If all derivatives of $f$ are uniformly bounded by a common constant and $f(1/n) = 0$ for all $n$, is $f$ identically zero? $$ f: \Re \longrightarrow \Re \ \in C^{\infty} \\
\exists \ L>0: \ \forall x \in \Re, \forall n\in N \\
|f^{(n)} (x)| \le L \\
f(\frac{1}{n})=0 \ \forall n\in N \\
f(x) \equiv 0  $$
Good morning,
Can you give me an help to make this proof?
This is my thought:


*

*$\lim_{n \rightarrow +\infty } \dfrac{1}{n}=0=\theta$

*$f(\theta)=0$

*$f^{(n)} (\theta) =0\\f(x)=\sum_{k=0}^n \dfrac{f^{(k)}(x-\theta)^k}{k!}$ 

*$f$ is analytic, the rest of Taylor's expansion is 0, because the derivatives are bounded,

*$f$ is null.
I have thought this reasoning, but I have difficulties to demonstrate part 3.
Can you help me?
Is there any mistake?
Can I use an easier way to make the proof?
Thanks.
 A: To fill in (iii) use the mean value theorem and induction. For $k=1$, there exists $c_n\in(0,\frac{1}{n})$ for each $n$ such that $f'(c_n)=\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=0$. Thus, $f'(0)=\lim_{n\to\infty} f'(c_n)=0$. You can make the same argument for $f''(0)$ now using the $c_n$ in place of $\frac{1}{n}$ so we can argue by induction.
A: This is how I would elaborate on your steps 3 and 4, avoiding the claim that $f$ is ananlytic and directly using the Lagrange formulas for the Taylor remainder:
Lemma. Let $f\colon\mathbb R\to\mathbb R$ be $C^\infty$ and let $x_n$ be a strictly monotonic sequence with $x_n\to a$ and $f(x_n)=0$ for all $n\in\mathbb N$. Then $f^{(k)}(a)=0$ for all $k\in \mathbb N_0$.
Proof. The claim for $k=0$ follows by continuity of $f$. Assume the claim is true for some $k$ (and arbitryry smooth $f$). By Rolle, for all $n\in\mathbb N$ there exists $x_n'$ between $x_n$ and $x_{n+1}$ such that $f'(x_n')=0$. As $x_n'$ is monotonic and $x_n'\to a$, we conclude from the induction hypothesis (applied to the functon $f'$ and the sequence $x_n'$) that $f^{(k+1)}(a)=(f')^{(k)}(a)=0$. $_\square$
By applying the lemma to our given $f$ and $x_n=\frac1n\to a=0$, we conclude that $f(0)=f'(0)=\ldots = 0$.
Then by Taylor's theorem with the Lagrange form of the remainder
$$\left|f(x)\right|=\left|\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k + \frac{f^{(k+1)}(\xi)}{(k+1)!}x^{k+1}\right|\le  \frac{L}{(k+1)!}|x|^{k+1}.$$
For any $x$, the right hand side converges $\to 0$ as $k\to\infty$. We conclude $f(x)=0$ for all $x$.
A: Here's an idea. Since, $f\left(\frac1n\right)=0$, $f(0)=0$ by continuity. Now, by the Mean Value Theorem, there is a sequence $\{a_n\}_{n=1}^\infty$ such that $a_n\in\left(\frac{1}{n+1},\frac1n\right)$ and $f'(a_n)=0$,for all $n$. So again by continuity of $f'$, $f'(0)=0$. So the argument can be completed by induction.
