The way I understand it is that because of this bound you can derive a solution by a fixed point. Set $u_0=0$ and $$ u_{n+1} = A + \int_0^t B |u_n|^{p-1}u_n ds $$ (A,B represents the various quantities appearing in the paper) The bound found gives $\|u_{n+1}\| \leq \alpha + \beta |I| \|u_{n}\|^3$.

If $\|u_n\|\leq K$, then $\|u_{n+1}\|\leq K$ provided $$ K\leq \alpha + \beta |I| K^3. $$ For example take $K=10\alpha$. Then this is true provided $|I|\leq \frac{9}{\beta 10^3\alpha^2}$.

So now you have a bounded sequence, and by the same token a contraction, provided $$ 3 \beta |I| K^2 <1. $$ so the solution you constructed converges and satisfies the $K$ bound, which is exactly what you wanted.

The way I understand it is that because of this bound you can derive a solution by a fixed point. Set $u_0=0$ and $$ u_{n+1} = A + \int_0^t B |u_n|^{p-1}u_n ds $$ (A,B represents the various quantities appearing in the paper) The bound found gives $\|u_{n+1}\| \leq \alpha + \beta |I| \|u_{n}\|^3$.

If $\|u_n\|\leq K$, then $\|u_{n+1}\|\leq K$ provided $$ K\leq \alpha + \beta |I| K^3. $$ For example take $K=10\alpha$. Then this is true provided $|I|\leq \frac{9}{\beta 10^3\alpha^2}$.

So now you have a bounded sequence, and by the same token a contraction, provided $$ 3 \beta |I| K^2 <1. $$ so the sequence you constructed converges and satisfies the $K$ bound, which is exactly what you wanted.