I had some major modifications in my previous question.
I first showed that $\tau$ satisfies Blackwell's conditions, which implies that $\tau$ is a contraction mapping.
$\tau$ takes a bounded continuous function $[0,1] \to [-M,M]$ as an input.
According to the definition of contraction mapping :
$$ d\big{(}\tau(v),\tau(u)\big{)} \leq \beta * d\big{(}v,u\big{)} $$ for some $\beta \in [0,1)$ and for any function $u$ and $v$. Here I will just assume $d$ is sup norm.
Now, I want to build up a function $v(x)$ likewise :
$$ v(x) \equiv u(f(x)) = u \bigg{(} \frac{ax}{ax+(1-a)(1-x)}\bigg{)}$$
where $a>1/2$.
Note that $f(x) \geq x$ for any $x\in[0,1]$ and $f(x)\in[0,1]$.
Then, I can conclude:
$$ d\big{(}\tau(u(f)),\tau(u)\big{)} \leq \beta * d\big{(}u(f),u\big{)} $$
However, what I really want to show is that, for any $x\in[0,1]$,
$$ \big{|}\tau(u(f(x)))-\tau(u(x))\big{|} \leq \beta * \big{|}u(f(x))-u(x)\big{|} $$
The latter implies the former (just by imposing sup). If I want to reach to the latter statement, what additional sufficient conditions should be satisfied?