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For an analysis question I have to show that f is the zero-function on $[0,1]$. The inequality $|f(x)| \leq M x_0 sup_{y \in [0, xo]} |f(y)|$ is provided as well as the hint that M should be chosen such that $x_0 < \frac{1}{M}$. Doing this, $|f(x)| < sup_{y \in [0, \frac{1}{M})} |f(y)|$ rolls out of the inequality but I'm not sure how it follows from here that $|f(x)| = 0$ $\forall x\in [0,1]$. Can someone help me out?

penguin131415
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  • Welcome to Math.SE! Can you give some extra context: for example, what is $x_0$ and what is the relation between $x$ and $x_0$? – Hrodelbert Jun 04 '15 at 10:11
  • Ah sorry forgot to mention that. $x \leq x_0 \in [0,1]$. Also $f(0) = 0$, $|f'(x) \leq M|f(x)|$ for $M > 0$ and $f$ is differentiable on $[0,1]$, but I used that earlier to show the inequality and I didn't think that was useful information anymore. – penguin131415 Jun 04 '15 at 10:12
  • Can anyone help me? – penguin131415 Jun 04 '15 at 10:41
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    Please put all the information you mentioned above in your question. You can edit your question using the edit button underneath your question. – Hrodelbert Jun 04 '15 at 11:29
  • Is $x_0$ fixed? What is going on with the $x$ and the $x_0$ and the $M$ and the $y$... Which ones are fixed and which are varying? – Eoin Jun 05 '15 at 03:33

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Note the inequality $|f(x)|< \sup\limits_{y\in [0, \frac{1}{M})} |f(y)|$ does not depend on $x$. If $M$ can be made arbitrarily large, then we would have $|f(x)|\leq 0$ and the result would follow.

It is difficult to claim that this is valid however without more information (see my comment above).

Eoin
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