In Theorem 3 on page 4 of this paper by Ozawa (for instance, equation (23)), he considers the entropy $H(T_G^{\mathrm{mix}}(1-\varepsilon))$ of a certain mixing time for a given graph $G$. My first question is why this even makes sense to consider: the setup in the first page states that the random walks on graphs will begin from some identified vertex, so that there is implied to be no randomness in initializing the walk, and $\varepsilon$ and $G$ are given -- so what exactly is the probability distribution whose entropy is being considered?
My second question is how he comes up with equation (24): how does knowing the entropy of this supposed distribution of mixing times tell anything about the difference of these two mixing times themselves? How do we end up inverting the lower bound in $H(t+1)-H(t)$ from the first line of the proof?
Thanks in advance for any answers.
The notation $H(t) := H(\mu^t)$ is established as (14). Now note that $T^{\mathrm{mix}}_G(1-\varepsilon)$ is a time, so $H(T^{\textrm{mix}}_G(1-\varepsilon))$ is well defined and equals the entropy of the distribution at the $1-\varepsilon$ mixture time.
For (24), I'll assume that $\varepsilon < 1-\varepsilon$ (the statement is otherwise). This in particular means $T_G^{\mathrm{mix}}(\varepsilon) \ge T_G^{\mathrm{mix}}(1-\varepsilon)$.
Now, it is shown (Lemma 2, and the definitino of $T^{\mathrm{mix}}$) that for $t \le T_G^{\mathrm{mix}}(\varepsilon)$, $H(t+1) - H(t) \ge (1-\rho)\varepsilon^2/16$. Summing this from $T_G^{\mathrm{mix}}(1-\varepsilon)$ to $T_G^{\mathrm{mix}}(\varepsilon)$ gives us that $$ \sum_{t = T_G^{\mathrm{mix}}(1-\varepsilon)}^{T_G^{\mathrm{mix}}(\varepsilon)} H(t+1) - H(t) \ge (1-\rho) \frac{\varepsilon^2}{16}(T_G^{\mathrm{mix}}(\varepsilon) - T_G^{\mathrm{mix}}(1-\varepsilon)).$$
But the left hand side above telescopes, and so is $$H(T_G^{\mathrm{mix}}(\varepsilon) + 1) - H(T_G^{\mathrm{mix}}(1-\varepsilon)) < \log |G| - (1-\delta)\log|G| = \delta \log |G|,$$ where the inequality follows since $H(t) \le \log|G|$ generically (entropy is at most log of support size) and $H(T_G^{\mathrm{mix}}(1-\varepsilon)) > (1-\delta)\log|G|$. Now (24) follows by just moving the appropriate terms across the inequality.