Does $f$ have a second derivative at $0$?

Question

Let $f\colon \mathbb {R}\rightarrow \mathbb{R}, \,f (x)=x^2$, when $x\in \mathbb{Q} $ and $f (x)=x^3$ when $x\in \mathbb {R}\setminus \mathbb{Q}$.

I know that $f $ is continuous at $0$. I know how to prove that $f $ is differentiable at $0$ and $f'(0)=0$ with the method of $ \epsilon $ and $\delta_{\epsilon} $.

Does $f $ have a second derivative at $0$?

Answer 1

Hint. If the second derivative of $f$ at $0$ exists then the following limit exists and it is finite. $$f''(0) = \lim_{h\to 0} \frac{f(2h)-2f(h)+f(0)}{h^2}.$$ Now consider two cases:

1) If $h\in \mathbb{Q}$ with $h\not=0$ then $$\frac{f(2h)-2f(h)+f(0)}{h^2}=\frac{(2h)^2-2(h)^2+0}{h^2}.$$

2) If $h\in \mathbb {R}\setminus \mathbb{Q}$ then $$\frac{f(2h)-2f(h)+f(0)}{h^2}=\frac{(2h)^3-2(h)^3+0}{h^2}.$$

Can you take it from here?