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Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $, $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.

$\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k>0$ and repeating the process $\vert f(x)\vert \leq K^n\vert f(z)\vert$ for $a<z<y<x$. If $K<1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea.

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    $\begingroup$ Have you tried anything in order to prove this? Can you think of some relevant facts or theorems? $\endgroup$ – user61527 Jan 18 '14 at 3:57
  • $\begingroup$ I have used mean value theorem to arrive to the inequality. $\endgroup$ – guerraufo Jan 18 '14 at 4:15
  • $\begingroup$ T. Bongers, thank you $\endgroup$ – guerraufo Jan 18 '14 at 4:53
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Consider the differentiable function $g$ defined on $[a,b]$ by $$\color{red}{g(x)=\mathrm e^{-2Mx}f^2(x)}.$$ For every $x$ in $[a,b]$, $$\color{red}{g'(x)}=2\mathrm e^{-2Mx}(f'(x)f(x)-Mf^2(x))\leqslant2\mathrm e^{-2Mx}(|f'(x)|-M\,|f(x)|)\,|f(x)|\color{red}{\leqslant0}.$$ Thus, the function $g$ is nonincreasing on $[a,b]$, that is, $g(x)\leqslant g(a)=0$. Since $g\geqslant0$ by definition, this proves that $g=0$ identically on $[a,b]$ hence $f=0$ identically on $[a,b]$.

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HINT: Remember that f differentiable in $\left[a,b\right]$ then it is continous in $\left[a,b\right]$.. and from there you use one of the big results such as mean value theorem for example.

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  • $\begingroup$ Not sure I see how MVT yields the result. You might want to be more specific. $\endgroup$ – Did Jan 18 '14 at 11:01
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We prove the claim by contradiction. Assume there is x(0) in (a, b] with f(x(0)) not equal to 0. We may assume that f(x(0)) > 0. Let A = { x : x in [a, x(0)] and f(x) = 0}. Since f(a) = 0, A is not empty. Further, it is bounded above so sup(A) exists. Let x(1) = sup(A). Thus there is a squence of numbers in A which converges to x(1). Since f is continuous, Limf(x(n)) = f(x(1)). And this means f(x(1)) = 0 since f(x(n)) = 0 for all n's. This shows x(1) < x(0). We also show that for all x's in (x(1), x(0)] we have that f(x) > 0. For if there is y in (x(1), x(0)] with f(y) < 0 or f(y) = 0, then if f(y) = 0, then y is in A and so y < x(1) or y = x(1), but y > x(1) a contradiction. If f(y) < 0, and coupled with f(x(0)) > 0 by Intermediate Value Theorem there is a z in (y, x(0)) such that f(z) = 0. so z is in A and then z < x(1), but z > y > x(1) meaning z > x(1) a contradiction again. Thus f(x) > 0 for all x's in (x(1), x(0)]. Next consider g(x) = ln(f(x)) on (x(1), x(0)]. g is differentiable on this interval. Let t be any number in (x(1), x(0)], and apply the mean value theorem for g on [t, x(0)] we have: /g(t) - g(x(0))/ = /g'(c)//t - x(0)/ < M/t - x(0)/ < M*/x(1) - x(0)/. Now let t--> x(1)+ we have /g(t) - g(x(0))/ ---> + infinity. A contradiction since it is bounded above by M*/x(1) - x(0)/.

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