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If $X=Y\oplus Z$ and the Banach spaces $Y$ and $Z$ are isomorphic, then the operator algebra $\mathcal{B}(X)$ has no characters. The two complementary projections have unequal character values, while mutually inverse off-diagonal maps force those values to be equal.

For a character of an algebra $\phi$ on a complex unital Banach algebra, $\phi (1)=1$ and $\phi (a)\in \sigma (a)$, so $|\phi (a)|\le \Vert a\Vert$ without any prior continuity assumption. Under the usual normalization $\Vert 1\Vert =1$, this gives $\Vert \phi \Vert =1$; for an unnormalized algebra norm, only the upper bound is automatic.

The character space ${\Phi }_A$ is the set of all characters on $A$.

The coordinate function has spectrum $[0,1]$: outside the interval its difference from a scalar has a continuously differentiable reciprocal, and on the interval it has a zero. Every character of an algebra therefore takes the coordinate to some $t\in [0,1]$. Polynomial density in the norm containing both function and derivative, together with automatic continuity of characters, forces evaluation at $t$ on every function. The maximal ideals are exactly $\{f:f(t)=0\}$.

The disk algebra consists of functions continuous on the closed unit disk and holomorphic in its interior, with the supremum norm. Polynomials are dense: dilations $f_r(z)=f(rz)$ converge uniformly to $f$ as $r\uparrow 1$, and each $f_r$ is uniformly approximated on the closed disk by its Taylor polynomials. A character of an algebra is therefore determined by its value $\lambda$ at the coordinate function; boundedness gives $|\lambda |\le 1$, and polynomial approximation yields $\phi (f)=f(\lambda )$. Consequently the disk algebra is a natural uniform algebra on the closed disk.

A uniform algebra is natural exactly when functions with no common zero generate the unit ideal. If a generated ideal is proper, it is contained in a maximal ideal. That maximal ideal is closed: the closure of a proper ideal stays proper, since an element within distance less than one of the identity is invertible by the Neumann series. The Gelfand-Mazur theorem identifies the maximal-ideal quotient with the complex numbers, yielding a character of an algebra. Naturality would make that character a common-zero evaluation, a contradiction. Conversely, the asserted unit-ideal property makes the zero sets of every finite collection in a character's kernel intersect. Compactness gives a point common to all these zero sets; applying this to $f-\phi (f)1$ proves $\phi (f)=f(x)$. For $C(K)$, one can directly take $g_j=\overline{f_j}/{\sum }_k|f_k{|}^2$.

The complex dual numbers $\mathbb{C}[\epsilon ]/({\epsilon }^2)$, with norm $\Vert a+b\epsilon \Vert =|a|+|b|$ and involution $a+b\epsilon \mapsto \bar{a}+\bar{b}\epsilon$, is a commutative unital Banach algebra with an isometric involution. Its sole character of an algebra kills $\epsilon$, so the Gelfand transform is neither injective nor isometric. The C-star identity excludes this example and gives the Commutative Gelfand--Naimark theorem. Even preservation of the involution on characters can fail without that identity: on ${\mathbb{C}}^2$, the involution $(a,b{)}^{\ast }=(\bar{b},\bar{a})$ exchanges the two evaluation characters.

In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the algebra character uniquely since the quotient is one-dimensional.

In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the character uniquely since the quotient is one-dimensional.

A uniform algebra on $K$ is natural if every character of an algebra is evaluation at a point of $K$. The evaluation embedding then identifies its character space homeomorphically with $K$. A natural uniform algebra is characterized by the finite-generator criterion for a natural uniform algebra. In particular, $C(K)$ is natural; the disk algebra is natural when its underlying space is the closed disk.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

Write $K={\sigma }_A(x)$ and $R_x(\zeta )=(\zeta 1-x{)}^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda \notin U$, the holomorphic function $1/(Z-\lambda )$ belongs to $\mathcal{O}(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda 1$. Thus existence forces $K\subset U$, even before continuity is used.

For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. DefineThe integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The Cauchy theorem shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying bounded linear functionals. We use the scalar Cauchy integral formula in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta )/(\zeta -z)$ to $2\pi if(z)$.

Linearity is immediate and, for a fixed cycle,This is a bound by one compact-uniform seminorm, proving continuity for local uniform convergence. To obtain ${\Theta }_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb{C}\setminus K$, to a large circle and integrate its Neumann series. Similarly $\zeta R_x(\zeta )=1+x{R}_x(\zeta )$ gives ${\Theta }_x(Z)=x$.

For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The resolvent identity isSubstitute it into the double integral for ${\Theta }_x(f){\Theta }_x(g)$. In the term containing $R_x(\eta )$, first integrate $f(\zeta )/(\zeta -\eta )$ around the outer cycle, obtaining $f(\eta )$. In the term containing $R_x(\zeta )$, first integrate $g(\eta )/(\zeta -\eta )$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. HenceThis proves the required continuous unital homomorphism, the holomorphic functional calculus.

For uniqueness use the following version of the allowed Runge theorem: on any open $U\subseteq \mathbb{C}$, rational functions with all finite poles outside $U$ approximate each holomorphic function locally uniformly; poles at infinity are allowed, giving polynomials. A unital homomorphism sending $Z$ to $x$ is forced on polynomials and on every inverse $(Z-\lambda {)}^{-1}$, $\lambda \notin U$, and hence on these rational functions. Continuity then forces its value on their limits. Thus ${\Theta }_x$ is unique. This argument works for disconnected $U$ as well and does not assume polynomials alone are dense. It proves continuity and uniqueness of holomorphic functional calculus.

We will use the character description of spectra, with a short justification. A character of an algebra $\phi$ is a nonzero complex multiplicative linear functional; it satisfies $\phi (1)=1$. Applying it to an inverse shows $\phi (a)\in {\sigma }_A(a)$, so $|\phi (a)|\le \Vert a\Vert$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda 1$, its principal ideal is proper because $A$ is commutative. Put it in a maximal ideal $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb{C}$ and its quotient map is a character taking $a$ to $\lambda$. This provesIt is the spectrum equals character values in a commutative Banach algebra statement and does not assume semisimplicity.

For any character, continuity permits passing it inside the contour integral. Since $\phi (R_x(\zeta ))=(\zeta -\phi (x){)}^{-1}$ and $\phi (x)\in K$, the scalar Cauchy formula givesApply the character description first to ${\Theta }_x(f)$ and then to $x$ to obtain the holomorphic spectral mapping theorem

On ${\Pi }_{+}$, use the analytic logarithm with argument in $(-\pi /2,\pi /2)$ and set $s(z)=\exp (\frac{1}{2}\mathrm{Log}z)$. It satisfies $s(z{)}^2=z$ and takes its values in ${\Pi }_{+}$. The principal square root in a commutative Banach algebra isFor uniqueness let $v$ be another root with spectrum in ${\Pi }_{+}$. For each character, $\phi (y)$ and $\phi (v)$ are scalar roots of $\phi (x)$ with positive real part, so both equal $s(\phi (x))$. Hence every character takes the nonzero value $2s(\phi (x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.

Let $h=h^{\ast }$. The exponential $u_t=e^{ith}$ is defined by its norm-convergent power series. Continuity and conjugate linearity of the involution give $u_t^{\ast }=e^{-ith}$, and multiplication of the commuting exponential series gives $u_t^{\ast }{u}_t=u_t{u}_t^{\ast }=1$. Thus $u_t$ is a Unitary element of a C-star algebra and, by the C-star identity, $\Vert u_t\Vert =1$.

A character of an algebra is automatically continuous with $|\phi (a)|\le \Vert a\Vert$, as proved in Question 5. Passing it through the exponential series givesThe left side is $e^{-t\Im \phi (h)}$. Considering both signs of $t$ forces $\Im \phi (h)=0$. So characters take real values on self-adjoint C-star elements.

Now write $x=h+ik$ withboth self-adjoint C-star elements. ThenThis proves that characters of a C-star algebra respect the involution. The argument itself does not require commutativity, although a noncommutative algebra need not have any characters.

The first assertion needs the C-star identity; a Banach-algebra involution alone is insufficient. For an explicit counterexample, take the dual-number algebraThis is a commutative unital Banach algebra, since it is finite-dimensional andIts involution is conjugate-linear, multiplicative, involutive and isometric. Every character of an algebra kills $\epsilon$, because $\varphi (\epsilon {)}^2=0$. Thus there is just one character, $a+b\epsilon \mapsto a$, and the Gelfand transform sends the nonzero norm-one element $\epsilon$ to zero. It is neither injective nor norm preserving. In particular $\Vert {\epsilon }^{\ast }\epsilon \Vert =0\ne \Vert \epsilon {\Vert }^2$, identifying the missing norm axiom.

Even the claimed compatibility with the involution is not automatic. On ${\mathbb{C}}^2$ with the maximum norm and pointwise multiplication, define $(a,b{)}^{\ast }=(\bar{b},\bar{a})$. This is another isometric involution, but evaluation at the first coordinate takes $(1,0{)}^{\ast }$ to zero rather than to the conjugate of its value $1$ at $(1,0)$. These examples establish involution alone does not imply an isometric Gelfand transform.

We now prove the intended assertion for a commutative complex unital C-star algebra. Let $\mathcal{M}$ be its maximal ideal space, identified with the compact character space of an algebra as in Question 3. The C-star identity and submultiplicativity giveand applying the same inequality to $a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the involution is continuous. For a self-adjoint element $h$, the exponential series defines $u_t=\exp (ith)$ in norm. Applying the involution term by term gives $u_t^{\ast }=\exp (-ith)$, and multiplying the absolutely convergent series shows $u_t^{\ast }{u}_t=1$. Hence $\Vert u_t\Vert =1$. Every character of an algebra is contractive by Question 3, and soTaking both signs of $t$ forces $\varphi (h)$ to be real. Writingwith $h,k$ self-adjoint, provesThis proves that characters of a C-star algebra respect the involution.

To prove the norm assertion, we also justify the spectral radius formula used here. Put $r(a)={\max }_{\lambda \in \sigma (a)}|\lambda |$. By spectrum equals character values in a commutative Banach algebra,For $R>\Vert a\Vert$, integrating the uniformly convergent Neumann series on $|z|=R$ givesOnly the coefficient of $z^{-1}$ survives. The resolvent is holomorphic outside the spectrum by Question 3, so the Cauchy integral theorem deforms this contour to any circle of radius $R>r(a)$. For algebra-valued integrals the same identity follows by applying every bounded linear functional and then using Hahn-Banach separation. Taking norms givesThe maximum is finite by resolvent continuity on the circle. Taking $n$th roots and then $R\downarrow r(a)$ proves ${\lim }_n\Vert a^n{\Vert }^{1/n}=r(a)$.

Every element is normal because $B$ is commutative. For self-adjoint $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. Therefore, for arbitrary $a$,Iterating yields $\Vert a^{2^n}{\Vert }^{1/2^n}=\Vert a\Vert$, so the proved spectral radius formula givesThe Gelfand transform is thus isometric and injective. Its range is closed, since $B$ is complete and an isometry takes norm-Cauchy sequences back to norm-Cauchy sequences. It contains the constants, separates the characters, and is closed under complex conjugation by the proved involution identity. The Stone-Weierstrass theorem says that such a unital self-adjoint point-separating subalgebra of $C(\mathcal{M})$ is uniformly dense: apply its real version to real and imaginary parts. Closedness then makes the range all of $C(\mathcal{M})$. Hence the Gelfand transform is an isometric star-isomorphism under the C-star hypothesis, establishing the Commutative Gelfand--Naimark theorem in the form required.

For the finite-dimensional operator application, assume first $V\ne 0$ and let $A$ be the unital algebra generated by $\alpha ,\beta ,{\alpha }^{\ast },{\beta }^{\ast }$. Their pairwise commutation makes $A$ commutative; it is closed under the adjoint and norm closed because it is finite-dimensional. With the operator norm it is a C-star algebra. Indeed inner product duality gives $\Vert T^{\ast }\Vert =\Vert T\Vert$, andThus the missing hypothesis in the first paragraph really does hold for this algebra, independently of any spectral theorem.

The proved result identifies $A$ with $C({\mathcal{M}}_A)$. This character space is finite. For if it had $N$ distinct points, point separation would give, for each pair, a continuous function taking values $1$ and $0$ at the chosen two points. Products of these functions give $N$ functions with Kronecker-delta values on the selected points, hence $N$ linearly independent functions. Taking $N>\dim A$ is impossible. Write ${\mathcal{M}}_A=\{{\varphi }_1,\dots ,{\varphi }_m\}$, a finite Hausdorff and therefore discrete space.

Let $p_j$ be the indicator of $\{{\varphi }_j\}$ and let ${\pi }_j$ be its inverse Gelfand transform in $A$. Since $p_j^2=p_j$, ${\bar{p}}_j=p_j$, $p_j{p}_k=0$ for $j\ne k$, and ${\sum }_j{p}_j=1$, the injective star-isomorphism givesThus the ${\pi }_j$ are mutually orthogonal linear projections, and in particular commute. Put ${\lambda }_j={\varphi }_j(\alpha )$ and ${\mu }_j={\varphi }_j(\beta )$. The pointwise identities $\hat{\alpha }={\sum }_j{\lambda }_j{p}_j$ and $\hat{\beta }={\sum }_j{\mu }_j{p}_j$ pull back toThe ranges are simultaneous eigenspaces: on ${\pi }_{j}V$, the two operators act as the scalars ${\lambda }_j,{\mu }_j$. This proves that finite commuting normal operators have joint spectral projections. If $V=0$, the same conclusion holds with empty sums.

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 $\boxed{\mathbb{C}\setminus {\sigma }_D(a)\text{ is connected}.}$

Use additive notation and fix Haar measure $m$ on $G$. A continuous unitary character is a continuous homomorphism $\chi :G\to \mathbb{T}$. With the conventionthe Fubini theorem and the character identity imply $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Also $|\hat{f}(\chi )|\le \Vert f{\Vert }_1$, and choosing $f=\chi a$ for a nonzero nonnegative $a\in C_c(G)$ shows that evaluation at $\chi$ is a nonzero functional. Thus each character gives a character of an algebra on the L1 convolution algebra. Here a multiplicative linear functional means a nonzero multiplicative linear functional; the zero functional must be excluded for the claimed bijection.

Conversely, let $\phi$ be such a functional. It is automatically bounded with $|\phi (f)|\le \Vert f{\Vert }_1$: extend it to the unitization of an algebra by $\tilde{\phi }(f+z1)=\phi (f)+z$. If $|z|>\Vert f{\Vert }_1$, the Neumann series makes $z1-f$ invertible, so its image under the unital homomorphism cannot be zero. Hence $\phi (f)$ cannot lie outside the norm disc.

Choose $a\in L^1(G)$ with $\phi (a)\ne 0$, and let $T_{x}a(t)=a(t-x)$. The Abelian translation identities $(T_{x}a)\ast b=a\ast (T_{x}b)$ giveTherefore, defining $c(x)=\phi (T_{x}a)/\phi (a)$, one has $\phi (T_{x}b)=c(x)\phi (b)$ for every $b$. Applying this to two successive translations gives $c(x+y)=c(x)c(y)$ and $c(0)=1$. By translation continuity in Lp on a locally compact group, $c$ is continuous. It is uniformly bounded by $\Vert a{\Vert }_1/|\phi (a)|$; applying this bound to $c(nx)=c(x{)}^n$ for positive and negative integers forces $|c(x)|=1$. Thus $\chi (x)=\overline{c(x)}$ is a continuous unitary character.

The Bochner integral identity $a\ast b={\int }_{G}b(y)T_{y}adm(y)$ holds in $L^1$, since $\Vert T_{y}a{\Vert }_1=\Vert a{\Vert }_1$. Applying the bounded functional yieldsCancel $\phi (a)\ne 0$ to obtain the required correspondenceThe recovery formula $\overline{\chi (x)}=\phi (T_{x}a)/\phi (a)$ proves uniqueness. These functionals have norm exactly one, since $b=\chi a$ with $a\ge 0$ and $\int a=1$ has $\Vert b{\Vert }_1=\phi (b)=1$.

For the topology, identify the Pontryagin dual group with these functionals and give it the Gelfand topology, namely pointwise convergence of $\hat{f}(\chi )$ for every $f\in L^1$. If a net ${\chi }_{\lambda }$ converges to $\gamma$ uniformly on every compact subset of $G$, then for $f\in C_c(G)$,Density of $C_c$ in $L^1$ and the uniform functional norm bound one extend this convergence to every $f\in L^1$.

Conversely, suppose convergence in the Gelfand topology and choose $a$ with $\hat{a}(\gamma )\ne 0$. For compact $K\subseteq G$, the set $\{T_{x}a:x\in K\}$ is compact in $L^1$ by translation continuity in Lp on a locally compact group. Evaluation convergence is uniform on this compact set: a finite $\delta$-net reduces its supremum to finitely many convergent evaluations plus $2\delta$, using the common functional norm one. ThusNow $\widehat{T_{x}a}(\chi )=\overline{\chi (x)}\hat{a}(\chi )$, and $\hat{a}({\chi }_{\lambda })\to \hat{a}(\gamma )\ne 0$. Dividing this identity proves ${\sup }_{x\in K}|{\chi }_{\lambda }(x)-\gamma (x)|\to 0$. Hence the Gelfand topology is exactly the compact-open topology, with the requested neighbourhood basisFor compact $K$, the continuous difference attains its maximum, so this strict pointwise inequality on $K$ is equivalent to a strict uniform bound. Finite intersections contain another such set by taking the union of the compact sets and the minimum of the tolerances.

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.

For every character of an algebra $\varphi$, the element $a-\varphi (a)1$ cannot be invertible, so $\varphi (a)\in {\sigma }_A(a)$. Conversely, if $\lambda \in {\sigma }_A(a)$, the principal ideal $A(a-\lambda 1)$ is proper: otherwise $b(a-\lambda 1)=1$ for some $b$, which in this commutative algebra gives an inverse. Part (b) puts that ideal inside a maximal ideal $M$, and the corresponding algebra character from part (c) has $\varphi (a)=\lambda$. ThereforeHere $\hat{a}(\varphi )=\varphi (a)$ is the Gelfand transform. The spectrum equals character values in a commutative Banach algebra result does not require the Gelfand transform to be injective: the algebra need not be semisimple.