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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 $M>0$ 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?

Lt. Commander. Data
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    Check this: https://math.stackexchange.com/q/642288/42969. Note that you can assume that $f(a)=0$. – Martin R Mar 12 '21 at 08:19
  • @KaviRamaMurthy: I think it is correct. – Other Q&As about the same topic: https://math.stackexchange.com/q/3303166, https://math.stackexchange.com/q/2046696, https://math.stackexchange.com/q/2780193, and probably many more. – Martin R Mar 12 '21 at 09:04

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