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Let $f:(-a,a) \longrightarrow R$, $a>0$.

Such that $$ |f(x)|≤x^2 $$

What I did was taking out the module bars so I get $-x^2≤f(x)≤x^2$ and I see that at $x=0$ the function must be zero.

I see why f'(0) must equal zero at that point, but I have no idea on how to prove it.

YoTengoUnLCD
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1 Answers1

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The inequality implies $f(0) = 0$. Further, $\left|\frac{f(x)}{x}\right| \le |x|$ for all $x\neq 0$. Taking the limit as $x\to 0$, we obtain $|f'(0)| \le 0$. Thus $f'(0) = 0$.

kobe
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    Why assume that $f'(0)$ exists? Doesn't that follow from the estimate $\left|\frac{f(x)}{x}\right| \le |x|$ ? – Martin R Apr 04 '15 at 14:49
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    The existence of $f'(0)$ is ensured by that same inequality. – Siminore Apr 04 '15 at 14:50
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    Indeed. In fact one deduces first that $\lim_{x\to0}|\frac{f(x)}x|$ exists and equals $0$; therefore $\lim_{x\to0}\frac{f(x)}x$ exists and equals $0$. – Bart Michels Apr 04 '15 at 14:57
  • How do you obtain $|f'(0)|\leq 0$ with that limit? – YoTengoUnLCD Apr 04 '15 at 15:54
  • @YoTengoUnLCD from $\lim_{x\to 0} \left|\frac{f(x)}{x}\right| \le \lim_{x\to 0} |x|$. The left hand side is $|f'(0)|$, and the right hand side is $0$. – kobe Apr 04 '15 at 15:56