Delay differential equation uniqueness - Please help :( I am having a really hard time understanding this problem. I know that for uniqueness we need that the derivative is continuous and that the partial derivative is continuous. I also know that the lipschitz condition gives continuity. I can't figure out what to do with this problem though. 

 A: Here is a hint.   Consider the operator $T$ defined below that maps continuous functions into continuous functions.  For any $h\in C(-\infty,r]$, let
$$
(Th)(t) := \phi(0) + \int_0^t f(\tau, h(\tau), h(g(\tau)))\  d\tau
$$
when $t>0$ and let $(Th)(t)= \phi(t)$ when $t\leq0$.  Notice that $T: C(-\infty,r] \rightarrow C(-\infty,r]$.  
Think about the fixed points of $T$.  If $h(t)$ is a fixed point of $T$, what differential equation would it satisfy?
To show that $T$ is a contraction for some value of $r>0$, you need to show that there exists an $r>0$ and a corresponding $\alpha\in (0,1)$ such that for any two functions $h_1$ and $h_2$ in $C(-\infty,r]$, 
$$
d(Th_1, Th_2) \leq \alpha \,d(h_1, h_2).
$$
That proof could be broken down into a few steps:
1) $d(Th_1, Th_2) := \sup_{t\in (-\infty, r]} | Th_1(t) - Th_2(t)|$.
2) $d(Th_1, Th_2) = \sup_{t\in (0, r]} \left| \int_0^t f(\tau, h_1(\tau), h_1(g(\tau))) - f(\tau, h_2(\tau), h_2(g(\tau)))\, d\tau \right|$.
3) $d(Th_1, Th_2) \leq \sup_{t\in (0, r]} \int_0^t |\,f(\tau, h_1(\tau), h_1(g(\tau))) - f(\tau, h_2(\tau), h_2(g(\tau)))|\, d\tau$.
4) $d(Th_1, Th_2) \leq 2 r L\, d(h_1, h_2)$.
5) $T$ is a contraction if $0\leq r < 1/(2 L)$.
Proving each of those statements would be most of the complete proof.  I have left some of the details for you to fill in.  Once you have done a dozen similar proofs, the whole process will seem much easier.
A: The problem has some inherent unboundedness in the way it is posed. You need to get rid of that in order to pursue. Some suggestions:
Pick first $r_0>0$ arbitrary and note that by compactness + continuity, $g([0,r_0])=[a,b]$ is a compact subset of $(-\infty,r_0]$. Therefore,
$$ R = \sup \left\{ |\phi(u)| : u \in g([0,r_0])\cap (-\infty,0])\right\} < +\infty$$
Intuitively, this gives a bound on the delayed $x$ when the delay goes negative.
Let $ M = R + 1+ |\phi(0)|$. 
Again by compactness we have that:
 $$ C = {\rm max}\; f([0,r_0]\times[-M,M]\times[-M,M]) +1 \in (0, + \infty)$$
This provides bounds upon $f$ on the 'relevant' domain. We now shrink $r$ to
take into account that $x$ should not become too big:
 $$ r = {\rm min} \{ r_0, (R+1)/C, 1/(4L)\} >0$$
The first is clear, the second because we don't want an integrated $x$ to go beyond $M$ and the last because we need a contraction in the end. We
define
$$ {\cal C} = \{ x\in C ((-\infty,r] ; {\Bbb R}): x(t)=\phi(t), t\leq 0; |x(t)|\leq M, t\in [0,r]\}$$
This is a closed subset of the Banach space where $M=+\infty$ and 
$\|x\| = \sup_{0\leq t\leq r} |x(t)|$ (note that the negative part is fixed, you don't have to consider it in the norm). Finally, we define as usual:
  $$ Tx(t) = \phi(0) + \int_0^t f(t,x(t),x(g(t)) dt, \; \; 0\leq t\leq r .$$
Our choice of $r$  (calculation...) ensures that for $x,x_1,x_2\in {\cal C}$:
$$ \|Tx\|\leq M \; \; {\rm and} \; \;  \|Tx_1-Tx_2\|\leq \frac12 \|x_1-x_2\|$$
so $T$ is a contraction on ${\cal C}$ and the rest is standard. One rather crucial point (which makes the problem tractable) is that the delay does not depend upon $x$, only upon $t$.
