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A norm-closed subalgebra of a Banach algebra is itself a Banach algebra with the inherited norm: completeness follows from closedness, and the submultiplicative norm bound restricts to it. A unital inclusion uses the same identity as the ambient algebra. Inverse membership may still differ between the two algebras; spectrum in a closed unital subalgebra describes that difference.

An algebra involution is a map $x\mapsto x^{\ast }$ satisfyingThus it is a conjugate-linear map, reverses multiplication, and has square equal to the identity map. These are algebraic axioms; norm continuity, when included in a convention for a Banach star algebra, is an additional requirement and is not used here. A Hermitian element of a star algebra satisfies $h^{\ast }=h$.

Conjugate linearity gives $0^{\ast }=0$. The map $\ast$ is surjective, so every $u$ is $v^{\ast }$ for some $v$. The identities $(v1{)}^{\ast }=1^{\ast }{v}^{\ast }=v^{\ast }$ and $(1v{)}^{\ast }=v^{\ast }{1}^{\ast }=v^{\ast }$ show that $1^{\ast }$ is a two-sided identity; uniqueness of the identity gives $1^{\ast }=1$. Therefore both zero and the identity are Hermitian algebra elements.

For the final unheaded request, first prove equality of the two spectra for a Hermitian element of a star algebra $h\in B$. Write $D=\mathbb{C}\setminus {\sigma }_A(h)$. The spectrum is a compact subset of the real axis by the assumption that $A$ is a Hermitian Banach algebra, so $D$ is connected: points above and below the real axis can be joined by paths passing beyond the endpoints of an interval containing the spectrum; points of the real axis outside the spectrum can first move a short distance vertically. DefineFor large $|\lambda |$, the Neumann series ${\lambda }^{-1}{\sum }_{n\ge 0}(h/\lambda {)}^n$ converges in the closed unital star-subalgebra $B$, so $E$ is nonempty. It is relatively open in $D$ by the same series applied to a perturbation of an already invertible element of $B$. It is relatively closed in $D$, since the resolvent of an element is continuous in $A$ and $B$ is closed. Connectedness gives $E=D$. Conversely an inverse in $B$ is an inverse in $A$, soThis proves the needed spectrum in a closed unital subalgebra equality for Hermitian algebra elements without presupposing that their spectra in $B$ are real.

Now suppose $b\in B$ is invertible in $A$. Both $b{b}^{\ast }$ and $b^{\ast }b$ are Hermitian algebra elements, since reversing products and applying $\ast$ twice fixes them. They are invertible in $A$, by part (ii), so the preceding equality of spectra of Hermitian algebra elements makes them invertible in $B$. Part (ii), applied within $B$, then makes $b$ invertible in $B$. Applying this to $b-\lambda 1$ for every scalar $\lambda$, and using the automatic converse for an inverse already in $B$, proves spectral permanence for Hermitian Banach algebras:Only closedness, a common identity, closure under the algebra involution, and real ambient spectra of Hermitian algebra elements were used; the C-star identity was not assumed.

For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the character space of an algebra of $A$, with its weak-star or Gelfand topology. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff character space and ${\sigma }_A(a)=\{\phi (a):\phi \in K\}$. In particular, character compactness follows from the Banach-Alaoglu theorem: the equations $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$ define a weak-star closed subset of the dual unit ball.

The Gelfand transform is the unital homomorphismEvery element of the commutative C-star algebra is normal. Part i and the character spectrum formula therefore giveThus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma (a^{\ast })=\overline{\Gamma (a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex Stone-Weierstrass theorem states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact Hausdorff space is uniformly dense. Its application here, together with closedness of the range, yieldsThis proves the Commutative Gelfand--Naimark theorem from the preceding parts and the stated approximation theorem.

The word “every” in the printed deduction must be read under the initial unital hypothesis. If nonunital commutative C-star algebras are included, the compact-space statement is false: the nonzero algebra $c_0$ has no multiplicative identity, whereas $C(K)$ for nonempty compact $K$ does. The general nonunital version is $C_0(K)$ on a locally compact Hausdorff space, with vanishing at infinity. This qualification preserves the proved unital result.

For the final real spectrum criterion for a normal C-star element, the ambient algebra may be noncommutative. Let $B$ be the norm-closed unital subalgebra generated by $x$ and $x^{\ast }$. Normality makes these two generators commute. The involution preserves their polynomial algebra and is continuous, so $B$ is a commutative C-star subalgebra.

We first show that $x$ has the same spectrum in $B$ as in $A$, without assuming spectral permanence. Put $\Omega =\mathbb{C}\setminus {\sigma }_A(x)$. A compact subset of the real axis has connected complement: upper and lower half-plane paths join around either end of a bounded interval containing it, and each omitted real point can be joined locally to the upper half-plane. DefineThis set is nonempty because the Neumann series puts the inverse in $B$ for $|\lambda |>\Vert x\Vert$. It is relatively open: from an inverse in $B$, the local inverse formulastays in $B$ for small $h$. It is relatively closed by continuity of the ambient resolvent and norm-closedness of $B$. Connectedness gives $V=\Omega$. Conversely an inverse in $B$ is an inverse in $A$, soThis is the connected-resolvent case of spectrum in a closed unital subalgebra.

Apply the just-proved commutative C-star representation to $B$. For every character $\psi$ of $B$, $\psi (x)$ lies in this real spectrum. Part ii makes $\psi (x^{\ast })=\overline{\psi (x)}=\psi (x)$. The Gelfand transform of $B$ is injective by its isometry, soFor a nonunital ambient C-star algebra, perform the same argument in its standard C-star unitization, using the usual unitization spectrum.

Normality cannot be dropped. In $M_2(\mathbb{C})$, takeSince $N^2=0$, its spectrum is $\{0\}$: for $\lambda \ne 0$, $(\lambda I-N{)}^{-1}={\lambda }^{-1}I+{\lambda }^{-2}N$, while $N$ is not invertible. Yet $N\ne N^{\ast }$ andThus a real spectrum alone does not force self-adjointness; normality is essential.

Take a nonzero complex unital Banach algebra, with identity $1$. The spectrum of an element isThe standard Banach algebra spectrum theorem states that it is a nonempty compact subset of $\mathbb{C}$, contained in the disk $|\lambda |\le \Vert a\Vert$. Conversely, every nonempty compact set $K\subseteq \mathbb{C}$ occurs: in the Banach algebra $C(K)$, the coordinate function $a(z)=z$ has Banach algebra spectrum exactly $K$, since $\lambda -a$ has a continuous reciprocal precisely when $\lambda \notin K$. Thus the possible Banach algebra spectra are exactly the nonempty compact subsets of the complex plane.

For a nonconstant polynomial $p$, factor $p(z)-\lambda =c{\prod }_{j=1}^n(z-z_j)$ over $\mathbb{C}$. Substitution givesThese factors commute. A finite commuting product is invertible exactly when every factor is invertible: if $P={\prod }_j{x}_j$ is invertible, its inverse commutes with each $x_j$, and $x_j^{-1}=({\prod }_{k\ne j}{x}_k)P^{-1}$ is a two-sided inverse. Hence $\lambda$ belongs to the Banach algebra spectrum of $p(a)$ exactly when some root $z_j$ belongs to ${\sigma }_A(a)$, which is equivalent to $\lambda \in p({\sigma }_A(a))$. Constant polynomials give the same result directly, since ${\sigma }_A(c1)=\{c\}$. This proves the polynomial spectral mapping theorem:For a closed unital Banach subalgebra $B\subseteq A$, use the common identity $1_A\in B$. An inverse in $B$ is an inverse in $A$, so $\boxed{{\sigma }_A(b)\subseteq {\sigma }_B(b)}$. Let $\lambda \in \partial {\sigma }_B(b)$. Choose ${\lambda }_n\notin {\sigma }_B(b)$ tending to $\lambda$. Then ${\lambda }_n\notin {\sigma }_A(b)$, and $({\lambda }_{n}1-b{)}^{-1}\in B$. If $\lambda$ were outside ${\sigma }_A(b)$, continuity of inversion in $A$ would givewhere membership follows from closedness. This would make $\lambda \notin {\sigma }_B(b)$, contrary to the closed Banach algebra spectrum having its boundary inside it. Therefore $\lambda \in {\sigma }_A(b)$ and is approached by points outside that Banach algebra spectrum, provingContinuity of inversion here follows, for example, by writing a nearby element as an invertible element times $1+$ a norm-small perturbation and using the Neumann series. This is spectrum in a closed unital subalgebra.

To retain the original Banach algebra spectrum in a commutative subalgebra, take the resolvent-generated commutative algebraEvery generator commutes with $a$ and with every other generator: commuting invertible elements have commuting inverses, and their inverses commute with all elements commuting with them. commutativity persists on the algebra they generate and on its norm closure. For every $\lambda \notin {\sigma }_A(a)$, the required inverse is in $C$ by construction, so ${\sigma }_C(a)\subseteq {\sigma }_A(a)$. The inclusion for unital subalgebras gives the reverse containment. Hence $\boxed{{\sigma }_C(a)={\sigma }_A(a)}$.

Finally, the Banach subalgebra generated by one element is $D=\overline{\{p(a):p\text{ a polynomial}\}}$, and it is commutative. Put $K={\sigma }_D(a)$ and suppose its complement has a bounded connected component $U$. The boundary of $U$ lies in $K$. For $\lambda \in U$, the maximum modulus principle on a bounded domain applied to any polynomial givesHere $r_D$ is the spectral radius, and the equality uses the already proved polynomial spectral mapping theorem. Define $\chi (p(a))=p(\lambda )$. The displayed bound makes this well defined, even when two polynomials represent the same algebra element, and bounded on the dense polynomial algebra. Its continuous extension to $D$ remains linear and multiplicative, with $\chi (1)=1$ and $\chi (a)=\lambda$, so it is a character of an algebra. If $\lambda 1-a$ were invertible in $D$, applying $\chi$ to its inverse equation would give $0=1$. Thus $\lambda \in K$, contradicting $\lambda \in U$.

There are therefore no bounded complementary components. A compact subset of the plane has exactly one unbounded complementary component: everything outside a sufficiently large disk is connected, and every unbounded component meets it. Consequently $\boxed{\mathbb{C}\setminus {\sigma }_D(a)\text{ is connected}.}$

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving

For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.