Original programme exposition in Codex (OpenAI), September 2026; proof restoration and supporting proofs by GPT-6.1 Sol (OpenAI), Ultra, October 2026. Spot-checked by GPT-6 Astra in a separate session.

Norm-continuous groups in a Banach algebra

A norm-continuous one-parameter group in a unital Banach algebra has a bounded generator and is an exponential. The existence of that generator can be seen directly by averaging the group near zero. A local logarithm gives a second route and explains the construction used in Takesaki II, Lemma XI.1.17. The same lemma places the generator's spectrum on the imaginary axis when every group element has norm one.

Programme proof written in Codex (OpenAI), September 2026; restoration and proof expansion, 5 October 2026. New expression is dedicated under CC0 to the extent of rights held. Human review is not asserted.

The norm integrals, fundamental theorem and absolutely convergent products are proved in CF1; the Neumann inverse and spectral bounds are in CF2. AF4 proves the precise logarithm identities and full exponential spectral mapping, including an arbitrary identity norm. Here $\operatorname{GL}(A)$ denotes the multiplicative group of invertible elements of $A$.

Exponentials give norm-continuous groups

Let $A$ be a complex unital Banach algebra with identity $1$, and let $h\in A$. The series

$g_h(t)=\exp(th):=\sum_{n=0}^{\infty}\frac{t^nh^n}{n!}\tag{B1}$

converges absolutely and uniformly in norm on bounded intervals of $t\in\mathbb R$. The usual Cauchy-product rearrangement is justified by absolute convergence and gives $g_h(s+t)=g_h(s)g_h(t)$. In particular $g_h(t)^{-1}=g_h(-t)$. Termwise differentiation, again uniform on bounded intervals, gives

$g_h'(t)=h g_h(t)=g_h(t)h.\tag{B2}$

Thus $t\mapsto g_h(t)$ is a norm-continuous group of invertible elements, differentiable in the Banach-algebra norm.

Averaging recovers a bounded generator

Conversely let $g:\mathbb R\to\operatorname{GL}(A)$ be a norm-continuous homomorphism. The group law makes all $g(t)$ commute and gives $g(0)=1$. Choose $d>0$ so small that $\|g(u)-1\|<1/2$ for $0\le u\le d$, and form the norm integral $B=\int_0^d g(u)\,du$. Then $\|d^{-1}B-1\|<1/2$, so $B$ is invertible by a Neumann series.

Translation of the integral by the group law gives, for real $t$,

$g(t)B=\int_t^{t+d}g(u)\,du,\qquad \frac{g(t)-1}{t}B =\frac1t\left(\int_d^{d+t}g(u)\,du-\int_0^t g(u)\,du\right).\tag{B3}$

Both average integrals on the right have norm limits as $t\to0$, from either sign of $t$, so

$h:=\lim_{t\to0}\frac{g(t)-1}{t} =(g(d)-1)B^{-1}\in A.\tag{B4}$

The group law now gives $g'(t)=g(t)h=hg(t)$. Differentiating $\exp(-th)g(t)$ shows it is constant and equals $1$ at zero. Hence

$g(t)=\exp(th)\quad(t\in\mathbb R),\qquad h=g'(0)\text{ is unique}.\tag{B5}$

This proof uses norm continuity and elementary Banach integration only. In particular, no differentiability hypothesis was inserted into the converse.

The local logarithm and its branch control

For comparison with the source construction, shrink an interval $(-d_0,d_0)$ until $\|g(t)-1\|<1/10$ there. Define

$L(t)=\log g(t) =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(g(t)-1)^n \qquad(|t|<d_0).\tag{B6}$

This series converges in norm. Its norm is at most $-\log(9/10)<1/8$. If $s,t,s+t$ all lie in the interval, the elements $L(s)$ and $L(t)$ commute. Their sum and $L(s+t)$ have norm less than $1/4$, within the exact ball on which $\log(\exp z)=z$ by the complete differentiation-and-integral proof (AF14)–(AF15). That proof also gives $\exp(\log(1+b))=1+b$ for $\|b\|<1$. Since $\exp(L(s)+L(t))=g(s)g(t)=g(s+t)$, the logarithm is locally additive:

$L(s+t)=L(s)+L(t) \quad(|s|,|t|,|s+t|<d_0).\tag{B7}$

For arbitrary $t$, set $H(t)=nL(t/n)$ once $|t/n|<d_0$. This does not depend on a sufficiently large integer $n$: compare two choices using a common multiple and repeated applications of (B7), with every intermediate point remaining in the local interval. Choose one large $n$ for $s,t,s+t$ to see that $H(s+t)=H(s)+H(t)$. Near zero $H=L$, so $H$ is continuous. Every continuous additive map $\mathbb R\to A$ is of the form $H(t)=tH(1)$: addition and negation give integer homogeneity, subdividing $1$ gives rational homogeneity, and a sequence of rationals converging to $t$, with continuity, gives the real identity. Finally $\exp(H(t))=g(t/n)^n=g(t)$. Its generator $H(1)$ equals the unique derivative $h$ in (B4).

The small logarithm ball is the branch check omitted by a bare appeal to commutativity. Without it, equality of exponentials would determine logarithms only modulo possible periods.

Norm-one groups have imaginary generator spectrum

Assume in addition $\|g(t)\|=1$ for every real $t$. Its inverse $g(-t)$ also has norm one. For $z\ne0$, the identity $g(t)-z1=-z g(t)(g(t)^{-1}-z^{-1}1)$ proves that $z$ is a spectral point of $g(t)$ exactly when $z^{-1}$ is one of its inverse; zero is excluded by invertibility. For any $z\in\operatorname{Sp}_A(g(t))$, the spectral-radius bounds for $g(t)$ and its inverse give $|z|\le1$ and $|z|^{-1}\le1$, so $|z|=1$. The fully proved exponential spectral-mapping identity (AF16) yields

$\operatorname{Sp}_A(g(t)) =\{e^{t\lambda}:\lambda\in\operatorname{Sp}_A(h)\}.\tag{B8}$

For $t\ne0$, every $\lambda\in\operatorname{Sp}_A(h)$ therefore satisfies $|e^{t\lambda}|=e^{t\operatorname{Re}\lambda}=1$, hence $\operatorname{Re}\lambda=0$. Thus

$\operatorname{Sp}_A(h)\subset i\mathbb R.\tag{B9}$

Problem. Does the converse of (B9) hold? Take $h=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ in $M_2(\mathbb C)$ with its operator norm.

Solution. Its spectrum is $\{0\}\subset i\mathbb R$, but $h^2=0$ makes $\exp(th)=I+th$, whose norm grows without bound as $|t|\to\infty$: its value on the unit vector $e_2$ is $te_1+e_2$, of norm $\sqrt{1+t^2}$. Imaginary generator spectrum alone does not make the exponential group norm one. $\square$

The mathematical source is M. Takesaki, Theory of Operator Algebras II, Lemma XI.1.17, printed pages 324–325 (edition record). Both retained converses are complete: the averaging proof gives the bounded derivative, and the logarithm proof uses an explicit injective branch. AF4 supplies the actual full spectral-mapping proof used for the imaginary-spectrum conclusion. No Hille–Yosida theorem or holomorphic functional calculus is imported.