Prove that f is differentiable at x=0

Question

Suppose $0 < f(x) < x^2, \ \forall x$. Prove that $f$ is differentiable at $x=0$ and $f'(x)=0$.

I don't even know how to start with the first one. For the second question, I'm guessing we can use squeeze theorem to show that the limit must also go to $0$ as the limit of the bounds goes to 0.

Much appreciated!

That is not possible. At $x=0$ we would have $0<0$, a contradiction. — Gustavo, Mar 24 '16 at 04:11
Wouldn't $f(x)=x^2\times\sin\left(\frac{1}{x}\right)^2$ (with $f(0)=0$ added) not be differentiable? Reference: http://math.stackexchange.com/questions/514809/is-there-a-tangent-of-x-sin1-x-at-x-0 — Ian Miller, Mar 24 '16 at 04:30
@IanMiller Unless I am mistaken, using first principles and simplifying the limit, we get $ f'(0) = \lim_{h \to 0} \sin^2(1/h) \cdot h$ which should exist by squeezing. — MathematicsStudent1122, Mar 24 '16 at 04:47
Oh right! Hmm $$f(x)=\begin{cases} x^2 & x\in\mathbb{Q} \ 0 & x\notin\mathbb{Q} \end{cases}$$ ? — Ian Miller, Mar 24 '16 at 04:55
@IanMiller The limit (of the difference quotient) should exist for that as well.
It ends up being something like $$g(h) = \begin{cases} h,&\text{if }h\in\Bbb Q\ 0,&\text{if }h\in\Bbb R\setminus\Bbb Q;. \end{cases}$$

And you take the limit as $h \to 0$. — MathematicsStudent1122, Mar 24 '16 at 05:03
@IanMiller Though it does blow my mind how these monsters are differentiable.... — MathematicsStudent1122, Mar 24 '16 at 05:07

score 1 · Answer 1 · answered Mar 24 '16 at 04:23

I take it you mean $0 \leq f(x)$. It is easy to see that $f(0) = 0$.

For $h\neq 0$, we have $0 \leq f(h) \leq h^2$.

If $$h>0 \Longrightarrow 0 \leq \frac{f(h)}{h} \leq h$$ $$h<0 \Longrightarrow h \leq \frac{f(h)}{h} \leq 0$$

Observe that these are the difference quotients for the three functions at $x=0$. Now squeeze.

score -1 · Answer 2 · answered Mar 24 '16 at 04:21

Once we have $ \ f: \mathbb{R} \to \mathbb{R} \ $ such that $ \ 0 \leq f(x) \leq x^2$, $\forall x \in \mathbb{R}$, is straightforward that $ \ f(0)=0$. Then, $\forall t \in \mathbb{R} \setminus \{ 0 \}$, we have that $$\displaystyle 0 \leq \frac{f(t)}{t} \leq \frac{t^2}{t} = t$$ By the Sandwich theorem, we conclude that $$\displaystyle f'(0) = \lim_{t \to 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \to 0} \frac{f(t)}{t} = 0$$

Prove that f is differentiable at x=0

2 Answers2