Let $K_n$ be complete graph with $n$ vertices and $$d^{\infty }(t) := \max_{x,y\, \in V} \left|\frac{P^{t}(x,y)}{\pi(y)}-1\right|$$ and $$t^{\infty}_{\text{mix}} := \min \{t \geq 0 : d^{\infty }(t) \leq \frac{1}{2e} \}$$ we want show that for lazy random walk on complete graph $K_n$, $t^{\infty}_{\text{mix}} \asymp \log n$.
the upper bound can be easliy get by directly plug into the eigenvalue bound $$ t^{\infty}_{\text{mix}} \leq \frac{1}{\gamma_{\ast}} \log \frac{1}{\pi_{\min}} \tag 1$$ where $\gamma_{\ast}$ is the spectral gap.
but how to see the lower bound, what I have attempted is:
for any vertices $x \neq y$ , and let $\tau_y$ be the hitting time of $y$ and
\begin{align} 1-P^{t}(x,y ) & \geq \text{Pr}(\tau_y >t) \\ & = (1-\frac{1}{2(n-1)})^t \end{align}
and
\begin{align} \max_{x,y\, \in V} \left|\frac{P^{t}(x,y)}{\pi(y)}-1\right| &\geq 1- \frac{P^{t}(x,y)}{\pi(y)} \\ & = 1- n\left(1- \left(1-\frac{1}{2(n-1)}\right)^t\right) \end{align}
and I think its don't make sense ...