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Suppose that we have a function $f$ that $f(0)=0$ and for any $x$,

$$|f(x)|>=\sqrt{|x|}$$

How could we prove that $f(x)$ is differentiable at zero or not?

P. S. My first attempt was to prove something like a discontinuity for $f$ at zero that not being differentiable would be quickly implied from it, but haven’t reached to anything interesting. However, any solution is acceptable.

Jigsaw
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    You can't: $f(x)=x^\alpha$ satisfies this inequality if $\alpha\ge \frac12$, yet it is differentiable at $0$ if and only if $\alpha\ge 1$. – Bernard Nov 15 '19 at 09:49

2 Answers2

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Hint : \begin{align*} \left| f(x) \right| \geq \sqrt{\left|x\right|} &\Leftrightarrow \left| f(x)\right| - \left| f(0)\right| \geq \sqrt{\left| x\right|} - \left|f(0)\right| \\ &\Leftrightarrow \frac{\left|f(x)\right| - \left|f(0)\right|}{\left|x\right|-0} \geq \frac{\sqrt{\left|x\right|} - \left|f(0)\right|}{\left|x\right|-0} \\ &= \frac{\left|f(x)\right|}{\left|x\right|} \geq \frac{\sqrt{\left|x\right|}}{\left|x\right|} \\ &= \frac{\left|f(x)\right|}{\left|x\right|} \geq \frac{1}{\sqrt{\left|x\right|}} \end{align*}

Rebellos
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Suppose $f'(0)$ exists. Call $l$. There exists $\delta$ such that $|\frac {f(x)-f(0)} x-l|<\epsilon$ for $|x| <\delta$. Hence $\frac {|f(x)|} {|x|} <|l|+\epsilon$ for $|x| <\delta$. This implies $1 <\sqrt |x| (|l|+\epsilon)$ for $|x| <\delta$. You get a contradiction by letting $x \to 0$

Kavi Rama Murthy
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