Can we bound from above sub-solutions of Volterra integral equations? (Nonlinear Gronwall's Lemma) Gronwall's lemma says the following. Assume that $v\in C^0([t_0, T])$ is a nonnegative function. If $u \in C^0([t_0, T])$ satisfies the integral inequality 
$$u(t) \le c + \int_{t_0}^t u(s)v(s)\, ds,\qquad t \in [t_0, T]$$
where $c\in\mathbb{R}$, then 
$$u(t) \le c \exp\left(\int_{t_0}^t v(s)\, ds\right), \qquad t \in [t_0, T].$$
In other words, sub-solutions of the linear integral equation 
$$w(t)=c+ \int_{t_0}^t v(s)w(s)\, ds, \qquad t \in [t_0, T]$$
are dominated by solutions of the same equation, provided that the coefficient $v$ is nonnegative.

Question What can we say about the general Volterra equation 
  $$\tag{1}\ w(t) = c + \int_0^t F(s, w(s))\, ds,\qquad t \in [0, T]?$$
  Under what conditions on $F$ is a sub-solution of (1) dominated by a solution?

 A: This is not an answer but a (long) comment to Did's answer. In a nutshell: I think that we need to add a Lipschitz condition on $F$ (assumption 2 below). Did probably assumed it implicitly. 
Here's a possible way to argue. Assume that: 


*

*for all $s\in[0, T]$, one has that $x\le y\Rightarrow F(s, x)\le F(s, y)$;

*there exists a constant $L>0$ such that $\lvert F(s, x)-F(s, y)|\le L\lvert x-y\rvert$.$^{[1]}$


Let
$$u(t)\le c+\int_0^tF(s, u(s))\, ds,\qquad w(t)=c+\int_0^tF(s, w(s))\, ds.$$
We claim that 
$$\tag{1}u(t)\le w(t).$$ 
Proof. Consider the set 
$$
X=\left\{t\in [0, T]\ :\ u(t)-w(t)>0\right\}.$$
Assume by contradiction that $X$ is nonempty and set 
$$
t_0=\inf X.$$
Since $u$ is a subsolution we have that $t_0>0$. By assumption 1 we have that 
$$
F(s, u(s))-F(s, w(s))\le 0\qquad \forall s\in[0, t_0].
$$
So for $t\in X$ we have by assumption 2
\begin{equation}
\begin{split}
0<u(t)-w(t)&\le \int_0^tF(s, u(s))-F(s, w(s))\,ds \\
&\le \int_{t_0}^t F(s, u(s))-F(s, w(s)\, ds \\
&\le L\int_{t_0}^tu(s)-w(s)\, ds,
\end{split}
\end{equation}
and Gronwall's inequality gives the contradiction $0<u(t)-w(t)\le 0$. $\square$
I don't know if condition 2 can be weakened, but surely it cannot be dropped altogether. One must have at least a uniqueness result for the integral equation, and condition 2 gives such a result (that's the standard Picard's existence and uniqueness theorem). 
For example, consider this problem: 
$$w(t)=\int_0^t \big(w(s)\big)^{1/3}\, ds.$$
We know that there exist more than one solution to this equation, and of course every solution is a subsolution. If our claim were true in this case we would have for a pair of distinct solutions $w_1, w_2$ the inequalities $w_1(t) \le w_2(t)$ and $w_2(t)\le w_1(t)$, that is, $w_1(t)=w_2(t)$, a contradiction. 
An open question remains, and it it necessity of the monotonicity condition on $F(s, \cdot)$. This is intuitively reasonable, but should be proved. 

$^{[1]}$ It is enough to assume those conditions for almost all values of $s$. 
A: A sufficient condition is that every function $F(s,\cdot)$ is nondecreasing. That is:

For every $s$ in $[\tau,T]$, $v\le v'$ implies $F(s,v)\le F(s,v')$. 

In full generality, this condition is probably also necessary.
