I'm trying to understand the proof of theorem 1.6 from the paper "A Matrix Expander Chernoff Bound".

In the proof they say: "Iterating this construction on the remainder a total of $T ≤ k$ times" and later they choose a value for $T$.

$k$ is the length of a given walk on the graph $G$ which has $n$ vertices.

They choose $$T = \frac{2log(\frac{F}{ε})}{1 − λ}$$

If we take $f$ to be a function that gets $1$ on half of the vertices and $-1$ on the other half then $F=\sqrt{n}$ by definition. So $T\gt log(n)$. But $k$ might be much smaller than $log(n)$. So if $T>k$, how can they choose this $T$?

Thanks!