@GiorgioMetafune Later on they choose $t=\frac{1}{2L_{\sigma}||K||_{L^{\infty}}}$ which is not necessarily $\leq 1$ right? Even later they switch from $||X||_{L^{\infty}([0,t])} \leq 2L_\sigma ||b||_{L^\infty} + 2||x||$ to $||X||_{L^\infty ([0,1])} \leq 2(2^N - 1)L_{\sigma} ||b||_{L^\infty} + 2^N||x||$ with $N=\left \lceil{\frac{1}{t}}\right \rceil $. Maybe if I understand how they switch there, I see why $t\leq 1$?

@GiorgioMetafune So you're saying $t\leq 1$?