Prove that $f'(x)=0$ if $|f'(x)|\leq M|f(x)-f(a)|$ where $M>0$ and $f:[a,b]\rightarrow\mathbb{R}$ is continuous I came across this question in my Real Analysis class:  Prove that $f'(x)=0$ if $|f'(x)|\leq M|f(x)-f(a)|$ where $M&gt;0$ and $f:[a,b]\rightarrow\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$.  The immediate first step is to use the mean value theorem which would simplify the expression to: $$|f'(x)|\leq M(x-a)|f'(c)|$$ Where $c\in[a,x]$. However, I'm not sure where to go on from here, but this is an idea that I have in mind:  We consider only $[a,\alpha]$ where $\alpha-a=\frac{1}{M}$. Take $x$ such that $f'(x) = \sup |f'|$ on the interval. Then $\exists c$ such that $|f'(x)|\leq|f'(c)|$. This is a contradiction?  These are my half-baked ideas, but in general I'm a bit stumped as to how to proceed. Any hints?
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 $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea.
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