inequality in finite group representation theory I would like to understand the first inequality of the proposition below.
Suppose $Q$ is constant on conjugacy classes of $G$, i.e. for all $g,h\in G$, $Q(hgh^{-1}) = Q(g)$. For an
irreducible representation $\rho$ let
\begin{equation}
    \tilde{Q}(\rho) = \frac{1}{d_\rho} \sum_{g \in G}Q(g)\chi_\rho(g)
\end{equation}
where $\chi_\rho(g) = \operatorname{Tr}(\rho(g))$ is the trace of $\rho(g)$. Then for each $g \in G$,
$$
1 - |G|Q(g) \le \sum_\rho d_\rho |\chi_\rho(g) \tilde{Q}(\rho)| \le \sum_\rho d_\rho^2 |\tilde{Q}(\rho)|
$$
where $\sum_\rho$ denotes summation over all non-trivial irreducible representations.
Moreover, in the article the authors say the inequality is a consequence of the inverse Fourier transform formula for $Q$. The Fourier transform of $Q$ is given by
$$
\hat{Q}(\rho) = \sum_g\rho(g)Q(g),
$$
and the inverse Fourier transform is given by
$$
Q(g) = \frac{1}{|G|}\sum_{\rho}d_{\rho}\operatorname{Tr}\left(\hat{Q}(\rho)\rho(g^{-1})\right),
$$
where $\operatorname{Tr}$ is the trace.
This inequality is in the article by Diaconis and Aldous, Proposition 6.3, which can be seen at this link:
Diaconis and Aldous
 A: First of all, notice that the Fourier inversion formula is given by
$$Q(g) = \frac{1}{|G|}\sum_{\hat G}d_{\rho}\operatorname{Tr}\left(\hat{Q}(\rho)\rho(g^{-1})\right),$$
where the summation is taken over all (inequivalent) irreducible representations of $G$.
Second, if $Q$ is constant on conjugacy classes and $\rho$ is an irreducible representation, then a corollary of Schur's lemma says that
$$\hat Q(\rho)=\tilde Q(\rho)I$$
(for example, see Serre (GTM 42), $\S$ 2.5, Proposition 6).
Thus,
\begin{align}
1 - |G|Q(g)&\leq |1-|G|Q(g)|\\
&=|1-\sum_{\hat G}d_{\rho}\operatorname{Tr}\left(\hat{Q}(\rho)\rho(g^{-1})\right)|\\
&=|1-\sum_{\hat G}d_{\rho}\operatorname{Tr}\left(\tilde Q(\rho)I\rho(g^{-1})\right)|\\
&=|1-\sum_{\hat G}d_{\rho}\tilde Q(\rho)\chi_\rho(g)|\\
&=|1-d_{\rho_1}\tilde Q(\rho_1)\chi_{\rho_1}(g)-\sum_\rho d_\rho \chi_\rho(g) \tilde{Q}(\rho)|,
\end{align}
where $\rho_1$ denotes the trivial representation. We have $d_{\rho_1}=1=\chi_{\rho_1}(g)$. Moreover, $$\tilde Q(\rho_1)=\sum_{g \in G}Q(g)\chi_\rho(g)=\sum_{g \in G}Q(g)=1.$$
The desired inequality follows.
