One parameter group of elements of a banach algebra

Question

Let $A$ be a complex Banach algebra with identity $1_A$. Suppose we have a continuous one parameter group of elements of $A$, i.e. a continuous map $U:\mathbb R\to A$ such that:

• $U(0)=1_A$
• $U(t+s)=U(t)U(s)$ for all $t,s\in\mathbb R$

Is it true that there exists some $x\in A$ such that $U(t)=e^{tx}$ for all $t\in\mathbb R$?

It's not hard to prove that the claim is equivalent to the differentiability of $U$ at 0, i.e. to the existence of the limit $\lim_{t\to 0} \frac{U(t)-1_A}{t}$, in which case $x$ is equal to the limit. I tried to come up with some counterexamples but always ended up constructing $U$ in such a way that it is not continuous.

Answer 1

$\newcommand\ex[1]{\exp\left(#1\right)}$

The answer is yes. Rather than proving that $U$ is differentiable, we will construct $x$ explicitly.

Denoting by $S$ the "square" map on $A$, that is, $$ S(a)=a^2, \quad (a\in A), $$ notice that the differential of $S$ at 1 is the map $$ DS_1(h)=2h, \quad\forall h\in A, $$ hence an invertible map. Using the Inverse Function Theorem we then have that there exists a neighborhood $V$ of 1 restricted to which, $S$ is injective.

Next consider the exponential map $$ \exp:A\to A, $$ and observe that its differential at zero is the identity map so, again by the Inverse Function Theorem there are open neighborhoods $Y\ni 0$ and $Z\ni 1$, such that $\exp$ maps $Y$ homeomorphically onto $Z$.

By shrinking both $Y$ and $Z$, we may suppose that $Y=B_r(0)$, namely the open ball centered at zero, for some radius $r>0$, and that $Z\subseteq V$.

Schematically $$ \exp: B_r(0)\to Z\subseteq V. $$ Using that $U$ is norm continuous, pick $\delta >0$ such that $$ |t|<\delta \ \Rightarrow \ U(t)\in Z. $$ For each nonzero $t$ with $|t|<\delta $, set $$ y(t)=(\exp|_Z)^{-1}\big (U(t)\big ), $$ so that $y(t)\in B_r(0)$, $\ex{y(t)}=U(t)$, and we claim that $$ \ex{y(t)/2} = U(t/2). \qquad (*) $$ To see this, notice that $$ \ex{y(t)/2}^2 = \ex{2y(t)/2} = \ex{y(t)} = U(t)= U(t/2)^2, $$ so, to conclude the proof of $(*)$, it is enough to check that both $\ex{y(t)/2}$ and $U(t/2)$ lie in $V$, where the square map $S$ is injective. Indeed, notice that $y(t)\in B_r(0)$, so $$ y(t)/2\in B_r(0) \ \Rightarrow \ex{y(t)/2}\in Z\subseteq V. $$ Also $U(t/2) \in V$, because $$ |t/2| < \delta \ \Rightarrow \ U(t/2) \in Z\subseteq V. $$

As a consequence of $(*)$ we have that $$ \ex{y(t)/2} = U(t/2) = \ex{y(t/2)}. $$ Moreover, since both $y(t)/2$ and $y(t/2)$ lie in $B_r(0)$, where the exponential is injective, we deduce that $$ y(t)/2=y(t/2), $$ and by induction we get that $$ y(t)/{2^n}=y(t/{2^n}),\quad \forall n\geq 0. $$

Choosing any integers $n$ and $p$, with $n\geq 0$, and $|p|<2^n$, we claim that $$ py(t)/{2^n}=y(pt/{2^n}). $$ To prove this notice that the two terms above lie in $B_r(0)$, so it is enough to show that $\exp$ takes on the same values on them. In fact $$ \ex{py(t)/{2^n}} = \ex{py(t/2^n)} = \ex{y(t/2^n)}^p = $$$$= U(t/2^n)^p = U(pt/2^n) = \ex{y(pt/2^n)}. $$ This shows that $ ry(t)=y(rt), $ for every diadic rational $r$ with $|r|<1$, and since $y$ is continuous, the same holds for every real number $r$, with $|r|\leq 1$.

Choosing any $t_0$ with $|t_0|<\delta $, and setting $x=y(t_0)/t_0$, we then claim that $U(s) = \ex{sx}$, for every real $s$. To prove this, find an integer number $p$ large enough so that $t:=s/p$ satisfies $|t|<|t_0|$, and then $$ U(s) = U(pt)=U(t)^p = $$$$ = \ex{y(t)}^p = \ex{y((t/t_0)t_0)}^p = $$$$ = \ex{(t/t_0)y(t_0)}^p = \ex{tx}^p = $$$$ = \ex{ptx} = \ex{sx}. \qquad\text{QED} $$

Answer 2

For an element $T$ of a unital Banach algebra, such that $\sigma(T)\subset \mathcal{C}\setminus (-\infty, 0]$ we can define $\log (T)$ by the analytic functional calculus. For $T=I+V$ and $\|V\|<1$ we have a power series representation of $\log(T)$ as $$\log(T)=\log(I+V)=\sum_{k=1}^\infty {(-1)^{k+1}\over k} V^k$$

Assume two elements $T=I+V,\,S=I+U$ are commuting and $$\|V\|\le {1\over 4},\quad \|U\|\le {1\over 4}$$ Then $TS=I+(V+U+VU)$ and $$\|V+U+VU\|\le {7\over 16}$$ Therefore all three elements $\log(I+V),\ \log(I+U)$ and $\log((I+V)(I+U))$ can be represented by the absolutely convergent power series.

We are going to show that $$\log(TS)=\log(T)+\log (S)\qquad (*)$$ The formula $(*)$ takes the form $$\log[(I+V)(I+U)]=\log(I+V)+\log(I+U) \quad (**)$$ For $z,w\in \mathbb{C}$ such that $|z|<{1\over 4}$ and $|w|<{1\over 4}$ we have $$\log((1+z)(1+w))=\log(1+z)+\log(1+w)$$ hence $$\sum_{k=1}^\infty {(-1)^{k+1}\over k} (z+w+zw)^k=\sum_{k=1}^\infty {(-1)^{k+1}\over k} z^k+\sum_{k=1}^\infty {(-1)^{k+1}\over k} w^k $$ The series on RHS are absolutely convergent. Also the series on LHS is absolutely convergent as $$|z+w+zw|\le |z|+|w|+|z|\,|w|\le {7\over 16}$$ Therefore both sides are of the form $$\sum_{k,l=1}^\infty a_{k,l}z^kw^l$$ and the double series is absolutely convergent. The coeefficients $a_{k,l}$ are uniquely determined. Thus we may substitute $z:=V$ and $w:=U$ to get $$\sum_{k=1}^\infty {(-1)^{k+1}\over k} (V+U+VU)^k=\sum_{k=1}^\infty {(-1)^{k+1}\over k} V^k+\sum_{k=1}^\infty {(-1)^{k+1}\over k} U^k $$ which gives $(**)$.

Let $$B=\{z\in \mathbb{C}\,:\,|z-1|\le {1\over 4}$$ By continuity of the semigroup, there exists $t_0>0$ such that $$\|U_t-I\|\le {1\over 4},\qquad 0<t\le t_0$$ Define $$A={1\over t_0}\log U_{t_0}$$ For $1\le k\le n$ we have ${kt_0\over n} \le t_0.$ By $(*)$ we get $$t_0A=\log U_{t_0}=\log U_{t_0/n}^n=n\log U_{t_0/n} $$ Hence $$ \log U_{t_0/n}={t_0\over n}A$$ Next for $1\le k\le n$ we get (again by applying $(*)$) $$\log U_{kt_0/n}=\log( U_{t_0/n}^k) =k\log U_{t_0/n}={k\over n}\log U_{t_0}$$ By continuity of $(0,t_0]\ni t\mapsto U_t$ and of $\log$ (for example by the power series representation) we obtain that $$\log U_{at_0}=a\log U_{t_0}=at_0A,\qquad 0<a\le 1$$ which implies $$\log U_t=tA,\qquad 0<t\le t_0\qquad (**)$$ Thus $$ U_t=e^{tA},\qquad 0<t\le t_0\qquad \quad(***)$$ By the group property the identity extends to the whole line.

Remark The implication $(**)\implies (***)$ is based on the following. Let $U$ be an element of a unital Banach algebra and $f$ be an analytic function in the open neighborhood of $\sigma(U).$ Then for any entire function $g$ we have $$g(f(U))=(g\circ f)(U)\quad\ (****)$$ In our case $U=U_t,$ $f(z)=\log z$ and $g(w)=e^w.$ Thus $$(g\circ f)(z)=z,\qquad e^{\log U_t}=(g\circ f)(U_t)=U_t$$

The formula $(****)$ can be proved straightforward. Let $g(z)=\sum g_nz^n.$ Then $$g(f(U))=\sum g_n f(U)^n =\sum g_n f^n(U)=\left (\sum g_n f^n\right )(U)=(g\circ f)(U)$$