Let $T:X\to X$ be a bounded sublinear operator. How can we show that the following inequality is true:
$$ T^{k+1}f(x) \le 2 \|T\|\, T^{k}f(x), $$ where $\|T\|$ denotes the operator norm and $T^{k}=T\circ T\circ \dots\circ T$ is is $k$ iterations of the operator $T$.
My attempt is:
$$ \|T^{k+1}f\|\le \|T\|\|T^{k}f\| $$ since $T$ is a bounded operator. But i can not continue.
This is wrong or your statement might not be completed.
Example: $T f := (-1)\, f $ for $X = L^{\infty}(\mathbb{R}).$
Consider any positive function $f_0,$ for instance $f_0(x) \equiv 1.$
$$T^k f_0 = (-1)^k \,\,\,\text{and}\,\,\, \|T\|=1.$$ When $k$ is odd, your claim is false.