'Test Functions' to Lower Bound the Norm of Elements of Dual Quantum Group There may well be an answer to this question in a simpler category than that of finite dimensional quantum groups and in that case this question is more suitable to math.stack and I apologise in advance if this is the case.
Consider a finite dimensional quantum group $A=F(\mathbb{G})$ with a Haar state $h:A\rightarrow\mathbb{C}$. In the finite dimensional case (and in more generality in fact), this allows us to define '$p$-norms' on $A$ via
$$\|a\|_p:=\sqrt[p]{h((a^*a)^{p/2})}.$$
In particular, $\|a\|_1=h((a^*a)^{1/2})$ and $\|a\|_2=\sqrt{h(a^*a)}$. There is a Cauchy-Schwarz Inequality:
$$\|ab\|_1\leq\|a\|_2\|b\|_2.$$
We can define a dual quantum group $\hat{A}$ via the map $\mathcal{F}:A\mapsto A'$, $\mathcal{F}(a)(b)=h(ba)$. In this, finite dimensional case, $\hat{A}=A'$, and the multiplication in $\hat{A}$ is given by the convolution:
$$\nu\star\mu=(\nu\otimes\mu)\Delta,$$
the Haar state $\hat{h}:\hat{A}\rightarrow\mathbb{C}$ is given by
$$\hat{h}(\mathcal{F}(a))=\varepsilon(a),$$
and the involution is
$$\nu^*(a)=\overline{\nu(S(a)^*)}.$$
I am interested in finding bounds for
$$\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2}).$$
Via $\varepsilon\star \nu=\nu$, we have the following upper bound on $\|\nu\|_1$:
$$\|\nu\|_1=\|\varepsilon\star \nu\|_1\leq\|\varepsilon\|_2\|\nu\|_2=\sqrt{\dim A}\cdot\|\nu\|_2.$$
For my application, I have a way of calculating and bounding $\|\nu\|_2$ above but I am also interested in bounding below:
$$?\leq \|\nu\|_1\leq \sqrt{\dim A}\|\nu\|_2.$$
In the classical case where $G$ is a finite group, a 1-norm on $F(G)$ might be given by
$$\|f\|_1'=\sum_{t\in G}|f(t)|,$$
and this fits quite well into this framework:
$$\|f\|_1=h\left((f^*f)^{1/2}\right)=h(|f|)=\frac{1}{|G|}\sum_{t\in G}|f(t)|.$$
Now I am more interesting in looking at $\|\cdot\|_1:\hat{A}\rightarrow[0,\infty)$. Now classically one might define the 1-norm on $\mathbb{C}G\supset M_p(G)$ as
$$\|\nu\|_1'=\sum_{t\in G}|\nu(\delta_t)|.$$
An advantage of working with this norm is that we have
$$\|\nu\|_1'=\sup_{\underset{f\in F(G)}{\|f\|_{\infty}\leq1}}|\nu(f)|,$$
so that we can generate lower bounds by looking at test functions $\phi\in F(G)^1$ and so we have
$$|\nu(\phi)|\leq \|\nu\|_1'\leq \sqrt{|G|}\|\nu\|_2'.$$
The problem with using $\|\cdot\|_1:\hat{A}\rightarrow[0,\infty)$ via the Haar state $\hat{h}$:
$$\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2}),$$
is that even in the classical case I don't quite have something like
$$\|\nu\|_1=\frac{1}{|G|}\sum_{t\in G}|\nu(\delta_t)|.$$
When we are in $A=F(G)$ the involution is simply
$$f^*(s)=\overline{f(s)},$$
and with pointwise multiplication and positivity in the C*-algebra equivalent to positivity of the coefficients, we have
$$(f^*f)^{1/2}(s)=|f(s)|,$$
so the 1-norm works quite nicely in there.
Things are more complicated however in $\hat{A}=\mathbb{C}G$ (as is alluded to in this question of mine). In general, even for symmetric probability measures, we don't have
$$(\nu^*\nu)^{1/2}=\nu.$$
Now in the classical case I can just use $\|\cdot\|_1'$ and the ordinary C-S to get my upper bounds. However in the truly non-commutative case I want to use $\|\cdot\|_1$... if I can get myself some lower bounds! Otherwise I can just use and bound $\|\cdot \|_2$ above.

In the not-necessarily-commutative, quantum group case, is there a way to generate lower bounds on $\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2})$ via 'test functions':
$$\|\nu\|_1\geq\sup_{s\in S}F(s,\nu)?$$
Perhaps $F(s,\nu)$ involves convolving $\nu$ with some element of $s\in S\subset\hat{A}$ with a small norm or maybe something like hitting $\nu$ with an element of $s\in S\subset A$ with small norm: $\nu(s)$.

Thank you for your help.
 A: In a more general setting than that of the original question: suppose we have a faithful normal state $h$ on a von Neumann algebra $M$. Suppose furthermore that $h$ is tracial, meaning that $h(xy)=h(yx)$ for all $x,y\in M$. (Warning! there are important examples of compact quantum groups where the Haar state is faithful but not tracial.)
In this setting we may define $\Vert x\Vert_{L^1(M,h)}$ to be $h(|x|)$. With this definition it is not clear that we have a norm; however, it is known that one has
$$ h(|x|) = \sup\{ | h(xy) | \colon y \in M, \Vert y\Vert_M \leq 1 \} \tag{$*$}$$
and the proof has been given by Martin Argerami on MathStackExchange. The motivating example to keep in mind is the commutative case $M=L^\infty[0,1]$ with usual ess.sup norm and usual weak-star topology, with $h(x) = \int_0^1 x(t)\,dt$.
Historical note: if one looks at Definition 3.2 of I. Segal's paper

MR0054864 (14,991f)
  I. E. Segal, A non-commutative extension of abstract integration.
  Ann. of Math. (2) 57 (1953). 401–457

one sees that Segal took the RHS of $(*)$ as the definition of the $L^1$-norm on $M$ given by $h$. He then, in part (d) of Corollary 10.1, shows that the RHS of $(*)$ is equal to the LHS of $(*)$.
