I have been studying real analysis, and I am on the section of differentiation. I am doing some problems to check understanding. I found this problem and attempted to solve it. My solution seems simple, and I am afraid that I am not understanding the concept.

Problem: Suppose that $f:\mathbb{R} \to \mathbb{R}$ satisfies $0\leq f(x) \leq x^2$ for every $x\in\mathbb{R}$. Prove that $f$ is differentiable at $x=0$.

My solution: We want to show that the limit of $\frac{f(x)-f(0)}{x-0}$ as $x$ goes to zero exists. Hence we have that

$$\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}$$ (Can we conclude that f(0)=0? If so, why? Assuming that we can, I proceeded in the following way.)

But note that for $x \neq 0$ $$0\leq\frac{f(x)}{x}\leq \frac{x^2}{x}=x.$$ Hence by taking limits we obtain

$$0\leq \frac{f(x)}{x} \leq 0.$$ In particular, this implies that

$$\lim_{x\to 0} \frac{f(x)}{x} =0.$$ Thus f is differentiable at $x=0$.

  • $\begingroup$ You can conclude $f(0) = 0$ because $0 \leq f(0) \leq (0)^2 = 0$. You have the essential idea, but note that dividing both sides of the inequality by $x$ does not necessarily preserve the inequality --- what if $x<0$? $\endgroup$ – AlohaSine May 1 '16 at 3:16
  • $\begingroup$ Didn't the x come from the limit definition ? I am using the fact that $0\leq f(x)\leq x^2$. I am not dividing by negative x, so the inequality doesn't change, right? $\endgroup$ – An P. May 1 '16 at 3:26
  • $\begingroup$ You are not dividing by negative $x$, but you are dividing by a number less than $0$ (if you're taking the limit from the left). This reverses the inequality. This is not a huge problem. Simply squeeze $\min(0,x) \leq \frac{f(x)}{x} \leq\max{(x,0)}$ as $x \to 0$. $\endgroup$ – AlohaSine May 1 '16 at 3:36
  • $\begingroup$ thanks! I think i got it now. $\endgroup$ – An P. May 1 '16 at 3:42

$0 \leq \dfrac{f(x)}{x} \leq \dfrac{x^2}{x} = x$ only works for $x > 0$. You also need to consider $x < 0$. Fortunately, the definition of limit lets you not care about the value of $f(x)$ at $x = 0$.

If $x < 0$, then $0 \le f(x) \le x^2$ implies $x \le \dfrac{f(x)}{x} \le 0$

So, for $x \ne 0$, you can say that $0 \le \left| \dfrac{f(x)}{x} \right| \le |x|$

The rest of your argument should proceed the same.


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