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](https://math.stackexchange.com/questions/1320060/positive-elements-of-mathbbcg-as-functionals-versus-as-elements-of-the-c)). 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.



'Test Functions' to Lower Bound the Norm of Elements of Dual Quantum Group
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](https://math.stackexchange.com/questions/1342359/the-1-norm-on-a-quantum-group-as-a-supremum). 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)](http://www.ams.org/mathscinet-getitem?mr=54864)
>  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 $(\*)$.


