Can we extend this proposition to complex-valued functions? I'm reading
Markov Chains and Mixing Times by David A. Levin, Yuval Peres, Elizabeth L. Wilmer, James G. Propp, David B. Wilson
and I'm trying to figure out if this Proposition can be applied when $f: \Omega \to \mathbb{C}$.
First recall that
if $\mu$ and $\nu$ are two probability distributions defined on a state space $\Omega$. The total variation distance between these two probability distributions is defined as
\begin{equation*}
    d_{TV}(\mu,\nu)=||\mu-\nu||_{TV}:= \max_{A \subseteq \Omega} |\mu(A) - \nu(A)|
\end{equation*}
where $\mu(A) = \sum_{ x \in A}\mu(x) $.
Also recall that given a probability distribution $\mu$ defined on $\Omega$ and $f:\Omega \to \mathbb{R}$, denote by $E_\mu[X] = \int_{\Omega}X(\omega) \ \mu(d\omega)$ the expectation of $X$ with respect $\mu$. Likewise, $Var_\mu(X)$ denotes the variance of $X$ with respect to $\mu$.
Proposition.
Let $\mu$ and $\nu$ be two probability distributions on a finite state space $\Omega$ and let $f:\Omega \to \mathbb{R}$ be a real valued function on $\Omega$. Then,
\begin{equation}\label{PeresBound}
    ||\mu- \nu||_{TV} \geq 1 - 8 \  \frac{\max\{Var_\mu(f),Var_\nu(f)\}}{|E_\mu[f]-E_\nu[f]|^2}.
\end{equation}
Proof.
Assume without loss of generality that $E_\mu[f] \geq E_\nu[f]$ and let $$r:=\frac{|E_\mu[f]-E_\nu[f]|}{\sigma_*}=\frac{|E_\mu[f]-E_\nu[f]|}{\sqrt{\max\{Var_\mu(f),Var_\nu(f)\}}}.$$ If $I := (E_\nu[f] + \frac{r}{2} \sigma_*, \infty) \subseteq \mathbb{R}$, by Chebyshev's inequality,
\begin{equation*}
        \begin{split}
            \nu f^{-1}(I) & = \nu( f- E_\nu[f] >  \tfrac{r}{2} \sigma_*) \leq \nu( |f-E_\nu[f]| > \tfrac{r}{2} Var_\nu[f]) \leq \frac{4}{r^2}\\
            \mu f^{-1}(I) & = \mu( f > E_\mu[f] - |E_\nu[f] - E_\mu[f]| + \tfrac{r}{2} \sigma_*) = \mu( E_\mu[f] - f < \tfrac{r}{2} Var_\mu[f]) \geq 1 - \frac{4}{r^2}.
        \end{split}
    \end{equation*}
Thus,
\begin{equation*}
        ||\mu f^{-1} - \nu f^{-1}||_{TV} \geq | \mu f^{-1}(I) - \nu f^{-1}(I)| \geq 1 - \frac{8}{r^2}
    \end{equation*}
and by Lemma, since $f: \Omega \to f(\Omega)$ with $f(\Omega)$ finite,
\begin{equation*}
        ||\mu  - \nu||_{TV} = ||\mu f^{-1} - \nu f^{-1}||_{TV} \geq 1 - \frac{8}{r^2} =1- 8 \  \frac{\max\{Var_\mu(f),Var_\nu(f)\}}{|E_\mu[f]-E_\nu[f]|^2}.
    \end{equation*}
Lemma.
Let $\mu$ and $\nu$ be probability distributions on $\Omega$ and let $f: \Omega \to \Gamma$, where $\Gamma$ is a finite set. Then
\begin{equation*}
   ||\mu-\nu||_{TV} \geq ||\mu f^{-1} -\nu f^{-1}||_{TV},
\end{equation*}
where \begin{equation*}
\mu f^{-1} (I) := \mu( f^{-1}(I)) = \mu( f \in I).
\end{equation*}
Proof.
For $J \subseteq \Gamma$, since $$| \mu f^{-1} (J) - \nu f^{-1} (J) | = | \mu (f^{-1} (J)) - \nu (f^{-1} (J)) |,$$ it follows that
$$||\mu-\nu||_{TV} = \max_{ I \subseteq \Omega}| \mu (I) - \nu (I) | \geq \max_{J \subseteq \Gamma} | \mu f^{-1} (J) - \nu f^{-1} (J) | = ||\mu f^{-1} -\nu f^{-1}||_{TV}.$$

My question is if I can use Proposition when $f: \Omega \to \mathbb{C}$. It would be very usefull for me to estimate the total variation distance using a complex-valued random variable. However, I look at the proof of the Proposition and it seems that it wouldn't be true when $f: \Omega \to \mathbb{C}$... Is there an equivalent result of this proposition for complex-valued functions?

 A: Yes, we just have to change it to
\begin{equation}
    ||\mu- \nu||_{TV} \geq 1 - 8 \  \frac{\max\{Var_\mu(f),Var_\nu(f)\}}{|\ |E_\mu[f]|-|E_\nu[f]|\ |^2}.
\end{equation}
The proof is then similar and uses the fact that for two complex numbers $w, z,$
$$|z|-|w| \leq |\ |z|-|w|\ | \leq |z-w|.$$
Assume without loss of generality that $|E_\mu[f]| \geq |E_\nu[f]|$ and let $$r:=\frac{|\ |E_\mu[f]|-|E_\nu[f]|\ |}{\sigma_*}=\frac{|\ |E_\mu[f]|-|E_\nu[f]| \ |}{\sqrt{\max\{Var_\mu(f),Var_\nu(f)\}}}.$$ If $I := (|E_\nu[f]| + \frac{r}{2} \sigma_*, \infty) \subseteq \mathbb{R}$ and $g:= |f|$, by Chebyshev's inequality,
\begin{equation*}
            \begin{split}
                \nu g^{-1}(I) & = \nu( |f|- |E_\nu[f]| >  \tfrac{r}{2} \sigma_*) \leq \nu( | f|-|E_\nu[f] | > \tfrac{r}{2} Var_\nu[f]) \\
& \leq \nu( |f-E_\nu[f]| > \tfrac{r}{2} Var_\nu[f])  \leq \frac{4}{r^2}\\
                \mu g^{-1}(I) & = \mu( |f| > |E_\mu[f]| - |\ |E_\nu[f]| - |E_\mu[f]|\ | + \tfrac{r}{2} \sigma_*) = \mu( |E_\mu[f]| - |f| < \tfrac{r}{2} \sigma_*)\\
                & \geq  \mu( |E_\mu[f]| - |f| < \tfrac{r}{2} Var_\mu[f]) 
                \geq \mu( |E_\mu[f] - f| < \tfrac{r}{2} Var_\mu[f]) \geq 1 - \frac{4}{r^2}.
            \end{split}
\end{equation*}
Thus,
\begin{equation*}
            ||\mu g^{-1} - \nu g^{-1}||_{TV} \geq | \mu g^{-1}(I) - \nu g^{-1}(I)| \geq 1 - \frac{8}{r^2}
\end{equation*}
and by Lemma, since $g: \Omega \to |f(\Omega)|$ with $f(\Omega)$ finite,
\begin{equation*}
        ||\mu  - \nu ||_{TV} \geq ||\mu g^{-1} - \nu g^{-1}||_{TV} \geq 1 - \frac{8}{r^2} =1- 8 \  \frac{\max\{Var_\mu(f),Var_\nu(f)\}}{|\ |E_\mu[f]|-|E_\nu[f]|\ |^2}.
\end{equation*}
