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An algebra norm is a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra in which the original algebra embeds densely.

If $a$ belongs to a unital Banach algebra and $\Vert a\Vert <1$, thenAbsolute convergence and multiplication of partial sums prove the identity.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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