contradicting properness. Its closure is therefore proper and maximality gives equality. The quotient $A/M$ is a complex Banach division algebra. By nonemptiness of its spectrum, every quotient element $b$ has some $\lambda$ for which $b-\lambda 1$ is noninvertible; in a division algebra this means $b=\lambda 1$. Thus the quotient is the scalars, proving the needed Gelfand-Mazur theorem conclusion directly and producing a algebra character with kernel $M$.

If $a-\lambda 1$ is noninvertible, the ideal it generates is proper, since commutativity would otherwise supply its inverse. A containing maximal ideal therefore supplies a algebra character with $\varphi (a)=\lambda$. The converse inclusion was already proved. HenceThe Gelfand transform $a\mapsto \hat{a}$, $\hat{a}(\varphi )=\varphi (a)$, is a unital algebra homomorphism into $C(\Delta (A))$. If $h=h^{\ast }$, the reality of its spectrum makes $\varphi (h)$ real. Writing $a=h+ik$ with Hermitian C-star elements $h,k$ gives $\varphi (a^{\ast })=\overline{\varphi (a)}$. Thus the transform respects the algebra involution. Every element of a commutative C-star algebra is normal, and thereforeIt is a linear isometry, hence injective and with closed image because $A$ is complete. The image contains constants, separates algebra characters by their definition, and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that image dense in $C(\Delta (A))$, and closedness makes it all of that space. We have proved the Commutative Gelfand--Naimark theorem:These steps show exactly how completeness, the algebra involution and the C-star norm identity enter the theorem. A general Banach algebra need not have an injective or norm-preserving Gelfand transform.

The general commutative theorem also has a nonunital form. Here is an elementary construction of the necessary C-star unitization, without assuming the representation theorem. Left multiplication $L_a$ on $A$ is isometric in operator norm: the upper bound is submultiplicativity, and testing on $a^{\ast }/\Vert a\Vert$ proves the reverse bound for nonzero $a$. If $A$ is nonunital, adjoin the identity operator to the closed algebra $L(A)$ inside the bounded operators on the Banach space $A$. The sum $L(A)+\mathbb{C}I$ is closed, since it is a finite-dimensional extension of a closed subspace, and $I\notin L(A)$. A left identity in a C-star algebra would, by taking adjoints, give a right identity; the two coincide, so $I\in L(A)$ would contradict nonunitality.

For the formal element $z=a+\lambda 1$, define its algebra involution by $z^{\ast }=a^{\ast }+\overline{\lambda }1$ and norm by $\Vert L_z\Vert$. For every $b\in A$,Taking suprema over the unit ball and using submultiplicativity gives $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Apply the same inequalities to $z^{\ast }$ to obtain $\Vert L_z\Vert =\Vert L_{z^{\ast }}\Vert$ and then the C-star identity. This is the C-star unitization by left multiplication. The original algebra is a closed ideal in it.

For commutative nonunital $A$, apply the proved unital theorem to $\tilde{A}$. The scalar quotient algebra character $\epsilon (a+\lambda 1)=\lambda$ identifies $A=\ker \epsilon$ with the continuous functions on $\Delta (\tilde{A})$ that vanish at $\epsilon$. Removing this one algebra character gives a locally compact Hausdorff space; those functions are exactly its continuous functions vanishing at infinity. Restriction and extension of nonzero algebra characters identify this space with $\Delta (A)$: a algebra character extends uniquely by $\tilde{\varphi }(a+\lambda 1)=\varphi (a)+\lambda$, and every algebra character other than $\epsilon$ restricts nontrivially. Thus the full nonunital conclusion is $A\cong C_0(\Delta (A))$.

Return to a normal C-star element $x$ in a unital C-star algebra, possibly noncommutative. Let $B=C^{\ast }(1,x)$, the closed unital star algebra generated by $x$ and $x^{\ast }$. It is commutative because the two generators commute. We first prove the spectral permanence for C-star algebras needed to use the ambient spectrum rather than a potentially larger subalgebra spectrum.

If $h=h^{\ast }\in B$, both its ambient and subalgebra spectra are real. The complement of the compact real set ${\sigma }_A(h)$ is connected. Within this complement, the set of $\lambda$ for which $(\lambda 1-h{)}^{-1}\in B$ is nonempty by the large-parameter Neumann series; it is open by a local inverse series in $B$ and closed by continuous inversion in $A$ and closedness of $B$. It is therefore the entire complement, proving ${\sigma }_B(h)={\sigma }_A(h)$. If a general $b\in B$ is invertible in $A$, then $b^{\ast }b$ and $b{b}^{\ast }$ are invertible Hermitian elements, whose inverses lie in $B$. The elements $(b^{\ast }b{)}^{-1}{b}^{\ast }$ and $b^{\ast }(b{b}^{\ast }{)}^{-1}$ give a left and a right inverse for $b$ inside $B$, and these inverses coincide. Applying this to every $b-\lambda 1$ proves ${\sigma }_B(b)={\sigma }_A(b)$ for all $b\in B$.

By the proved commutative theorem, $B\cong C(\Delta (B))$. The mapis onto by the character-spectrum correspondence and spectral permanence. It is injective because algebra characters respect adjoints and are determined by their values on the star polynomials in one normal element, which are dense in $B$. It is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Consequently the inverse Gelfand transform, under this identification, suppliesIt is a continuous unital star homomorphism and a linear isometry. Its image is exactly $B$. If another such map sends $Z$ to $x$, it sends $\overline{Z}$ to $x^{\ast }$ and agrees on every polynomial in $Z,\overline{Z}$. These functions contain constants, separate points of the compact spectrum and are closed under conjugation; Stone-Weierstrass gives their uniform density. Continuity proves uniqueness on all of $C({\sigma }_A(x))$. This is the requested continuous functional calculus, including its isometry and image assertion.

Two conventions in the printed statement need to be made explicit. A unital map into $A$ requires $A$ to have an identity. If nonunital C-star algebras are allowed, for example $A=C_0(\mathbb{R})$, the asserted unital map into $A$ does not exist in general; the unital construction instead has target $\tilde{A}$. Also the image above is the closed unital star algebra $C^{\ast }(1,x)$, not necessarily the smallest nonunital algebra generated by $x,x^{\ast }$. For $x=0$ in $A=\mathbb{C}$, the former is $\mathbb{C}1$ and the latter is $\{0\}$. Thus the intended assertion is correct if “generated” includes the identity; without that convention, this is a counterexample. The distinction is the unital versus nonunital generation by a normal element.

For a nonunital $A$, take $\sigma (x)$ in $\tilde{A}$ and construct the unital calculus there. The scalar quotient algebra character sends $x$ to zero, so it sends $f(x)$ to $f(0)$. Functions with $f(0)=0$ therefore give elements of $A$. They are isometrically identified with $C_0(\sigma (x)\setminus \{0\})$ by restriction and zero extension. Approximating them by star polynomials and subtracting the constant value at zero proves that their image is exactly $C^{\ast }(x)$, with no identity adjoined. This proves the nonunital continuous functional calculus as well as specifying the corrected form of the original claim.

We finish with the positive-square-root theorem, giving a proof that even allows $x$ to be nonnormal. Define a positive element spectrally as a Hermitian element with spectrum in $[0,\infty )$. The calculus of a Hermitian element shows its square is positive. Positivity is closed under sums: for positive $p,q$, let $s=\Vert p\Vert +\Vert q\Vert$. The calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the analogous bound for $q$, hence $\Vert s1-p-q\Vert \le s$. Since $p+q$ has real spectrum, the spectral norm bound forces it into $[0,2s]$. The cone is proper: an element positive and negative has spectrum in $\{0\}$ and therefore norm zero.

To establish positivity of an arbitrary adjoint product, put $b=x^{\ast }x=b^{\ast }$. Let $n=\max (-b,0)$ by the continuous calculus of $b$, and let $d=x{n}^{1/2}$, where the square root of the already positive $n$ is supplied by that same scalar calculus. Multiplicativity givesThus $d^{\ast }d$ has nonpositive spectrum. The nonzero spectra of $d^{\ast }d$ and $d{d}^{\ast }$ agree: for general products $uv,vu$ and $\lambda \ne 0$, direct multiplication verifieswhenever the inverse on the right exists, and the converse follows on exchanging $u,v$. Hence $d{d}^{\ast }$ is also nonpositive. Write $d=u+iv$ with Hermitian C-star elements $u,v$. Thenis positive, as a sum of Hermitian squares, and nonpositive, as a sum of the two nonpositive elements. Properness makes it zero. Therefore $d^{\ast }d=-d{d}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $d=0$, and $n^2=0$ then implies $n=0$. Thus $b=x^{\ast }x$ is positive. This proof does not use positivity of adjoint products in establishing an earlier step.

Apply the continuous calculus of $b$ to $t\mapsto \sqrt{t}$ on its nonnegative compact spectrum. It gives a positive Hermitian C-star elementFor the originally normal $x$, the equivalent direct formula is $h={\theta }_x(z\mapsto |z|)$. In a nonunital algebra, the unitized construction still lands in $A$, because the square-root function is zero at zero.

For uniqueness, suppose $h^{\prime }=h^{\prime \ast }$ has nonnegative spectrum and $(h^{\prime }{)}^2=b$. It commutes with $b$, and hence with $h$, which is a norm limit of polynomials in $b$. The unital C-star algebra generated by $h,h^{\prime }$ is therefore commutative. Its Gelfand transform, together with spectral permanence, gives two nonnegative continuous functions whose squares are equal. Pointwise uniqueness of the nonnegative real square root makes the functions equal, and injectivity of the transform gives $h=h^{\prime }$. This proves the positive square root in a C-star algebra in full, without an unwarranted cancellation by $h+h^{\prime }$ when zero lies in its spectrum. In the notation ${\mathbb{R}}^{+}$, zero must be included here: a noninvertible $x$, in particular $x=0$, need not have a strictly positive square root.

For $\lambda \ne 0$, suppose $R=(\lambda 1-ba{)}^{-1}$ exists. The identities $(\lambda 1-ab)a=a(\lambda 1-ba)$ and $b(\lambda 1-ab)=(\lambda 1-ba)b$ show, by multiplication on either side, thatFor example, multiplying on the left gives $1-{\lambda }^{-1}ab+{\lambda }^{-1}a(\lambda 1-ba)Rb=1$, and the right product gives the same cancellation. Thus invertibility of $\lambda 1-ba$ implies invertibility of $\lambda 1-ab$. Interchanging $a,b$ proves the converse. The nonzero spectra of products in opposite orders are therefore equal:In particular the requested inclusion follows. The exclusion of zero matters: for the unilateral shift operator $S$, $S^{\ast }S=1$ has Banach algebra spectrum $\{1\}$, whereas $S{S}^{\ast }=1-P_0$ has Banach algebra spectrum $\{0,1\}$.

Use the spectral definition of a Positive element of a C-star algebra: $h$ is positive if $h=h^{\ast }$ and ${\sigma }_A(h)\subseteq [0,\infty )$. We use the standard continuous functional calculus for a Hermitian element of a C-star algebra, including its real Banach algebra spectrum and the equality of its norm with its spectral radius. To prove $a^{\ast }a$ positive without assuming this as part of the definition, first establish two elementary cone facts.

If $p,q$ are positive and $s=\Vert p\Vert +\Vert q\Vert$, the continuous functional calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the corresponding bound for $q$, henceSince $p+q$ is self-adjoint, every spectral value $t$ is real and satisfies $|s-t|\le s$, so $t\ge 0$. Thus the positive cone of a C-star algebra is closed under addition. It is proper: if both $h$ and $-h$ are positive, their Banach algebra spectra force $\sigma (h)\subseteq \{0\}$, and the self-adjoint norm equality gives $h=0$. It is also norm closed: if $p_n\ge 0$ converges to $p$, choose $r\ge {\sup }_n\Vert p_n\Vert$; then $\Vert r1-p_n\Vert \le r$ passes to the limit and forces every spectral value of $p$ to be nonnegative.

Now put $h=a^{\ast }a$, and form its negative part $d=\max (-h,0)$ by continuous functional calculus. Let $c=a{d}^{1/2}$. Because $d$ is a function of $h$, the scalar identity $t\max (-t,0)=-\max (-t,0{)}^2$ givesThis has nonpositive Banach algebra spectrum. The previously proved equality of nonzero product Banach algebra spectra shows that $c{c}^{\ast }$ also has nonpositive Banach algebra spectrum; it is self-adjoint, so both $c^{\ast }c$ and $c{c}^{\ast }$ are negative elements. Write $c=u+iv$, where $u=(c+c^{\ast })/2$ and $v=(c-c^{\ast })/(2i)$ are self-adjoint. Direct multiplication givesBy the spectral mapping theorem, the squares $u^2,v^2$ are positive, so their sum is positive by the cone fact above. But the same sum is negative, since both its original summands are negative. Properness forces $c^{\ast }c+c{c}^{\ast }=0$. Then $c^{\ast }c=-c{c}^{\ast }$ is both negative and positive, so $c^{\ast }c=0$, whence $d^2=0$. The C-star identity applied to the self-adjoint $d$ gives $\Vert d{\Vert }^2=\Vert d^2\Vert =0$. Thus $d=0$, provingThis establishes positivity of adjoint products from spectral positivity. It also proves positivity under congruence: for $p\ge 0$ and any $z$, write $p=r^2$ using its functional-calculus square root; then $z^{\ast }pz=(rz{)}^{\ast }(rz)\ge 0$.

For the comparison of squares, we first derive inversion reverses the order of strictly positive elements. If $0<x\le y$ with both positive and invertible, thenThe continuous functional calculus gives $k^{-1}\le 1$, because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (k)\subseteq [1,\infty )$. Congruence then givesNo commutation of $x,y$ was assumed.

Next derive the square-root resolvent integral in a C-star algebra. For $s\ge 0$,For $s>0$, substituting $t=s{u}^2$ reduces the integral to $\sqrt{s}{\pi }^{-1}{\int }_0^{\infty }2/(1+u^2)du=\sqrt{s}$; for $s=0$ it is zero. For a positive element $x$, the norm of $x(t1+x{)}^{-1}$ is at most $\min (1,\Vert x\Vert /t)$. Thus the integral converges in norm, with integrable bounds $t^{-1/2}$ near zero and $\Vert x\Vert t^{-3/2}$ near infinity, uniform on $\sigma (x)$. The continuous functional calculus consequently givesPut $\alpha =a^2$ and $\beta =b^2$. The assumption $b^2-a^2\ge 0$ says $\alpha \le \beta$. For every $t>0$, inversion reverses the order of the strictly positive elements $t1+\alpha$ and $t1+\beta$. Subtracting the two square-root formulas therefore yieldsHere ${\alpha }^{1/2}=a$ and ${\beta }^{1/2}=b$ because $a,b$ themselves are positive. Every truncated integral has a positive integrand, so is positive; norm convergence and closedness of the positive cone of a C-star algebra make its limit positive. This proves order preservation by the positive square root in the required case:It is important that this argument does not factor $b^2-a^2$ as $(b-a)(b+a)$, which would be invalid for noncommuting elements.

For $p=b-a\ge 0$, apply the continuous function $t\mapsto \sqrt{t}$ on its nonnegative compact Banach algebra spectrum. The resulting element $p^{1/2}$ is self-adjoint, has nonnegative Banach algebra spectrum by continuous functional calculus, and satisfies $(p^{1/2}{)}^2=p$. Thus $b-a$ has a positive square root in a C-star algebra.

Finally, if the positive element $a$ is invertible, its compact Banach algebra spectrum is bounded away from zero. Let $\epsilon =\min \sigma (a)>0$; continuous functional calculus gives $a\ge \epsilon 1$. Since $b\ge a$, we also have $b\ge \epsilon 1$, so $\sigma (b)\subseteq [\epsilon ,\infty )$. The reciprocal function is continuous on this Banach algebra spectrum and supplies $b^{-1}$. Therefore $\boxed{a\text{ invertible}⟹b\text{ invertible}}$ under the given positivity assumptions.

The dense commutant orbit makes $\Omega$ separating for $M$, so $S_0(a\Omega )=a^{\ast }\Omega$ is well defined on the dense subspace $M\Omega$. Put $\tau (a)=⟨a\Omega ,\Omega ⟩$. The assumed identity says that this positive vector functional is a tracial positive functional. HenceIt extends to an antiunitary involution $J$ on $H$, with $J\Omega =\Omega$. This is the modular conjugation, and the modular operator in this tracial case is $\Delta =I$.

On the dense subspace $M\Omega$,Thus it acts by right multiplication and commutes with all left multiplications; $JMJ\subseteq M^{\prime }$. We prove the reverse inclusion rather than assuming the general Tomita commutation theorem.

Let $T\in M^{\prime }$, $C=\Vert T\Vert$, and $\eta =T\Omega$. The setis weakly compact and convex. Indeed, trace-class duality and the Banach-Alaoglu theorem make the bounded-operator ball compact in a topology stronger than the weak operator topology. The algebra $M$ is weak-operator closed, and $a\mapsto a\Omega$ is weakly continuous, so its image is weakly compact.

For $b\in M$ use the polar decomposition of a bounded operator $b=uh$, $h=|b|$. Both $h$ and $u$ belong to $M$: continuous functional calculus gives $h$, and $b(h+\epsilon I{)}^{-1}$ tends strongly to $u$. The polar decomposition is proved independently in Question 8. Since $T$ commutes with $M$,The trace identity givesTherefore $|⟨\eta ,b\Omega ⟩|\le C\tau (h)$. Meanwhilewith the supremum attained by $a=Cu$. For the upper bound, traciality gives $\tau (b^{\ast }a)=\tau (h^{1/2}{u}^{\ast }a{h}^{1/2})$; evaluating this on $\Omega$ bounds its absolute value by $\Vert a\Vert \tau (h)$.

These support inequalities imply $\eta \in \mathcal{B}$. Otherwise the Hahn-Banach separation theorem would separate $\eta$ from this closed convex set by a real inner-product functional. Since $M\Omega$ is dense, its separating vector could be approximated by some $b\Omega$, preserving strict separation, contrary to the displayed inequality. Thus $T\Omega =a\Omega$ for an $a\in M$ of norm at most $C$.

The bounded operator $R_a=J{a}^{\ast }J$ commutes with $M$ and sends $\Omega$ to $a\Omega$. It agrees with $T$ on the cyclic vector and therefore on all of $M\Omega$, hence on $H$. This proves $T\in JMJ$, and so

For the regular representation, take $\Omega ={\delta }_e$. It is cyclic for the left action and for the commuting right action. On finite sums of left translations, its vector functional is the identity coefficient, so $\tau (ab)=\tau (ba)$. This extends to the whole group von Neumann algebra by the bounded strong-star approximations of Question 6, taking limits of the vector functional and products. The conjugation is explicitlyConsequentlyThus the commutant in (b) is exactly the Von Neumann algebra of right regular translations, with the inverse in $\rho (g)$ fixing the representation convention.

Pass to $B={\overline{A}}^{\Vert \cdot \Vert }$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ is a star-algebra, every closed invariant subspace reduces it.

Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element $b$; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection $e\in B$ by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb{C}p$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb{C}v$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in \mathbb{C}v$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ is irreducible.

Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts $H$.

Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\Vert bv\Vert >0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.

Since $a-1$ and $b-1=(b-a)+(a-1)$ are positive, their spectra are nonnegative. Translation of the spectrum of an element gives $\sigma (a),\sigma (b)\subseteq [1,\infty )$, so both elements are invertible. Their inverses and inverse square roots are positive by the continuous functional calculus. Also $1-a^{-1}$ is positive because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (a)$.

For the order reversal, setThe inequality uses congruence preserves positivity in a C-star algebra. continuous functional calculus gives $k^{-1}\le 1$. Sinceanother congruence showsConsequentlyHere $b^{-1}$ is strictly positive in the stronger invertible sense: $b^{-1}\ge \Vert b{\Vert }^{-1}1>0$, because its spectrum is the set of reciprocals of the positive spectral values of $b$. This is inversion reverses the order of strictly positive elements; no commutativity of $a,b$ is required.