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For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

For a compact Hausdorff space $K$, every character of the Banach algebra $C(K)$ is an evaluation characterat a unique $x\in K$. The map $x\mapsto {\delta }_x$ is a homeomorphism from $K$ onto ${\Phi }_{C(K)}$ with its Gelfand topology.

Let $B$ be the completion under the assumed algebra norm. Restriction sends every $\psi \in {\Phi }_B$ to a character of the dense subalgebra $C(\mathbb{R})$, hence by part i to ${\delta }_x$ for some $x$. Continuity of the restriction says exactly that $x\in K$. Conversely, every ${\delta }_x$ continuous for the algebra norm extends uniquely through the completion, and continuity of multiplication makes the extension a character of $B$. Restriction and extension therefore give a bijection

For $b\in B$, choose $h_n\in C(\mathbb{R})$ converging to $b$ in the algebra norm. Characters on the unital Banach algebra $B$ are uniformly norm bounded, so $h_n{|}_K$ converges uniformly to the function $x\mapsto {\tilde{\delta }}_x(b)$. This function is continuous. Hence $x\mapsto {\tilde{\delta }}_x$ is continuous into the Gelfand topology. Its inverse is again $\psi \mapsto \psi (t)$, so the bijection is a homeomorphism. In particular, $K$ is compact.

Suppose an algebra norm made $C(\mathbb{R})$ a Banach algebra. Its unital character space of an algebra would be compact in the Gelfand topology. By part i it consists of the evaluations ${\delta }_x$. The mapis continuous from the usual topology because every $h\in C(\mathbb{R})$ is continuous, and its inverse is the continuous map $\phi \mapsto \phi (t)$ defined by the coordinate function $t$. Thus the character space is homeomorphic to the noncompact space $\mathbb{R}$, a contradiction. No complete algebra norm exists.

In a complex unital Banach algebra, the spectral radius formula makes the root norms of $(x/t{)}^n$ tend to $r(x)/t$. If $t>r(x)$, those norms are eventually bounded above by a geometric sequence and the powers vanish. Conversely, vanishing powers eventually have norm less than one; the infimum version of the radius formula then forces $r(x)/t<1$. Consequently $r(x)=\inf \{t>0:(x/t{)}^n\to 0\}$, including the case $r(x)=0$.

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

An element of a complex unital Banach algebra is quasinilpotent if its spectral radius is zero, equivalently its spectrum of an element is $\{0\}$. A nilpotent element is quasinilpotent, but the converse need not hold. The Volterra convolution algebra on a finite interval gives powers which never vanish, yet have factorially decreasing norms, making the inverse series for $\lambda e-x$ converge for every $\lambda \ne 0$.

Let $S$ be a bounded multiplicatively closed subset of a unital Banach algebra. The displayed auxiliary norm $p$ is equivalent to the original norm, and each left multiplication $L_s$ is a contraction for $s\in S$. Taking the operator norm of the faithful left regular representation produces an equivalent algebra norm with $\Vert 1{\Vert }_S=1$ and $\Vert s{\Vert }_S\le 1$. If $M=\max (\Vert 1\Vert ,{\sup }_S\Vert s\Vert )$, then $\Vert a\Vert /M\le \Vert a{\Vert }_S\le M\Vert a\Vert$. The auxiliary norm is itself submultiplicative, since $p(ab)\le p(a)\Vert b\Vert \le p(a)p(b)$, but $p(1)=M$ can exceed one. The operator norm enforces the required unital normalization.

A representation of a complex Banach algebra is a complex-linear multiplicative action on a vector space. For a unital action require $\pi (1)=I$; a nonunital algebra can be handled by extending the action to its unitization. A nonzero action is algebraically irreducible when it has no nonzero proper invariant vector subspace, equivalently when it defines a simple module.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For a closed unital subalgebra $B$ of a unital Banach algebra $A$ and $x\in B$, the smaller-algebra algebra resolvent set is relatively open and closed in the larger one. Openness follows from the Neumann series; closedness follows from continuity of inversion in $A$ and norm closedness of $B$. Every larger-algebra resolvent component is therefore either retained whole or absorbed into the smaller-algebra Banach algebra spectrum. The unbounded component is always retained, since inverses at large modulus are norm-convergent series in $x$.

For a closed linear operator $A$, the resolvent is the bounded inverse $(zI-A{)}^{-1}$ when $z$ lies in the resolvent set of an operator. This definition includes unbounded $A$. It specializes to the resolvent of an element when $A$ is an element of a Banach algebra.

A right inverse of an element $a$ is an element $b$ with $ab=1$. A left inverse of an algebra element and a right inverse of an algebra element, when both exist, coincide. For the unilateral shift operator $S$, $S^{\ast }S=1$ but $S{S}^{\ast }\ne 1$, exhibiting the distinction in the Banach algebra of bounded operators.

In a complex unital Banach algebra, right multiplication by $a$ induces a bounded module endomorphism $T$ of $A/L$ when $La\subseteq L$. Every nonzero module endomorphism of this simple module is algebraically bijective, since its kernel and range are submodules. Its inverse is bounded by the bounded inverse theorem. Choosing $\lambda$ in the nonempty spectrum of an element of $T$ forces $T-\lambda I=0$. Applying this identity to $1+L$ gives $a-\lambda 1\in L$, with uniqueness because $1\notin L$.

A Banach algebra is semisimple when its Jacobson radical is zero. This does not assert that every quasinilpotent element is zero. For example, the algebra of bounded operators on a Banach space of dimension greater than one is semisimple, but contains nonzero square-zero rank-one operators. This notion also does not require the finite-dimensional direct-sum decomposition associated with a semisimple algebra.

A commutative Banach algebra is semisimple when its Jacobson radical is zero. For a complex unital Banach algebra this is equivalent to injectivity of the Gelfand transform. This condition concerns the Jacobson radical; it does not assert a finite product decomposition as in finite-dimensional semisimple algebra theory.

For a finite commuting family in a unital Banach algebra, divide each $x_j$ by $r(x_j)+\epsilon /2$. The spectral radius formula makes each normalized element's powers bounded. Commutativity lets every product be reordered into one power of each generator, so the entire generated semigroup is bounded. Renorming a bounded multiplicative semigroup gives the displayed bounds simultaneously. Applying this to two commuting elements proves submultiplicativity and subadditivity of their spectral radius.

For an element $a$ of a complex Banach algebra,