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For a positive element $a$, its positive square root in a C-star algebra gives $c^{\ast }ac=(a^{1/2}c{)}^{\ast }(a^{1/2}c)$, which is positive. Applying this to $b-a$ shows that $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. The map need not preserve multiplication, but it preserves order; if $c$ is invertible, the inverse congruence also preserves order.

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.

For an element $a$ of a complex unital Banach algebra, its spectrum of an element isThe nonzero spectra of products in opposite orders agree:For $\lambda \ne 0$, if $R=(\lambda 1-ab{)}^{-1}$ exists, direct multiplication on both sides verifiesInterchanging $a,b$ proves the converse. Zero may differ. On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, let $S$ be the unilateral shift operator $S{e}_n=e_{n+1}$ and take $a=S^{\ast }$, $b=S$. Then $ab=S^{\ast }S=I$, whereas $ba=S{S}^{\ast }=I-P_0$, where $P_0$ is projection onto the first coordinate. HenceBoth values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.

We use the spectral definition of a Positive element of a C-star algebra: $h$ is positive when $h=h^{\ast }$ and $\sigma (h)\subseteq [0,\infty )$. The continuous functional calculus, supplied by the Commutative Gelfand--Naimark theorem for the algebra generated by a Normal element of a C-star algebra and its adjoint, sends $F\in C(\sigma (h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma (F(h))=F(\sigma (h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.

For example, $F(t)=\sqrt{t}$ gives the positive square root in a C-star algebra of a Positive element of a C-star algebra. It satisfies $(h^{1/2}{)}^2=h$. Its uniqueness follows from the composition rule: if $d\ge 0$ and $d^2=h$, then the continuous functional calculus in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\le h$ and $r\ge \Vert h\Vert$, the function $r-t$ gives $r1-h\ge 0$; the same calculus givesThese transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.

To prove closure under addition without assuming an order theorem, let $a,b\ge 0$, set $\alpha =\Vert a\Vert$, $\beta =\Vert b\Vert$ and $s=\alpha +\beta$. If $s=0$, both elements are zero. Otherwise,The element $a+b$ is self-adjoint, so any $\lambda \in \sigma (a+b)$ is real and satisfies $|s-\lambda |\le s$. Thus $\lambda \ge 0$, provingWe will also use that the positive cone is proper: if $h\ge 0$ and $-h\ge 0$, then $\sigma (h)\subseteq \{0\}$, and the norm formula for self-adjoint elements gives $h=0$.

We next prove $a^{\ast }a\ge 0$ without presupposing that positivity is defined by such products. Put $h=a^{\ast }a$, which is self-adjoint, and form its negative part by the continuous functional calculus:This is positive. Let $c=a{h}_{-}^{1/2}$. Since $h_{-}$ commutes with $h$,The nonzero product-spectrum identity proved above gives $\sigma (c{c}^{\ast })\setminus \{0\}=\sigma (c^{\ast }c)\setminus \{0\}$. As $c{c}^{\ast }$ is self-adjoint, it too has spectrum in $(-\infty ,0]$, so $c{c}^{\ast }\le 0$.

Write $c=u+iv$ with $u=u^{\ast }$ and $v=v^{\ast }$. The continuous functional calculus makes $u^2$ and $v^2$ positive, and the addition result givesBut both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^{\ast }c+c{c}^{\ast }=0$. Now $c^{\ast }c=-c{c}^{\ast }$ is positive as well as nonpositive, so $c^{\ast }c=0$. The C-star identity gives $c=0$, whence $h_{-}^2=0$ and $h_{-}=0$. Thus $h$ has no negative spectral part:This proves the positivity of adjoint products from spectral positivity and the equivalence with the usual definition by elements of the form $d^{\ast }d$: the reverse direction is the factorization $h=(h^{1/2}{)}^{\ast }{h}^{1/2}$ for $h\ge 0$.

Finally, if $a\ge 0$ and $c\in A$, writeThe preceding result therefore provesIn particular, congruence preserves positivity in a C-star algebra, and $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. These facts will justify the noncommutative order manipulations in the numbered parts.

A character of an algebra is a nonzero algebra homomorphism $\phi :A\to \mathbb{C}$. The character space isFor a unital algebra, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$, since applying $\phi$ shows that $a-\phi (a)1$ cannot be invertible. In a Banach algebra,Thus $\Vert \phi \Vert \le 1$, while $\phi (1)=1$ gives equality.

If $A$ is commutative and $\lambda \in {\sigma }_A(a)$, the proper ideal generated by $a-\lambda 1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb{C}$, and its quotient map is a character taking $a$ to $\lambda$. ThereforeSince every Banach-algebra element has nonempty spectrum, ${\Phi }_A$ is nonempty.

For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. ThenMultiplicativity would give $\phi (P)=\phi (S)\phi (T)=\phi (T)\phi (S)=\phi (Q)$, a contradiction. Hence ${\Phi }_{\mathcal{B}(X)}$ is empty. This is the absence of characters on an operator algebra with isomorphic complementary summands.

If $a=a^{\ast }$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda \in \sigma (a)$, the spectral mapping theorem gives $e^{it\lambda }\in \sigma (e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\mathrm{Im}\lambda =0$, so $\sigma (a)\subset \mathbb{R}$.

If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal elementhas spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda {)}^{-1}\in B$. Thus ${\sigma }_B(x)={\sigma }_A(x)$.

The Commutative Gelfand--Naimark theorem says that the Gelfand transform is an isometric unital star-isomorphismfor every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto \sqrt{t}$ on $\sigma (a)\subset [0,\infty )$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^{\ast }(1,a)\cong C(\sigma (a))$ gives the unique positive square root. This is the positive square root in a C*-algebra.