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 $x=0$.

Answer 1

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 $$f(x) = \begin{cases}x^2 & \text{if} \ x \in \mathbb{Q} \\ 0 & \text{if} \ x \in \mathbb{R}\setminus\mathbb{Q}\end{cases}.$$

Answer 2

It is only true at $x=0$

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

Answer 3

It doesn't even have to be continuous.

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