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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!

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  • $\begingroup$ So, with $A=\mathbb C$, the case of discrete groups is easy (use a unitary operator argument). For compact groups look at the coefficient functions, and use a similar "unitary" trick. So the first thing I can't see how to do is $G=\mathbb R$. Any ideas here? $\endgroup$ – Matthew Daws Nov 13 '13 at 8:52
  • $\begingroup$ @MatthewDaws: Would you mind elaborating on the unitary-operator argument for discrete groups? My impression is that unitary group representations always lead to representations of $ ({C_{c}}(G,\mathscr{A}),\star,^{*}) $ that are bounded by the $ L^{1} $-norm. $\endgroup$ – user36116 Nov 16 '13 at 17:24
  • $\begingroup$ Btw, in your definition of the product, you need to replace $a\in A$ by a group element. $\endgroup$ – Matthew Daws Nov 16 '13 at 21:20
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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.

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  • $\begingroup$ Thank you for your answer Matthew. Unfortunately, I’m seeking a $ * $-representation that is not contractive for the $ {C^{*}}(G) $-norm (which would automatically be contractive for the $ {L^{1}}(G) $-norm). $\endgroup$ – user36116 Nov 18 '13 at 12:00
  • $\begingroup$ Yes... the point of my comment was that you cannot find such a counter-example for G discrete or compact. So I suggested you look at $\mathbb R$, this being an example I couldn't immediately see... $\endgroup$ – Matthew Daws Nov 18 '13 at 14:28
  • $\begingroup$ So this is definitely not an answer... just a comment that was too long for a 2comment"... $\endgroup$ – Matthew Daws Nov 18 '13 at 14:28
  • $\begingroup$ I understand your point now! I misunderstood your comment as an answer to my question. I’ll accept your answer, but I’d like to keep the bounty open to attract more attention to the problem. Thank you for your help! $\endgroup$ – user36116 Nov 18 '13 at 16:07
  • $\begingroup$ Well, I wouldn't accept! Let's hope someone comes along with some more thoughts (as I think this is a nice question!) $\endgroup$ – Matthew Daws Nov 18 '13 at 20:50
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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 $

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