If $f: \mathbb{R}^{n} \to \mathbb{R}$ is such that $|f(x)|\leq x^{3}$, is $f$ second-differentiable in $0$?

Question

Let $f: \mathbb{R}^{n} \to \mathbb{R}$ be a function such that $|f(x)|\leq x^{3}$. Is $f$ second-differentiable in $0$?

I dont know how to start this question. The trouble is: if $x \in \mathbb{R}^{n}$, $x^{3}$ doesnt make sense. Should not be $f: \mathbb{R} \to \mathbb{R}$?

Answer 1

The map

$$f(x) = \begin{cases} x^3 \sin(1/x) & \mbox{for} & x\neq 0\\ 0 & \mbox{for} & x= 0 \end{cases}$$ is such that $\vert f(x) \vert \le \vert x \vert^3$ for all $x \in \mathbb R$. However, it is not twice differentiable at $0$ as

$$f^\prime(x) = \begin{cases} 3x^2 \sin(1/x)-x\cos(1/x) & \mbox{for} & x\neq 0\\ 0 & \mbox{for} & x= 0 \end{cases}$$

and $\frac{f^\prime(x) - f^\prime(0)}{x}$ doesn't have a limit as $x \to 0$.

Answer 2

To be twice differentiable at $0$, $f'$ must be defined in a neighborhood of $0$. The function $$f(x) = \left\{ \begin{array}{ll} x^3 & x \text{ rational} \\ 0 & x \text{ irrational} \end{array} \right.$$ satisfies the condition on the line, but is clearly not even continuous at any point other than $x = 0$.