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For a self-adjoint C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $x$ is a normal C-star element, thenAll powers are normal C-star elements, so induction gives $\Vert x^{2^j}\Vert =\Vert x{\Vert }^{2^j}$. The general spectral radius formula now gives $r(x)=\Vert x\Vert$.

One can also see the needed growth estimate directly. For $R>r(x)$ the resolvent Cauchy coefficient formula givesThe formula follows by the Neumann series on a larger circle and contour deformation through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to \infty$ to get $\Vert x\Vert \le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the norm bound on the algebra spectrum. ThusThis proves spectral radius norm equality for normal elements from the norm identity.

Work in a nonzero complex unital Banach algebra $B$, with identity $1$. Write $R(x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is invertible}\}$ for the resolvent set, $\sigma (x)=\mathbb{C}\setminus R(x)$ for the spectrum of an element, and $Q(\lambda )=(\lambda 1-x{)}^{-1}$ for its Banach algebra resolvent. Distinguishing the set from the inverse-valued function avoids a notational ambiguity.

The printed formula omits an essential modulus. Complex spectral values cannot be ordered to take the indicated supremum. Even for $x=-1$ in $B=\mathbb{C}$, the spectrum is $\{-1\}$ while the spectral radius is one. The correctly interpreted assertion, proved below, isAll required analytic facts about the Banach algebra resolvent will be derived from norm-convergent series and scalar complex analysis.

For $\Vert a\Vert <1$, the Neumann series ${\sum }_{n\ge 0}{a}^n$ converges in $B$ and is a two-sided inverse of $1-a$: multiplying its finite sums gives $1-a^{N+1}$ and passing to the limit gives $1$. Therefore for $|\lambda |>\Vert x\Vert$,The series converges locally uniformly on that exterior region and tends to zero in norm as $|\lambda |\to \infty$. This statement does not require the optional normalization $\Vert 1\Vert =1$; the zeroth term is $1/\lambda$, and the other terms have the usual geometric bound.

At any ${\lambda }_0\in R(x)$, factor ${\lambda }_{0}1-x+h1=({\lambda }_{0}1-x)(1+hQ({\lambda }_0))$. If $|h|\Vert Q({\lambda }_0)\Vert <1$, the Neumann series givesThus the resolvent set is open, $Q$ is norm-continuous there, and the series proves norm differentiability with $Q^{\prime }(\lambda )=-Q(\lambda {)}^2$. Repeated differentiation gives $Q^{(j)}(\lambda )=(-1{)}^{j}j!Q(\lambda {)}^{j+1}$. Direct subtraction of the inverse equations also yields the resolvent identityIn particular, every continuous linear functional $\varphi$ makes $\varphi (Q(\lambda ))$ a scalar holomorphic function; this has been proved, not assumed as an algebra-valued analytic theorem.

The spectrum of an element is closed, and is contained in $\{\lambda :|\lambda |\le \Vert x\Vert \}$ by the exterior series, so it is compact. It is nonempty. Otherwise $Q$ would be defined and norm-continuous on the entire plane, bounded on every fixed closed disk and tending to zero at infinity. Each $\varphi (Q)$ would be a bounded entire scalar function, hence identically zero by Liouville's theorem. Continuous complex linear functionals separate points of $B$, so $Q(\lambda )=0$, impossible for an inverse. This separation follows from the real Hahn-Banach theorem: a continuous real functional $h$ not vanishing on a given vector gives the complex linear functional $\varphi (v)=h(v)-ih(iv)$, which also does not vanish there. We have therefore proved thatis well-defined and finite.

Next establish the existence of the power-norm limit. If $x^m=0$ for some $m$, the limit is zero. Otherwise fix $m$ and write $n=qm+r$, $0\le r<m$. Submultiplicativity givesThe finitely many remainder factors have bounded size, and $q/n\to 1/m$. Hence ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$ for every $m$. The opposite lower bound by the infimum is automatic, soFor $|\lambda |>r_0$, the root test proves convergence of $\sum x^n/{\lambda }^{n+1}$ even if $|\lambda |\le \Vert x\Vert$. Its finite sums telescope against $\lambda 1-x$, and their remainder tends to zero, so this sum is an inverse. Thus $s\le r_0$.

For the reverse inequality, fix $R>s$ and first choose $T>\max (R,\Vert x\Vert )$. The exterior Neumann series converges uniformly on $|\lambda |=T$, so termwise vector integration givesOnly its $x^n{\lambda }^{-n-1}$ term contributes. The annulus between radii $R$ and $T$ is in the resolvent set. For every continuous complex linear $\varphi$, the scalar Cauchy integral theorem applied to ${\lambda }^n\varphi (Q(\lambda ))$ makes the two contour integrals equal. Continuous linear functionals commute with vector integration and separate points, so the vector integrals themselves agree. This proves the resolvent Cauchy coefficient formula without presupposing any Banach-algebra-valued analytic-function theorem:Let $M_R={\max }_{|\lambda |=R}\Vert Q(\lambda )\Vert$, finite by continuity. The contour-length estimate gives $\Vert x^n\Vert \le M_R{R}^{n+1}$. Taking $n$th roots gives $r_0\le R$, and letting $R\downarrow s$ gives $r_0\le s$, including $s=0$. Combined with the earlier inequality, this proves the spectral radius formula and the corrected printed identity.

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.