I must show that:
Let $f(\phi)$ = $F(\phi(0), \phi(-r))$, where $F: R^{2n} -> R^{n}$ is continuous , and thus the above becomes: $x'(t) = F(x(t), x(t-r))$. Show that $f$ is completely continuous.
My attempt:
By definition, we say that f is completely continuous if it maps bounded subsets of $C$ =$C([-r,0];R^{n})$ to bounded subsets of $R^{n}$. Since $F$ is a continuous operator from $R^{2n}$ to $R^{n}$, then this is the same as saying that it is bounded and linear. So $F(\phi(0), \phi(-r))$ = $A\phi(0) + B\phi(-r)$, for some matrices $A,B$.
We will be done once we show that $f$ is Lipschitz. However, that follows since $|f(\phi) - f(\psi)|= A|\phi(0)-\psi(0)|+B|\phi(-r)-\psi(-r)|\leq (A+B)||\phi -\psi||_{\infty}$ Thus, f is clearly Lipschitz and so the claim follows.
Is this fine for a proof? Or, am I assuming something I shouldn't?