Inequality using integrals and absolute values Let $u,v$ be continous functions in $[a,b]$ a compact interval and let $c > 0$. Suppose that $\forall x\in [a,b]$, the following inequality is true:
$$|u(x)-v(x)|\leq c\int^x_a|u(t)-v(t)|dt$$
Prove that $u(x) = v(x)$ $\forall x\in [a,b]$.
Ok so I have no idea for where to break into this problem. My first guess is to assume there is some point $x_0$ such that $u(x_0) > v(x_0)$ (w.l.o.g), from there I have the inequality:
$$|u(x_0)-v(x_0)|\leq c\int^{x_0}_a|u(t)-v(t)|dt$$
But what about that? I can't assume anything about the sign of the integral in the interval $[a,x_0]$, even if I could do that, I know the left part of the inequality would be $\frac{1}{c}$ times the derivative of the right part, but I'm not sure about what information I can get from there either.
 A: 
We going to prove that $|u(x)|\leq c\int_{0}^b |u(t)|dt\Rightarrow u\equiv 0$. For $c=1$.

Define for $x\in[0,b]$, $V(x)=\max_{0\leq t\leq x}u(t)$. Suppose by contradiction $u(x)>0$ for $0<x\leq b$. We have that $u(t)\leq V(t)$, for all $t\in [0,b]$. Fix $t\in[0,b]$, since $u$ is continuous, there is $z\in [0,t]$ such that $u(z)=\max_{0\leq s\leq t}u(s)=V(t)$. Therefore,
$$V(t)=u(z)\leq\int_{0}^z |u(s)|ds\leq\int_{0}^t |V(s)|ds$$
Let $\overline{V}(x)=\int_{0}^x |V(t)|dt$ for $x\in [0,b]$. Clearly $\overline{V}$ is continuous and:


*

*$\overline{V}(0)=0$.

*$V(x)\leq \overline{V}(x)$.

*$\overline{V}'(x)=|V(x)|\leq |\overline{V}(x)|$. 


Let $\delta\in (0,b)$ then $$b\geq b-\delta=\int_{\delta}^b 1dx\geq \int_{\delta}^b \frac{\overline{V}'(x)}{|\overline{V}(x)|}dx=\int_{\overline{V}(\delta)}^{\overline{V}(b)}\frac{1}{|z|}dz=\int_{\varepsilon}^c\frac{1}{|z|}dz$$ for some $c\geq \varepsilon>0$. If we do $\delta\to 0^+$ then $\overline{V}(\delta)=\varepsilon\to 0^+$.
So we have 
$$b\geq \lim_{\varepsilon\to 0+}\int_{\varepsilon}^c\frac{1}{|z|}dz=\infty$$
Which is absurd, therefore $u\equiv 0$. 
