Proof involving some inequality

[Extropy]

Difficulty with some inequality

Suppose that one has three numbers, $$v_j^0$$ for j=1, 2, and 3, and let their absolute values be less than or equal to 1. Suppose that one has nine continuous functions $$c_{jk}$$ all of which are bounded, and thus all of which are bounded by an amount $$\tfrac{k}{3}$$.

Define

$$v_j^{(1)}(s)=v_j^0+\int_0^s \sum_{k=1}^3 c_{jk}(\sigma) v_k^0 d\sigma$$

and

$$v_j^{(n)}(s)=v_j^0+\int_0^s \sum_{k=1}^3 c_{jk}(\sigma) v_k^{(n-1)}(\sigma) d\sigma$$

It is perfectly understandable why

$$|v_j^{(1)}-v_j^0| \le ks$$,

but not so clear why

$$|v_j^{(n)}-v_j^{(n-1)}| \le \frac{k^n s^n}{n!}$$.

Somehow, one is supposed to be able to divide by n, but I do not see how is able to.

 

[fresh_42]

If we set $v_j^{(0)}=1$ and $c_{ji}=\dfrac{k}{3}$ then we get $v_j^{(1)}(s)=1+ks$ and the one is the culprit for the denominator $n!$. This is because in the second step we get $v_j^{(2)}(s)=1+ks+\dfrac{k^2s^2}{2}$ and the difference will be $|v_j^{(2)}(s)-v_j^{(1)}(s)| =\dfrac{k^2s^2}{2}$ and so on.

This example is somehow the worst case scenario, but it illustrates what happens. Now you have to generalize this for arbitrary $c_{ji}(\sigma)$ and $v_j^{(0)}$, i.e. within the given constraints.