4

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.

psmears
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jggarita
  • 459

3 Answers3

7

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 $H(a)=0$.

Julián Aguirre
  • 76,354
5

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 $x \in [a, b]$.

Robert Lewis
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4

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 $[a,b]$.

r9m
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  • Just one question! What assures me I can take x1 that close to a?? – econ_ugrad Jul 10 '15 at 03:16
  • @JuanManuelCastro $x\in [a,b]$. – Mark Viola Jul 10 '15 at 03:19
  • What I meant, is how may I extend this result in the whole interval!? – econ_ugrad Jul 10 '15 at 03:21
  • @JuanManuelCastro $a,c$ are fixed, that gives us an interval $[a,x_1]$ (with $x_1 = a+\frac{1}{2c} < a+\frac{1}{c}$) of positive $\frac{1}{2c}$ length where, $h(x)$ vanishes. Next we take the interval $[x_1,b]$ and choose $x_2$ in it similarly (so that $x_2 - x_1 = \frac{1}{2c}$) and eventually cover $[a,b]$ with subintervals of length $1/{2c}$ where $h(x)$ vanishes. – r9m Jul 10 '15 at 03:33