I got stuck in this problem and would appreciate some useful hints without ruining me the work to do. First, assume that we have a function $f$ such that $|f(x)|$ $\le$ $x^2$. The exercise has two parts:
$i$) prove that $f$ is differentiable at $0$.
$ii$)generalize this result by replacing $x^2$ with $|g(x)|$ and adding the necessary properties that $g$ must have.
First, I think I have solved the first part correctly, giving the following answer:
By hypothesis, we have that $-0^2 \le f(0) \le 0^2$ which means that $f(0) = 0$. Also, since $-x^2 \le f(x) \le x^2$ for all x and both $-x^2$ and $x^2$ approach zero near zero then, $f$ is continuous at $0$. Moreover, $|f(0+h)/h| = |f(h)|/|h| \le h^2/|h| = |h|$. Therefore, for every number $\epsilon \gt 0$ if $0 \lt |h| \lt \epsilon$ it follows that $|f(h)/h - 0| \le |h| \lt \epsilon$. This means that $f$ is differentiable at $0$ and $f'(0) = 0$.
Please let me know if this is a correct proof. Next, for solving ii) I have had difficulty solving it. I feel like the necessary conditions are that $g$ is continuous at $0$ with $g(0) = 0$. $g(0) = 0$ allows me to ensure that $f(0) = 0$.This, combined with $f$ being "trapped" between $|g|$ and $-|g|$ with both of these tending to $0$ near $0$(by property of limits with absolute values) is enough to say via Squeeze Theorem that $f$ is continuous at $0$ with $f(0) = 0$.I required g to approach 0 near zero since this felt like the only way in which I can guarantee that $f$ will be continuous at $0$, which is a requisite for being differentiable. However, from here I dont see how to conclude anything else about the limit of $[f(0+h)+f(0)]/h = f(h)/h$ using those assumptions, differently from how I succesfully did in the part $i)$ of the exercise. At this moment I have only that $|f(h)|/|h| \le |g(h)|/|h|$. Perhaps I could say that $g$ is differentiable at $0$ and then be able to say something else about the limit definiton for $f'(0)$?