Twice-differentiable function such that $f(\frac{1}{n})=0$ for every positive integer $n$.

Question

Q. Let $f :\mathbb R → \mathbb R$ be a twice-differentiable function such that $f(\frac{1}{n})=0$ for every positive integer $n$. Then

1. $f(0)=0$

2. $f'(0)=0$

3. $f''(0)=0$

4. $f$ is a non zero polynomial

We can say 4 is not correct by taking $f(x)=x^4 \sin \frac{1}{x}$.

How to prove the rest of them?

Answer 1

For a function in $\mathcal C^2(\mathbf R)$, including at $0$, you can take the function $$f(x)=\begin{cases}x^5\sin\frac\pi x&\text{ if }x\ne 0,\\ 0&\text{ if }x=0. \end{cases}$$ (With a smaller integer exponent for $x$, it is not twice continuously differentiable at $0$.)

Answer 2

$\forall n\in\Bbb N \,\exists x_n\in (1/(n+1),\,1/n): (\,f'(x_n)=\frac {f(1/n)-f(1/(n+1))}{1/n-1/(n+1)}=0\,).$

Now $f'$ is continuous because $f''$ exists, so $f'(0)=\lim_{n\to\infty}f'(x_n)=0.$

Now $f'(0)=0$, so $f''(0)=\lim_{n\to\infty}\frac {f'(x_n)-f'(0)}{x_n-0}=0.$

And $f$ is continuous because $f'$ exists, so $f(0)=\lim_{n\to\infty \,:n\in \Bbb N}f(1/n)=0.$