Maybe the answer to the following question is known but I am unable to find it in the literature. Anyway, let me begin my question by fixing some notations and terms.

Let $G = (A, \Delta)$ be a compact quantum group in the sense of Woronowicz. For a finite dimensional unitary representation $U \in B(H) \otimes A$ of $G$, the contragredient $U^c$ of $U$ is the representation $(j \otimes \operatorname{Id})(U^*) \in B\bigl(\overline{H}\bigr) \otimes A$ of $G$, where the carrier space $\overline{H}$ is the conjugate Hilbert space of the finite dimensional Hilbert space $H$, and $j : B(H) \to B\left(\overline{H}\right)$ the $\ast$-anti-isomorphism sending $T$ to $\overline{T^*}$, the latter denotes the operator $\overline{\xi} \mapsto \overline{T^* \xi}$ on $\overline{H}$. Define $c(G)$ to be the supremum of all $\| U^c \|$, where $U$ runs through all finite dimensional unitary representations (of course, irreducible ones suffice) of $G$.

It is clear that if $G$ is of Kac-type, then $c(G) = 1$, as $U^c$ in the above is always unitary. With some effort, I can prove that $c(G) = +\infty$ for $G = SU_q(2)$ with $-1 < q \ne 0 < 1$, but whether $c(G) = +\infty$ for general non-Kac type $G$ seems more delicate, which prompts me into asking the following

**Question.** Does $c(G) < +\infty$ imply $G$ being of Kac type?

As there are already many results concerning characterization of Kac type compact quantum groups, it may well be possible that this question is already settled, in which case, I appreciate a reference to the literature.