Expectation of a function evaluated on a random walk in a group Let $G$ be a finite group, and let $Q$ be a probability measure on $G$. Suppose that $Q$, as a function on $G$, is a class function. We denote by $Q^{*k}$ the $k$-fold convolution of $Q$ with itself - again a class function. We denote by $U$ the uniform probability measure on $G$. 
One is often interested in bounds on the L1-norm $||Q^{*k}-U||=\sum_{g \in G} |Q^{*k}(g)-U(g)|$. A famous bound of Diaconis and Shahshahani, 'the upper bound lemma', is useful in this case.
Suppose instead that we have a class function $f:G\to \mathbb{C}$, and suppose that we know how to explicitly write $f$ as a linear combination of characters. Is there a general upper bound for
$$\sum_{g \in G}|f(g) (Q^{*k}(g)-U(g) )| = |E_{Q^{*k}}(f) - E_{U}(f)|?$$
One approach is to use Cauchy-Schawrz and reduce the problem to an estimate on $||Q^{*k}-U||_2$ (which may be estimated by following the proof of the upper bound lemma) and on $||f||_2$. This feels too naive. Are there any other useful bounds?
 A: Following a paper of Berestycki and Kozma (beginning of Section 2), I am able to answer my question.
We begin by writing $f$ as a linear combination $\sum_{\rho} a_{\rho} \chi_{\rho}$, wher $\chi_{\rho}$ is the irreducible character of the irreducible representation $\rho$. Then
$$\sum_{g \in G} f(g) (Q^{*k}(g)-U(g)) = \sum_{\rho} a_{\rho} \sum_{g \in G} \chi_{\rho}(g) (Q^{*k}(g)-U(g))$$
$$=\sum_{\rho} a_{\rho} \mathrm{Tr}(\sum_{g \in G} \rho(g) (Q^{*k}(g)-U(g)) ).$$
Claim 1: The trivial representation does not contribute to the sum. 
Proof: For $\rho=1$, we have $\mathrm{Tr}(\sum_{g \in G} \rho(g) (Q^{*k}(g)-U(g)) ) = \sum_{g \in G} Q^{*k}(g)-\sum_{g \in G}U(g) = 1-1 = 0$.
Claim 2: For non trivial $\rho$, we have $\mathrm{Tr}(\sum_{g \in G} \rho(g) U(g))=0$.
Proof: This trace is just the mean value of $\chi_{\rho}$, which is $0$ by orthogonality of characters.
Claim 3: $\sum_{g \in G} \rho(g) Q^{*k}(g) = (\sum_{g \in G} \rho(g) Q(g))^k$.
This follow from the definitions of $Q^{*},\rho$.
Claim 4: The linear map $A_{\rho,Q}:=\sum_{g \in G} \rho(g) Q(g)$ is a scalar map.
Proof: This follows from Schur's lemma, by noticing, that since $Q$ is a class function, then $A_{\rho,G}$ is a $G$-invariant map, that is, it commutes with $\rho(g)$ for any $g\in G$.
Let $\lambda_{\rho,Q} \in \mathbb{C}$ be defined via $A_{\rho,Q} = \lambda_{\rho,Q} I$. 
Claim 5: We have $$\mathrm{Tr}(\sum_{g \in G} \rho(g) Q^{*k}(g) ) = \lambda_{\rho,Q}^k \mathrm{dim}(\rho).$$
Proof: Follows from Claims 3 and 4.
Combining the claims, we obtain:
$$\sum_{g \in G} f(g) (Q^{*k}(g)-U(g)) = \sum_{\rho} a_{\rho} \sum_{g \in G} \chi_{\rho}(g) (Q^{*k}(g)-U(g))$$
$$=\sum_{\rho \text{ non trivial}} a_{\rho}\lambda_{\rho,Q}^k \mathrm{dim}(\rho).$$
Thus, a useful bound for the sum in question is
$$|E_{Q^{*k}}(f) - E_{U}(f)| \le \sum_{\rho \text{ non trivial}} |a_{\rho}| |\lambda_{\rho,Q}|^k \mathrm{dim}(\rho).$$
The value of $\lambda_{\rho,Q}$ depends strongly on $Q$. In the simplest interesting case, where $Q$ is the uniform probability measure on a conjugacy class $C$ of $G$, then 
$$\lambda_{\rho,Q} =\frac{\mathrm{Tr}(A_{\rho,Q})}{\mathrm{dim}(\rho)} = \frac{\chi_{\rho}(c)}{\chi_{\rho}(1)},$$
where $c$ is any element from $C$.
