# Real analysis derivate problem. How can I prove this?

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.

• Have you tried anything in order to prove this? Can you think of some relevant facts or theorems? – user61527 Jan 18 '14 at 3:57
• I have used mean value theorem to arrive to the inequality. – guerraufo Jan 18 '14 at 4:15
• T. Bongers, thank you – guerraufo Jan 18 '14 at 4:53

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 $[a,b]$.
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.