TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Gleason Kahane Zelazko

Statement

Let $A$ be a complex unital Banach algebra (Unital Banach algebra) and let $φ : A \to C$ be a complex linear map with $φ (1) = 1$ which is nonzero on every invertible element: $φ (a) \neq 0$ whenever $a \in A$ is invertible. Then $φ$ is continuous and multiplicative:

$∣ φ (a) ∣ \leq ∥ a ∥ and φ (a b) = φ (a) φ (b) (a, b \in A) .$

No commutativity of $A$ is assumed, and the argument is choice-free. The hypothesis is that $φ$ is nonzero on invertible elements, not that it is nonzero on $A$ ; together with $φ (1) = 1$ it is the exact hypothesis used.

Facts & Assumptions

Given: A complex unital Banach algebra $A$ and a complex-linear $φ : A \to C$ with $φ (1) = 1$ that is nonzero at every invertible element.

[F1]

$A$ is complete, $∥ 1 ∥ = 1$ , the multiplication is associative and bilinear with $∥ x y ∥ \leq ∥ x ∥ ∥ y ∥$ , and $1 \neq 0$ (Unital Banach algebra).

[F2]

If $∥ y ∥ < 1$ then $1 - y$ is invertible with inverse $\sum_{n \geq 0} y^{n}$ ; hence $λ 1 - a$ is invertible whenever $∣ λ ∣ > ∥ a ∥$ (Neumann series).

[F3]

A normed space is a Banach space if and only if every absolutely convergent series in it converges (Series criterion for Banach spaces).

[F4]

For every $ζ \in C$ the series $\sum ζ^{n} / n!$ converges absolutely (The complex exponential series converges absolutely for every complex argument, The complex exponential by its power series).

[F5]

The radius is defined by the real absolute-value coefficient series. The sum on its open disc is analytic, hence holomorphic; there all derivatives are given by termwise differentiation. Nonnegative series admit the direct comparison test, and absolutely convergent complex series converge. (Complex series, absolute convergence, complex power series, and radius of convergence, The sum of a complex power series is analytic throughout its open disc of convergence, Every complex analytic function is holomorphic, A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation, If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , Every absolutely convergent complex series converges, and rearrangements preserve its sum).

[F6]

If $f$ is entire and zero-free with constants $M > 0$ , $C \geq 0$ satisfying $∣ f (z) ∣ \leq M e^{C ∣ z ∣}$ for all $z$ , then $f (z) = f (0) e^{a z}$ with $a = f^{'} (0) / f (0)$ (Zero free entire function of exponential type is an exponential).

[F7]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[F8]

The complex exponential satisfies $e^{z + w} = e^{z} e^{w}$ , agrees with the real exponential on reals and has modulus $∣ e^{z} ∣ = e^{Re z}$ . ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Proof

technique · direct

1.1F1F2algebra

$∣ φ (a) ∣ \leq ∥ a ∥$ for every $a$ : otherwise put $λ : = φ (a)$ , so $∣ λ ∣ > ∥ a ∥$ and $λ 1 - a = λ (1 - a / λ)$ is invertible by [F2], while $φ (λ 1 - a) = λ - φ (a) = 0$ by linearity and $φ (1) = 1$ ; this contradicts the hypothesis that $φ$ vanishes at no invertible element.

1.2F1F3F4F5algebra

Put $E_{a} (z) = \sum_{n \geq 0} z^{n} a^{n} / n!$ . Comparison with $\sum (∣ z ∣ ∥ a ∥)^{n} / n!$ and completeness prove convergence and $∥ E_{a} (z) ∥ \leq e^{∣ z ∣ ∥ a ∥}$ . For $S_{N} (z) = \sum_{k = 0}^{N} z^{k} a^{k} / k!$ , the terms of total degree $n \leq N$ in $S_{N} (z) S_{N} (- z)$ sum to $a^{n} z^{n} \sum_{k = 0}^{n} (- 1)^{n - k} / (k! (n - k)!)$ , namely 1 for $n = 0$ and 0 otherwise by the finite binomial formula. Writing $t = ∣ z ∣ ∥ a ∥$ , the remaining terms have norm at most $\sum_{n = N + 1}^{2 N} t^{n} \sum_{k = \max (0, n - N)}^{\min (n, N)} 1 / (k! (n - k)!) \leq \sum_{n > N} (2 t)^{n} / n! \to 0$ . Multiplication is continuous by submultiplicativity. Passing to the limit, and then exchanging the two factors in the same computation, gives $E_{a} (z) E_{a} (- z) = E_{a} (- z) E_{a} (z) = 1$ . No commutativity beyond the powers of the single element $a$ is used.

1.3F4F5algebra

The scalar analytic estimate used below is direct: if $∣ d_{n} ∣ \leq D r^{n} / n!$ with $D, r \geq 0$ , then for every $R \geq 0$ , $\sum ∣ d_{n} ∣ R^{n} \leq D e^{r R} < \infty$ by comparison. The absolute-value coefficient series therefore converges at every real argument, and its radius, hence the complex radius, is infinite. By [F5] its sum is entire and its derivative at zero is $d_{1}$ . This includes $D = 0$ and $r = 0$ , using the constant-term convention $0^{0} = 1$ .

2.1step 1.1step 1.2step 1.3F5algebra

$φ (E_{a} (z)) = \sum_{n \geq 0} z^{n} φ (a^{n}) / n!$ for every $z$ , by continuity of $φ$ from [step 1.1] applied to the partial sums of the absolutely convergent series of [step 1.2]; consequently $f_{a} (z) : = φ (E_{a} (z))$ is an entire function of $z$ , with $∣ f_{a} (z) ∣ \leq ∥ E_{a} (z) ∥ \leq e^{∣ z ∣ ∥ a ∥}$ , $f_{a} (0) = φ (1) = 1$ , derivative $f_{a}^{'} (0) = φ (a)$ by termwise differentiation, and $f_{a}$ is zero-free because $E_{a} (z)$ is invertible by [step 1.2] and $φ$ vanishes at no invertible element.

3.1step 2.1F6

By [F6] applied to the zero-free entire $f_{a}$ of [step 2.1] with $M = 1$ and $C = ∥ a ∥$ : $f_{a} (z) = e^{z φ (a)}$ , that is, $φ (E_{a} (z)) = e^{z φ (a)}$ for all $z \in C$ and all $a \in A$ .

4.1step 1.1step 1.2step 1.3step 3.1F5algebra

Fix $a, b \in A$ and put $F (z, w) : = φ (E_{a} (z) E_{b} (w))$ . For fixed $w$ , the function $z \mapsto F (z, w) = \sum_{n} z^{n} φ (a^{n} E_{b} (w)) / n!$ is entire with $∣ F (z, w) ∣ \leq e^{∣ z ∣ ∥ a ∥} e^{∣ w ∣ ∥ b ∥}$ , the bound coming from [step 1.2] and [step 1.1]; $F (0, w) = φ (E_{b} (w)) = e^{w φ (b)}$ by [step 3.1]; $F (\cdot, w)$ is zero-free because $E_{a} (z) E_{b} (w)$ is a product of invertibles [step 1.2]; and $\partial_{z} F (0, w) = φ (a E_{b} (w))$ by termwise differentiation of this scalar power series, whose coefficient bound is $∥ a ∥^{n} ∥ E_{b} (w) ∥ / n!$ .

5.1step 1.1step 4.1F6F8algebra

Applying [F6] to $z \mapsto F (z, w) / F (0, w)$ (zero-free, value $1$ at $0$ , growth $M_{w} e^{∥ a ∥ ∣ z ∣}$ with $M_{w} : = e^{2 ∥ b ∥ ∣ w ∣}$ ) gives $F (z, w) = F (0, w) e^{z c (w)}$ for every $z$ , where $c (w) : = \partial_{z} F (0, w) / F (0, w) = φ (a E_{b} (w)) e^{- w φ (b)}$ .

6.1step 1.1step 1.3step 4.1step 5.1F5F8algebra

The numerator $φ (a E_{b} (w)) = \sum_{n \geq 0} w^{n} φ (a b^{n}) / n!$ is entire by step 1.3, since its coefficients are bounded by $∥ a ∥ ∥ b ∥^{n} / n!$ . The exponential factor in $c (w) = φ (a E_{b} (w)) e^{- w φ (b)}$ is entire by the same estimate; the product is holomorphic by the product rule, obtained directly by splitting its difference quotient. Thus $c$ is entire. Fix $w$ . If $c (w) \neq 0$ , set $z = t \overline{c (w)} / ∣ c (w) ∣$ , $t > 0$ . The identity in step 5.1 and [F8] give $∣ F (0, w) ∣ e^{t ∣ c (w) ∣} \leq e^{t ∥ a ∥ + ∣ w ∣ ∥ b ∥}$ . Since $∣ F (0, w) ∣ = e^{Re (w φ (b))}$ , this implies $t (∣ c (w) ∣ - ∥ a ∥) \leq 2 ∣ w ∣ ∥ b ∥$ . Divide by $t$ and let $t \to \infty$ to obtain $∣ c (w) ∣ \leq ∥ a ∥$ . If $c (w) = 0$ this bound is immediate.

7.1step 1.2step 5.1step 6.1F7algebra

By [F7] the bounded entire function $c$ is constant. At zero, $E_{b} (0) = 1$ , so $c (0) = φ (a)$ and $c (w) = φ (a)$ for all $w$ .

8.1step 5.1step 7.1algebra

Consequently $F (z, w) = e^{w φ (b)} e^{z φ (a)}$ for all $z, w \in C$ by [step 5.1] and [step 7.1].

9.1step 4.1step 6.1step 8.1F4F5algebra

Differentiate the scalar identity in step 8.1 with respect to $z$ at zero, using the derivative established in step 4.1 and the exponential power series. It gives $φ (a E_{b} (w)) = φ (a) e^{w φ (b)}$ . Differentiate this identity with respect to $w$ at zero, using the numerator series in step 6.1 and [F5]. Its left derivative is $φ (a b)$ and its right derivative is $φ (a) φ (b)$ . Thus $φ (a b) = φ (a) φ (b)$ without invoking any double-series interchange.

10.1step 1.1step 9.1∎

By [step 1.1] $φ$ is bounded, hence continuous, and by [step 9.1] it is multiplicative; both assertions of the theorem are proved.

Remarks

The hypothesis is used twice. It gives continuity through the spectrum argument [step 1.1] and zero-freeness of the functions $f_{a}$ and $F (\cdot, w)$ in [steps 2.1 and 4.1]; no other invocation occurs.
The two-variable step is not a formal consequence of the one-variable step, which is why the function $F$ and its $w$ -dependent constant $c (w)$ are introduced: the one-variable theorem applied for fixed $w$ produces a constant that has to be shown independent of $w$ .

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Sources

Vahid Shirbisheh, Lectures on C-star Algebras, v2 — Theorem 3.1.11, printed pp. 59–61 (standard reference, not scraped)
Theo Bühler and Dietmar A. Salamon, Functional Analysis — §5.5.1, printed pp. 258–262 (standard reference, not scraped)