Inequality of continuous functions Let $u, v$ be continuous functions on $[a,b]$ and let $c>0$. Suppose that for all $x \in [a,b]$  we have the following inequality:
$$|u(x)-v(x)| \leq c \int_a^x |u(t)-v(t)| dt$$
Show that $u(x)=v(x)$ for all $x \in [a,b]$
My first thought was to consider $h(x)=|u(x)-v(x)|$ and try to show that $h=0$, but I got stuck. Also, I proved the inequality considering the case $c(b-a) \leq 1$, but I'm not sure how to continue.
 A: Let $H(x)=\int_a^xh(t)\,dt$. Then $H'=h$, $H'\le c\,H$ and $H(a)=0$.
$$
(e^{-ct}H)'=e^{-ct}(H'-c\,H)\le0.
$$
$e^{-ct}H$ is positive, decreasing and $H(a)=0$.
A: An integral-equation flavored approach:
Set
$g(x) = \vert u(x) - v(x) \vert; \tag{1}$
then the given inequality may be written
$g(x) \le c\int_a^x g(t) dt = \int_a^x cg(t) dt; \tag{2}$
now apply the integral form of Gronwall's inequality, which, as far as the present purposes are concerned, may be taken to state that continuous $w(x)$ satisfying
$w(x) \le b + \int_a^x cw(t) dt, \tag{3}$
$b$ constant, also satisfy
$w(x) \le be^{\int_a^x cdt} = be^{c(x - a)}; \tag{4}$
note $g(x) \ge 0$ and (4) shows, since $b = 0$,
$g(x) \le 0.  \tag{5}$
We conclude that
$g(x) = 0, \tag{6}$
i.e.,
$u(x)  = v(x) \tag{7}$
for $x \in [a, b]$.
A: Fix an $x_1 \in [a,b]$, then the function $h(x) = u(x) - v(x)$, satisfies,
$$|h(x)| \le c\int_a^x |h(x)|\,dx$$
Consider, $\displaystyle \sup\limits_{x \in [a,x_1]} |h(x)| = M_{x_1}$
Then, $$M_{x_1} = \sup\limits_{x \in [a,x_1]} |h(x)| \le c\sup\limits_{x \in [a,x_1]} \int_a^x |h(x)|\,dx \le M_{x_1}c(x_1-a)$$
Hence, if we choose an $x_1$ such that, $x_1-a < 1/c$ it must imply $M_{x_1} = 0$, thus $h(x) = 0$ in the interval $[a,x_1]$. This gives us the same scenario as in $[x_1,b]$ and we may conclude that $h(x) = 0$ on $[a,b]$.
