# Let f be a function such that $|f(x)| \leq x^2$ for all x. Prove f is differentiable at 0. [duplicate]

This question is straight from Spivak Chapter 9, Question 15. I have attempted the previous problem, which is another differentiable function I had to prove.

I want to attempt this problem using the definition of a derivative as a limit. The issue in this problem for me is the $$\leq$$ sign. How do you work with this restriction? Any hints are appreciated.

Surely I can take $$f$$ to be $$x^2$$ since this satisfies the condition that $$|f(x)| \leq x^2$$? Is this sufficient to prove the condition?

Proof. Let a function $$f$$ be defined as $$|f(x)| < x^2$$. Since $$x^2$$ is such a function, we know that the derivative of $$x^2 = 2a$$, and since $$x^2$$ is differentiable at 0, the (half-assed) proof is complete.

I have skipped many steps in the proof. I merely asserted that $$x^2$$ is differentiable at 0 and meets the restriction specified.

• Shouldn't that be $|f(x)| \le x^2$? Otherwise, you get $|f(x)|<0$. – HiMatt Nov 21 '19 at 1:34
• The inequality does not define anything. You have been asked to prove that all the functions that exist (be them defined by an explicit formula or otherwise) and which satisfy that inequality for all $x$ also satisfy the property of being differentiable at $0$. – Gae. S. Nov 21 '19 at 1:40
• Or equivalently (but highly not recommended) that all functions fall into at least one of these categories: (a) those which are differentiable at $0$; (b) those for which there is some $c\in\Bbb R$ such that $\lvert f(c)\rvert>c^2$. – Gae. S. Nov 21 '19 at 1:43

You cannot take $$f$$ to be $$x^2$$. You can't assume anything more than what you're given, which is that $$\vert f(x) \vert \leq x^2$$. This does not imply that $$f$$ is actually equal to $$x^2$$. Instead, just work directly with the definition of the derivative as the limit of the difference quotient. Since $$\vert f(x) \vert \leq x^2$$, in particular $$f(0) = 0$$. Therefore for $$h \neq 0$$ we have $$\frac{f(0 + h) - f(0)}{h} = \frac{f(h)}{h}.$$ Since $$\vert f(h) \vert \leq h^2$$, we have $$0 \leq \left\vert \frac{f(h)}{h} \right\vert \leq \left\vert \frac{h^2}{h} \right\vert = \vert h \vert.$$ Since the left and right-hand sides both go to $$0$$ as $$h \to 0$$, this implies that $$\lim_{h \to 0} \left\vert \frac{f(h)}{h} \right\vert = 0 \iff \lim_{h \to 0} \frac{f(h)}{h} = 0,$$ so $$f$$ is differentiable at $$0$$ and $$f'(0) = 0$$.
Hint. Draw a picture showing the graph of $$x^2$$ and look at where the graph of $$f$$ must be compared to that.
That should tell you what value $$c$$ the derivative of $$f$$ must have at $$0$$. Then try to show why the limit defining the derivative at $$0$$ is in fact $$c$$.