Let and .
Suppose .
Let be continuous
and for some we have

There there is a s.t. :

has a unique solution in .

Proof

Firstly, is cts on a closed set hence bounded, .
Pick .
Let .
The metric space (of continuous functions)
is complete in uniform metric.
Hence define

Note that is indeed cts whenever is cts (by FTA).
We do need to check that :

Now we check that is a contraction mapping on .

&\leq \sup_t\int_{t_0}^t|\phi(x,g(x))-\phi(x,h(x))|dx\\ &\leq \epsilon \sup_x|\phi(x,g(x))-\phi(x,h(x))|\\ &\leq \epsilon K|g(x)-h(x)|\\ &\leq |g(x)-h(x)|/2 \end{align}$$ Hence, we are done by [[Analysis/Contraction mapping\|contraction mapping theorem]] and FTA (i.e. $(Tg)=g\iff g$ is the solution to original problem).