Please, help out of the mind trap. In [this](https://projecteuclid.org/download/pdf_1/euclid.cmp/1103839770) prominent paper Kadison and Ringrose prove among other things the following


Corollary 8. Each norm-continuous representation of a connected topological
group by automorphisms of a $C^\*$-algebra has range consisting of
$\pi$-inner automorphisms.


By $\pi$-inner these authors mean permanently weakly inner, see the definition on page 4 of the same paper, for instance.


**Question:** Now suppose $A$ is a commutative $C^\*$-algebra, and let $\pi:A\to\mathcal{B}(H)$ be a faithful $\star$-representation. A commutative $A$ has obviously no non-trivial inner automorphisms. The strong/weak closure $\pi(A)''\subset\mathcal{B}(H)$ is also commutative, so $\pi(A)$ cannot have even non-trivial weakly inner automorphisms. By the above corollary this would mean that $A$ has no norm-continuous group of automorphisms. 


But let $A=C[0,1]$ for clarity (acting on $L^2[0,1]$), and let $y(x;t)=x^t$ for all $x\in[0,1]$ and $t\in(0,\infty)$. Now $y(.;t)$ is a homeomorphism of $[0,1]$ for every $t$, and therefore $\alpha\_t(f)=y\_\*f=f[x^t]$ is definitely $\|f\|\_{sup}$-preserving automorphism of $C[0,1]$, thus $\|\alpha\_t\|\_{\mathrm{op}}=1$. More generally, if $A\simeq C\_0(M)$ by Gelfand-Naimark then every homeomorphism group $\phi\_t\in\mathrm{Homeo}(M)$ gives such an automorphism group $\alpha\_{\phi\_t}$. In fact, the particular $\alpha\_t$ above is not even spatial. The map $U:L^2[0,1]\to L^2[0,1]$ which gives $\alpha\_t(f)=U\_t f U\_t^{-1}$ is also given by $U\_t[h](x)=h(x^t)$ for all $h\in L^2[0,1]$, and is not unitary, because Lebesgue measure is not invariant under $y(.;t)$. $U\_t$ is even unbounded for $t>1$.


So where does the source of confusion lie? Thank you.



Every norm-continuous group of $C^*$-algebra automorphisms weakly inner?
I think the source of confusion is in the definition of "norm-continuous representation". Probably you are interpreting this as "point-norm continuous", i.e., for each $x \in A$ the map $t \mapsto \alpha\_t(x)$ is continuous from the group into the norm topology on $A$. The correct definition is that the map $t \mapsto \alpha$ is continuous from the group into the norm topology on ${\rm Aut}(A)$.


Indeed, there are no nontrivial norm-continuous representations of connected topological groups on abelian C\*-algebras, because the distance between any two automorphisms of an abelian C\*-algebra is $2$. Proof: let $\alpha$ and $\beta$ be distinct automorphisms. Multiplying both by $\beta^{-1}$, we can assume $\beta = {\rm id}\_A$. If $A = C(X)$ then $\alpha$ arises by composition with a non-identity self-homeomorphism $\phi$ of $X$, so find $p \in X$ such that $\phi(p) \neq p$. Then find $f \in C(X)$ with $\|f\|\_\infty = 1$, $f(p) = 1$, and $f(\phi(p)) = -1$. Evaluating on this function yields $\|\alpha - {\rm id}\_A\| \geq 2$. (The reverse inequality is easy because $\|\alpha\| = \|{\rm id}\_A\| = 1$.)


