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Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.

We prove the forward direction of the finite-generator criterion for a natural uniform algebra. Suppose that $A$ is a natural uniform algebra, and let $f_1,\dots ,f_n$ have no common zero on $K$. The idealis all of $A$. Otherwise it is contained in a maximal ideal $M$, whose associated character of an algebra vanishes on all the $f_j$. Since every algebra character is an evaluation character under naturality, this would give a common zero, a contradiction. Therefore $1\in J$, meaning that

Here are the Banach-algebra details behind the algebra character associated with $M$. A proper ideal cannot have dense closure: if it contained an element within distance less than one of $1$, the Neumann series would make that element invertible, forcing the ideal to contain $1$. Thus the closure of a proper ideal is proper. A maximal ideal, obtained from Zorn's lemma, is therefore closed. Its quotient is a complex unital Banach algebra in which every nonzero element is invertible. The Gelfand-Mazur theorem identifies this quotient with $\mathbb{C}$, so the quotient map is the required algebra character. No closedness of the originally generated ideal $J$ is needed.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

A character of an algebra is a nonzero complex linear multiplicative map $\varphi :A\to \mathbb{C}$. It necessarily satisfies $\varphi (1)=1$. If $\lambda =\varphi (x)$, then $x-\lambda 1$ cannot be invertible, since applying $\varphi$ to an inverse identity would give $0=1$. Therefore $\varphi (x)\in \sigma (x)$. The Neumann series bound $\sigma (x)\subseteq \{z:|z|\le \Vert x\Vert \}$ givesproving automatic continuity of characters. In particular every algebra character belongs to the closed unit ball of $A^{\ast }$; no prior continuity was needed.

The character space of an algebra is ${\Phi }_A$, the set of all algebra characters. Its Gelfand topology is the weakest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous, equivalently the weak-star topology inherited from $A^{\ast }$. Within the weak-star compact ball $B_{A^{\ast }}$, it is the closed set specified byThese equations are closed conditions on finitely many evaluations at a time. By the Banach-Alaoglu theorem, ${\Phi }_A$ is compact, and it is Hausdorff because distinct characters differ on an evaluation.

To obtain every spectral value from a algebra character, take $\lambda \in \sigma (x)$. Commutativity makes the ideal $(x-\lambda 1)A$ proper, so it is contained in an algebraic maximal ideal $\mathfrak{m}$. Every such maximal ideal is norm closed: its closure is an ideal, and cannot be all of $A$, because an element of $\mathfrak{m}$ sufficiently close to $1$ would be invertible by the Neumann series. Maximality then makes its closure equal to itself.

The quotient $A/\mathfrak{m}$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. Composing the quotient map with this scalar identification gives a algebra character with $\varphi (x)=\lambda$. Together with the first inclusion,The same maximal-ideal argument shows that the algebra character space of a nonzero algebra is nonempty.

Now suppose $A$ is a commutative C-star algebra. All its elements are normal. The C-star identity givessince the self-adjoint element $h=x^{\ast }x$ satisfies $\Vert h^2\Vert =\Vert h{\Vert }^2$. Iterating yields $\Vert x^{2^n}\Vert =\Vert x{\Vert }^{2^n}$. The spectral radius formula along this subsequence provesThis is spectral radius norm equality for normal elements, applied to the commutative case.

For a self-adjoint $h$, the element $v_t=\exp (ith)$ is unitary for every real $t$, so $\Vert v_t\Vert =1$. Algebra character continuity and multiplicativity giveIts modulus is $\exp (-t\mathrm{Im}\varphi (h))$, forcing $\varphi (h)\in \mathbb{R}$. Write $x=h+ik$ with $h,k$ self-adjoint. ThenThis proves that characters of a C-star algebra respect the involution. The conjugation bar is present in the original PDF and must be retained; the TeX aid omits it.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra is isometrically star-isomorphic to $C(K)$ for a compact Hausdorff space, canonically $K={\Phi }_A$. Its Gelfand transform isIt is a unital algebra homomorphism; the conjugation identity just proved makes it preserve the involution. The algebra character description of the spectrum and the spectral-radius equality giveThus it is injective and isometric, and its range is closed because $A$ is complete. The range contains constants, is closed under complex conjugation, and separates points of ${\Phi }_A$: two distinct algebra characters differ on some element of $A$.

The complex Stone-Weierstrass theorem says that a self-adjoint unital subalgebra of $C(K)$ separating points is uniformly dense. For completeness, its familiar lattice argument explains the last step. The real part of its uniform closure is closed under absolute value, by polynomial approximation of $|t|$ on bounded real intervals, and hence under finite maxima and minima. Constants and point separation permit a real function matching any given real continuous $g$ at any chosen pair of points. Fixing the first point, take a maximum of finitely many such functions to obtain a function above $g-\epsilon$ everywhere and equal to $g$ at that first point. It is below $g+\epsilon$ in a neighborhood of that point. A finite cover by these neighborhoods and the minimum of their associated functions then lies between $g-\epsilon$ and $g+\epsilon$ everywhere. Real and imaginary parts give density for complex functions.

Apply this to $\mathcal{G}(A)$. Its range is both dense and closed, so it is all of $C({\Phi }_A)$. ThereforeThis proves the commutative theorem in the setting of the question. For the zero algebra the corresponding compact space is empty and the representation is $C(\varnothing )=\{0\}$.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

Invertibility is stable under sufficiently small perturbations, and inversion is continuous. Let $1$ be the identity of the complex unital Banach algebra $A$, with its submultiplicative algebra norm. For $\Vert h\Vert <1$, completeness makes the Neumann seriesconverge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.

Fix $a\in G(A)$, the group of invertible elements of a Banach algebra. If $\Vert b-a\Vert <\Vert a^{-1}{\Vert }^{-1}$, thenso the Neumann series makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local boundWe retain $\Vert 1\Vert$ here so the estimate does not silently assume a normalized identity. The inverse identitythen impliesThis proves continuity of inversion in a Banach algebra.

For a unital Banach algebra, the spectrum of an element isFor a nonunital Banach algebra, use its unitization of an algebra $A^{+}=A\oplus \mathbb{C}$ with multiplication and normIt is a unital Banach algebra, with identity $(0,1)$, and define ${\sigma }_A(x)={\sigma }_{A^{+}}((x,0))$. In this nonunital convention, $0$ belongs to the spectrum: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^{+}$.

The spectrum is nonempty and compact. It suffices to work in a nonzero unital Banach algebra, since unitization reduces the other case to this one. The group of invertible elements of a Banach algebra is open, so the complement of the spectrum of an element is open. For $|\lambda |>\Vert x\Vert$, the Neumann series givesThus ${\sigma }_A(x)$ is closed and lies in $\{\lambda :|\lambda |\le \Vert x\Vert \}$, hence is compact.

For nonemptiness of the Banach-algebra spectrum, suppose that $R$ were defined on all of $\mathbb{C}$. At any ${\lambda }_0$, factoring $\lambda 1-x=({\lambda }_{0}1-x)(1+(\lambda -{\lambda }_0)R({\lambda }_0))$ gives a locally convergent power seriesFor every bounded linear functional $\ell \in A^{\ast }$, the scalar function $\ell (R(\lambda ))$ is therefore entire. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the Liouville theorem, it is constant; since it tends to zero at infinity, it is identically zero. The version of the Hahn-Banach theorem used here says that bounded linear functionals separate points of a normed vector space: if $z\ne 0$, there is $\ell$ with $\ell (z)\ne 0$. Consequently $R(\lambda )=0$ for every $\lambda$, contradicting $(\lambda 1-x)R(\lambda )=1$. This proves the assertion.

Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb{C}$. This is the Gelfand-Mazur theorem; completeness is not needed in its statement. Let $D$ be a normed division algebra, and take its completion of a normed space using the algebra norm to obtain a unital Banach algebra $\hat{D}$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.

For $x\in D$, the spectrum of an element of $x$ in $\hat{D}$ contains some $\lambda$. If $x-\lambda 1\ne 0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\hat{D}$. This contradicts $\lambda \in {\sigma }_{\hat{D}}(x)$. Therefore $x=\lambda 1$. The map $\lambda \mapsto \lambda 1$ is a bijective complex algebra homomorphism, andIt and its inverse are continuous; when $\Vert 1\Vert =1$, it is an isometry.

A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known. Let $A$ be the given algebra of functions. Form the same artificial unitization $A^{+}$ even if $A$ already has an identity. For each $t\in K$, the algebraic mapis a unital multiplicative complex linear functional. No continuity has been assumed. Since ${\chi }_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying ${\chi }_t$ to an inverse equation would give $0=1$. Hence $f(t)\in {\sigma }_{A^{+}}(f)$, and the spectrum bound already proved yieldsTaking the supremum proves the supremum bound for a complete function-algebra norm:In particular every $f$ is bounded, and every point evaluation is a bounded linear functional of norm at most $1$. This is an instance of automatic continuity of characters: a character of an algebra on a Banach algebra is bounded because its value at any element belongs to that element's spectrum.

A character of an algebra is a nonzero complex-linear multiplicative map $\phi :A\to \mathbb{C}$. The character space of an algebra ${\Phi }_A$ carries the Gelfand topology, namely pointwise convergence on $A$, equivalently the subspace weak-star topology in $A^{\ast }$.

For a unital complex Banach algebra, $\phi (1)=1$: multiplicativity and nonzeroness force this. If $a-\phi (a)1$ were invertible, applying $\phi$ to its product with its inverse would give $0=1$. Thus $\phi (a)\in {\sigma }_A(a)$. The Neumann series shows ${\sigma }_A(a)\subset \{|z|\le \Vert a\Vert \}$, so $|\phi (a)|\le \Vert a\Vert$ and the character of an algebra is automatically bounded. The same conclusion for a nonunital algebra follows by extending $\phi$ to its unitization of an algebra via $\tilde{\phi }(a,t)=\phi (a)+t$ and using the standard norm $\Vert (a,t)\Vert =\Vert a\Vert +|t|$.

In the algebra $R(K)$ let $u(z)=z$ and fix $\phi \in {\Phi }_{R(K)}$. Write $\lambda =\phi (u)$. If $\lambda \notin K$, the rational function $(u-\lambda {)}^{-1}$ belongs to $R(K)$, which is impossible because a character of an algebra cannot vanish on an invertible element. Thus $\lambda \in K$. For a rational function $r=p/q$ without poles in $K$, choose a reduced denominator having no zeros in $K$. Multiplicativity gives $\phi (r)=p(\lambda )/q(\lambda )=r(\lambda )$. Density of these rational functions and boundedness of $\phi$ then yield $\phi (h)=h(\lambda )$ for every $h\in R(K)$.

Conversely, evaluation at each $\lambda \in K$ is a character of an algebra. The map $\lambda \mapsto {\delta }_{\lambda }$ is continuous in the Gelfand topology because every $h\in R(K)$ is continuous; its inverse is $\phi \mapsto \phi (u)$, also continuous. Hence the character space of R(K) isThe identification includes the topology, not merely a bijection of sets.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula