Extending a $ * $-Representation of $ ({C_{c}}(G,\mathscr{A}),\star,^{*}) $ to a $ * $-Representation of $ \mathscr{A} \rtimes_{\alpha} G $ 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!
 A: 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.
A: Here is a complete argument proving automatic continuity for any $ C^{\ast} $-dynamical system $ (G,A,\alpha) $ where $ G $ is discrete.

Haar’s Theorem essentially states that every Haar measure on $ G $ is a positive scalar multiple of the counting measure on $ G $, so let $ k \in \mathbb{R}_{> 0} $ and equip $ G $ with $ k $ times the counting measure on $ G $.
For each $ a \in A $ and $ g \in G $, let $ a \bullet \delta_{g} $ denote the function in $ {C_{c}}(G,A) $ defined by
$$
\forall x \in G: \qquad
(a \bullet \delta_{g})(x) \stackrel{\text{df}}{=}
\begin{cases}
a     & \text{if} ~ x = g; \\
0_{A} & \text{if} ~ x \in G \setminus \{ g \}.
\end{cases}
$$
We then have the following relations for the convolution $ \star $ and the involution $ ^{\ast} $ on $ {C_{c}}(G,A) $:


*

*$
    (a \bullet \delta_{g}) \star (b \bullet \delta_{h})
  = (k \cdot a {\alpha_{g}}(b)) \bullet \delta_{g h}
  $ for all $ a,b \in A $ and $ g,h \in G $.

*$
    (a \bullet \delta_{g})^{\ast}
  = {\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}
  $ for all $ a \in A $ and $ g \in G $.


Let us suppose that $ \pi $ is an algebraic $ \ast $-homomorphism from $ ({C_{c}}(G,A),\star,^{\ast}) $ to a $ C^{\ast} $-algebra $ C $. We will prove that $ \pi $ is automatically contractive with respect to the $ L^{1} $-norm on $ {C_{c}}(G,A) $ and the norm on $ C $. Pick any $ a \in A $ and $ g \in G $. Then
$$
  \pi(a \bullet \delta_{g})^{\ast}
= \pi((a \bullet \delta_{g})^{\ast})
= \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}),
$$
which yields
\begin{align}
    \pi(a \bullet \delta_{g})^{\ast} \pi(a \bullet \delta_{g})
& = \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}})
    \pi(a \bullet \delta_{g}) \\
& = \pi(
       ({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}) \star
       (a \bullet \delta_{g})
       ) \\
& = \pi(
       (k \cdot {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{g^{- 1}}}(a))
       \bullet \delta_{g^{- 1} g}
       ) \\
& = \pi(
       (k \cdot {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{g^{- 1}}}(a))
       \bullet \delta_{e}
       ) \\
& = \pi(
       (
       k \cdot
       {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{e}}({\alpha_{g^{- 1}}}(a))
       )
       \bullet \delta_{e e}
       ) \\
& = \pi(
       ({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e}) \star
       ({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
       ) \\
& = \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e})
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\
& = \pi({\alpha_{g^{- 1}}}(a)^{\ast} \bullet \delta_{e})
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\
& = \pi(
       {\alpha_{e^{- 1}}}({\alpha_{g^{- 1}}}(a)^{\ast})
       \bullet \delta_{e^{- 1}}
       )
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\
& = \pi(({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast})
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\
& = \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}).
\end{align}
Using the $ C^{\ast} $-norm identity, we thus obtain
\begin{align}
    \| \pi(a \bullet \delta_{g}) \|_{C}^{2}
& = \| \pi(a \bullet \delta_{g})^{\ast} \pi(a \bullet \delta_{g}) \|_{C} \\
& = \|
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
    \|_{C} \\
& = \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C}^{2},
\end{align}
so
$$
(\spadesuit) \qquad
  \| \pi(a \bullet \delta_{g}) \|_{C}
= \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C}.
$$
For each $ g \in G $, define a linear map $ \rho_{g}: A \to C $ by
$$
\forall a \in A: \qquad
{\rho_{g}}(a) \stackrel{\text{df}}{=}
\frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}).
$$
Then we have for all $ g \in G $ and $ a,b \in A $ that
\begin{align}
    {\rho_{g}}(a b)
& = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a b) \bullet \delta_{e}) \\
& = \frac{1}{k} \cdot
    \pi(({\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b)) \bullet \delta_{e})
    \\
& = \frac{1}{k^{2}} \cdot
    \left[
    k \cdot
    \pi(({\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b)) \bullet \delta_{e})
    \right] \\
& = \frac{1}{k^{2}} \cdot
    \pi(
       (k \cdot {\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b))
       \bullet \delta_{e}
       ) \\
& = \frac{1}{k^{2}} \cdot
    \pi(
       (k \cdot {\alpha_{g^{- 1}}}(a) {\alpha_{e}}({\alpha_{g^{- 1}}}(b)))
       \bullet \delta_{e e}
       ) \\
& = \frac{1}{k^{2}} \cdot
    \pi(
       ({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \star
       ({\alpha_{g^{- 1}}}(b) \bullet \delta_{e})
       ) \\
& = \frac{1}{k^{2}} \cdot
    \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
    \pi({\alpha_{g^{- 1}}}(b) \bullet \delta_{e}) \\
& = \left[
    \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
    \right]
    \left[
    \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(b) \bullet \delta_{e})
    \right] \\
& = {\rho_{g}}(a) {\rho_{g}}(b), \\
    {\rho_{g}}(a^{\ast})
& = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e})
    \\
& = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a)^{\ast} \bullet \delta_{e})
    \\
& = \frac{1}{k} \cdot
    \pi(
       {\alpha_{e^{- 1}}}({\alpha_{g^{- 1}}}(a)^{\ast})
       \bullet \delta_{e^{- 1}}
       ) \\
& = \frac{1}{k} \cdot
    \pi(({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}) \\
& = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}
    \\
& = \left[
    \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
    \right]^{\ast} \\
& = {\rho_{g}}(a)^{\ast}.
\end{align}
We thus see for all $ g \in G $ that $ \rho_{g}: A \to C $ is an algebraic $ \ast $-homomorphism, so it is automatically contractive with respect to the norms on $ A $ and $ C $. In other words,
$$
\forall g \in G, ~ \forall a \in A: \qquad
     \left\|
     \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})
     \right\|_{C}
\leq \| a \|_{A},
$$
or equivalently,
$$
(\clubsuit) \qquad
\forall g \in G, ~ \forall a \in A: \qquad
\| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C} \leq k \| a \|_{A}.
$$
Finally, let $ f \in {C_{c}}(G,A) $, and let $ S $ denote the compact/finite support of $ f $. Then
\begin{align}
       \| \pi(f) \|_{C}
& =    \left\|
       \pi \! \left( \sum_{g \in S} f(g) \bullet \delta_{g} \right)
       \right\|_{C} \\
& =    \left\| \sum_{g \in S} \pi(f(g) \bullet \delta_{g}) \right\|_{C} \\
& \leq \sum_{g \in S} \| \pi(f(g) \bullet \delta_{g}) \|_{C} \\
& =    \sum_{g \in S}
       \| \pi({\alpha_{g^{- 1}}}(f(g)) \bullet \delta_{e}) \|_{C} \qquad
       (\text{From $ (\spadesuit) $.}) \\
& \leq \sum_{g \in S} k \| f(g) \|_{A} \qquad
       (\text{From $ (\clubsuit) $.}) \\
& =    \| f \|_{1}.
\end{align}
Therefore, $ \pi $ is contractive with respect to the $ L^{1} $-norm on $ {C_{c}}(G,A) $ and the norm on $ C $. $ \quad \blacksquare $
