Let $ (\mathscr{A},G,\alpha) $ be a $ C^{\*} $-dynamical system, and consider the twisted convolution $ \* $-algebra $ ({L^{1}}(G,\mathscr{A}),\star,^{\*}) $ defined by
\begin{align\*}
\forall \phi,\psi \in {L^{1}}(G,\mathscr{A}), ~ \forall g \in G: \quad
(\phi \star \psi)(g) & \stackrel{\text{def}}{=}
\int\_{G} \phi(x) {\alpha\_{x}}(\psi(x^{-1} g)) \, d{{\mu\_{G}}(x)}, \\
{\phi^{\*}}(g) & \stackrel{\text{def}}{=}
\Delta(g^{-1}) \cdot {\alpha\_{g}}(\phi(g^{-1})^{\*}).
\end{align\*}
**Note:** $ \mu\_{G} $ is a Haar measure on $ G $ and $ \Delta: G \to \mathbb{R}\_{> 0} $ is the modular function of $ G $.


If $ \pi $ is an algebraic $ \* $-representation of $ ({L^{1}}(G,\mathscr{A}),\star,^{\*}) $ on some Hilbert space $ \mathcal{H} $, then $ \pi $ is automatically bounded by the $ L^{1} $-norm on $ {L^{1}}(G,\mathscr{A}) $, i.e., $ \| \pi(\phi) \|\_{B(\mathcal{H})} \leq \| \phi \|\_{L^{1}} $ for all $ \phi \in {L^{1}}(G,\mathscr{A}) $. (This is because an algebraic \*-homomorphism from a Banach $ \* $-algebra to a $ C^{\*} $-algebra is automatically norm-decreasing.) Consequently, we can extend $ \pi $ to a $ \* $-representation $ \tilde{\pi} $ of the crossed-product $ C^{\*} $-algebra $ \mathscr{A} \rtimes\_{\alpha} G $ on $ \mathcal{H} $.


If we replace $ {L^{1}}(G,\mathscr{A}) $ by $ {C\_{c}}(G,\mathscr{A}) $, then I suspect that the condition



> 
> $ \pi $ is bounded by the $ L^{1} $-norm on $ {C\_{c}}(G,\mathscr{A}) $
> 
> 
> 


is not for free and that we have to explicitly assume it if we want to extend $ \pi $ to a $ \tilde{\pi}: \mathscr{A} \rtimes\_{\alpha} G \to B(\mathcal{H}) $. My suspicion is substantiated by the following result, which is given in Dana P. William’s book *Crossed Products of $ C^{\*} $-Algebras* together with a ‘clever’ proof due to Iain Raeburn.



> 
> **Theorem:** If $ \pi: ({C\_{c}}(G,\mathscr{A}),\star,^{\*}) \to B(\mathcal{H}) $ is an algebraic $ \* $-representation that is continuous with respect to the inductive limit topology on $ {C\_{c}}(G,\mathscr{A}) $, then $ \pi $ is bounded by the universal norm $ \| \cdot \|\_{\mathscr{A} \rtimes\_{\alpha} G} $ on $ {C\_{c}}(G,\mathscr{A}) $ and so is bounded by the $ L^{1} $-norm on $ {C\_{c}}(G,\mathscr{A}) $.
> 
> 
> 


Hence,



> 
> **My question:** Is there a ‘simple’ example of an algebraic $ \* $-representation $ \pi: ({C\_{c}}(G,\mathscr{A}),\star,^{\*}) \to B(\mathcal{H}) $ that is *not* bounded by the $ L^{1} $-norm on $ {C\_{c}}(G,\mathscr{A}) $?
> 
> 
> 


To keep things simple, we can assume that $ \mathscr{A} = \mathbb{C} $. Thank you very much for your help!



Extending a $ * $-Representation of $ ({C_{c}}(G,\mathscr{A}),\star,^{*}) $ to a $ * $-Representation of $ \mathscr{A} \rtimes_{\alpha} G $
Here's the discrete group argument, when $A=\mathbb C$. Then $C\_c(G,A)$ is just the algebra of finitely supported functions $G\rightarrow\mathbb C$, which has a basis $(\delta\_g)\_{g\in G}$ say. That $\pi:C\_c(G,A)\rightarrow B(H)$ is a $\*$-representation means that firstly $\pi(\delta\_e)$ is a self-adjoint idempotent; so wlog we may assume $\pi(\delta\_e) = 1\_H$ (or else just restrict to the invariant subspace of $H$). Then, as $\delta\_g^\*=\delta\_{g^{-1}}$,
$$ 1\_H = \phi(\delta\_g^\* \delta\_g) = \phi(\delta\_g)^\* \phi(\delta\_g). $$
Similarly $\phi(\delta\_g) \phi(\delta\_g)^\* = 1\_H$. So $\phi(\delta\_g)$ is a unitary. It now follows that $\phi$ extends to a contraction from $\ell^1(G)$. I think a similar argument works for any $A$.


**Edit:** For a compact group, it's easier, by virtue of hiding the work in known structure results. Let $Pol(G)$ be the collection of matrix coefficients of finite dimensional (unitary) representations-- this a is a dense $\*$-subalgebra of $C(G)$. If we view $C^\*(G)$, the universal $C^\*$-completion of $L^1(G)$, as a direct sum of full matrix algebras, then $Pol(G)$ exactly corresponds to the algebraic direct sum of full matrix algebras. From this picture, you can see that any $\*$-representation of $Pol(G)$ is contractive for the $C^\*(G)$ norm.


