solution. If $$\label{inequal}
\dot w\leq f(t,w), \q w(0)=w_0\leq u_0,\q w_0\in D,$$ then $u(t)\geq w(t)$ for all $t$ for which $u(t)$ and $w(t)$ are defined. \[ineq\]

[**Proof**]{} Suppose first $f(t,w)<g(t,u)$, if $w\leq u$. Since $w_0\leq u_0$ and $\dot w (0)\leq f(t,w_0)<g(t,u_0)=\dot u (0)$, there exists $\delta
>0$ such that $u(t)> w(t)$ on $(0,\delta]$. Assume that for some $t_1> \delta$ one has $u(t_1)< w(t_1).$ Then for some $ t_2<t_1$ one has $$u(t_2)=w(t_2) \q\t{and}\q
u(t)<w(t) \q\t{for}\q  t\in (t_2,t_1].$$ One gets $$\dot w(t_2)\geq \dot u(t_2)=g(t,u(t_2))>f(t,w(t_2))\geq \dot w(t_2).$$ This contradiction proves that there is no point $t_2$ such that $u(t_2)=w(t_2)$.

Now consider the case $f(t,w)\leq g(t,u)$, if $w\leq u$. Define $$\dot u_n = g(t,u_n)+\ep_n, \q u_n(0)=u_0,\q \ep_n>0,\q n=0,1,...,$$ where $\ep_n$ tends monotonically to zero. Then $$\dot w\leq f(t,w)\leq g(t,u)<g(t,u)+\ep_n, \q w\leq u.$$ By Step 1 $u_n(t)\geq w(t)$, $n=0,1,...\, $. Fix an arbitrary compact set $[0,T_1]$, $0<T_1<T.$ $$\label{un}
u_n(t)=u_0+\il_0^tg(\tau,u_n(\tau))d\tau + \ep_n t.$$ Since $g(t,u)$ is continuous, the sequence $\{u_n\}$ is uniformly bounded and equicontinuous on $[0,T_1]$. Therefore there exists a subsequence $\{u_{n_k}\}$ which converges uniformly to a continuous function $u(t)$. By continuity of $g(t,u)$ we can pass to the limit in (\[un\]) and get $$\label{un1a}
u(t)=u_0+\il_0^tg(\tau,u(\tau))d\tau,\q t\in [0,T_1].$$ Since $T_1$ is arbitrary (\[un1a\]) is equivalent to 