For non-zero $q\in [-1,1]$, Woronowicz's quantum group $\operatorname{SU}_q(2)$ is given as the universal unital $\mathrm{C}^*$-algebra generated by elements $a,c\in C(\operatorname{SU}_q(2))$ subject to some relations. See here.
The $*$-subalgebra generated by $a,c$ is a Hopf*-algebra $\operatorname{Pol}(\operatorname{SU}_q(2))$ where the antipode $S:\operatorname{Pol}(\operatorname{SU}_q(2))\rightarrow \operatorname{Pol}(\operatorname{SU}_q(2))$ is given by:
$$S(a)=a^*,\qquad S(a^*)=a,\qquad S(c)=-qc,\qquad S(c^*)=-q^{-1}c^*.$$
(Assuming the norm is the norm in $C(\operatorname{SU}_q(2))$, I am actually not sure) Famously this antipode map is unbounded.
How can I see this?
I am presuming it will have something to do with the sequence $((c^*)^n)_{n\geq 1}$.
This is missing details but should be the basic idea. Let $|q|<1$.
The algebra of continuous functions on $\operatorname{SU}_q(2)$ may be identified with the algebra of bounded operators on the Hilbert space $\ell_2(\mathbb{Z}_+\times \mathbb{Z})$.
I understand that $c^*(e_{n,k})=q^ne_{n,k-1}$. I think it follows that both $\|c^*\|= 1$ and $\|(c^*)^r\|= 1$ or $r\in\mathbb{N}$.
The antipode for $\operatorname{Pol}(\operatorname{SU}_q(2))$ is an anti-morphism and so: $$\|S((c^*)^r)\|=\|S(c^*)^r\|=\left\|\frac{(-1)^r}{q^r}(c^*)^r\right\|=\frac{1}{q^r}\rightarrow \infty.$$
The punchline is that $S$ is not bounded, and although densely defined on $C(\operatorname{SU}_q(2))$, it does not extend to the whole of that algebra. This is in contrast to the comultiplication, which does.