Let $G$ be a real Lie group and $A$ a unital Banach algebra. Let us call a map $\varphi:G\to A$ a (norm-)continuous representation, if it is continuous 
$$
x\_i\to x\quad\Longrightarrow\quad ||\varphi(x\_i)-\varphi(x)||\to 0,
$$
and multiplicative
$$
\varphi(x\cdot y)=\varphi(x)\cdot\varphi(y), \quad \varphi(1)=1.
$$



> 
> Is every (norm-)continuous representation $\varphi:G\to A$ smooth?
> 
> 
> 


If yes, I would be grateful for the references.



Are norm-continuous representations smooth?
First an attempt at a counterexample. Theorem 5.2 of the paper


* Peter W. Michor: The moment mapping for a unitary representation. Annals of Global Analysis and Geometry 8, No 3 (1990), 299--313. [(pdf)](http://www.mat.univie.ac.at/~michor/moment.pdf)


shows (in a simple way) that for a unitary representation 
$\rho: G\to L(H,H)$ there is the the Frechet subspace $H\_\infty \subset C^\infty(G,H)$
consisting of those $v\in H$ such that $g\mapsto \rho(g)v$ is $C^\infty$ from $G$ to $H$,
and, in general, $H\_\infty\ne H$. For the regular representation of a compact Lie group $G$ on $L^2(G)$ we get $L^2(G)\_\infty = C^\infty(G)$.


However, for a unitary representation as above, the map $\hat\rho:G\times H\to H$ is only separately continuous, thus $\rho: G\to (L(H,H), \|\;\|)$ is not continuous, and we do not get a counterexample.


Attempt at a proof:
===================


For $ 1\le p\le\infty$, consider the space $W^{\infty,p}(G)$ of all $f\in C^\infty(G)$ such that $R\_Xf\in L^p(G)$ for all $X\in\mathfrak g$, where $R\_X$ is the right invariant vector field corresponding to $X$. Then $W^{\infty,2}(G)$ is a Frechet algebra under convolution, because $W^{\infty,2}\subset W^{\infty,1}$. 
Now we get an algebra homomorphism $W^{\infty,2}(G) \to A$, which is continuous by theorems about automatic continuity of algebra homomorphisms between Frechet algebras (Papers by Ernest Michael, if I remember correctly), thus smooth. 
We seem to be near smoothness of $\rho:G\to A$.


Moreover, $\rho: G \to L(W^{\infty,2}(G),W^{\infty,2}(G))$ is smooth since every vector is smooth (where $L$ denotes all bounded linear mappings).


Continuation of the proof attempt. Let $h\_\epsilon\in W^{\infty,2}$ be an approximate unit:
$h\_{\epsilon} \star f\to f$ in $W^{\infty,2}$ for $\epsilon\to 0$ (a suitable sequence of smooth bunmp functions). Then $(\rho(g)h\_\epsilon) \star f\to \rho(g)f$.
Then $g\mapsto \rho(g)h\_\epsilon$ is smooth $G\to W^{\infty,2}$ and thus smooth $G\to A$.
Now it remains to prove that for $\epsilon\to 0$ this converges in $C^\infty(G,A)$.


Lots of things remain to be checked. 


Second proof:
=============


If $\rho: G\to A$ is norm continuous, and if $t\mapsto g(t)=\exp(tX)$ is a 1-parameter subgroup in $G$, than the usual proofs in books on Lie groups (see 4.21 of [here](http://www.mat.univie.ac.at/~michor/dgbook.pdf)) show that 
the continuous 1-parameter subgroup $t\mapsto \rho(g(t))$ in the Banach Lie group $A^{inv}$ of invertible elements in $A$ (which is open in $A$, with exponential mapping $\exp:A\to A^{inv}$ which is a local diffeomorphism) is smooth. But this proves, as usual (see 4.2 as cited above), that $\rho:G\to A$ is smooth. 


