If $-x^2 \leq f(x) \leq x^2$ for all $x \in R$, prove that f is differentiable.

Question

If $-x^2 \leq f(x) \leq x^2$ for all $x \in R$ prove that f is differentiable.

I have found this link here for differentiability at 0, but what about the rest of the proof? Any ideas will be appreciated.

EDIT:

Here is the link:

Prove that if $|f(x)| \leq x^2$, then the function is continuous and differentiable at $x=0$.

It's not true outside of 0. Make a piecewise linear function which is y=0 around 0. — Matthew C, Nov 01 '18 at 03:01
I apologize for that comment about being a duplicate. I had two windows open at once and pasted it into the wrong window. — John Douma, Nov 01 '18 at 03:14

Mohammad Riazi-Kermani · Answer 1 · 2018-11-01T17:20:32.983

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It is only true at $x=0$

You can make a zig zag between $x^2$ and $-x^2$ Which is not differentiable.

edited Nov 01 '18 at 17:20

answered Nov 01 '18 at 03:13

Mohammad Riazi-Kermani

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score 1 · Answer 2 · answered Nov 01 '18 at 03:17

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It doesn't even have to be continuous.

Consider $$f(x) = \begin{cases} 0, & x\lt 2 \\ 1, & x\ge 2 \end{cases}$$

answered Nov 01 '18 at 03:17

John Douma

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score 1 · Accepted Answer · answered Nov 01 '18 at 07:06

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With just the given condition that $-x^2 \le f(x) \le x^2$ for all $x \in \mathbb{R}$, it is even possible that $f$ is discontinuous everywhere except $x = 0$, and thus, not differentiable anywhere except $x = 0$. Consider the function $$f(x) = \begin{cases}x^2 & \text{if} \ x \in \mathbb{Q} \\ 0 & \text{if} \ x \in \mathbb{R}\setminus\mathbb{Q}\end{cases}.$$

answered Nov 01 '18 at 07:06

JimmyK4542

54,331

okay great..... do I have to assume that $x \neq 0$ to continue the proof as that given in the link above? – Idonotknow Nov 01 '18 at 07:16
I am wondering from where we get the line before last in the link mentioned above? – Idonotknow Nov 01 '18 at 07:22

If $-x^2 \leq f(x) \leq x^2$ for all $x \in R$, prove that f is differentiable.

3 Answers3