# Proving that for a smooth function if $f(\frac 1 k)=0 :\forall k\in \mathbb N$ then $f(x)=0 :\forall x\in[-1,1]$

Let $$f\in C^{\infty} ([-1,1])$$ and suppose there's a constant $$M>0$$ such that $$|f^{(j)}(x)|\le M:\forall j\in\mathbb N$$ (including zero) and for all $$x\in [-1,1]$$.

Prove that if $$f(\frac 1 k)=0 :\forall k\in \mathbb N$$ then $$f(x)=0 :\forall x\in [-1,1]$$

I understand that the function is always smooth and Lipschitz, so it's always uniformly continuous too. So it can't be that $$f(\frac 1 k)=0$$ and for the closest irrational number say $$y$$ we will have $$f(y)\neq0$$ because that will contradict uniform continuity. It's a bit similar to Dirichlet function.

If I'm right then it can be proved without using the fact that the function is smooth and I should be able to find for $$f^{(0)}$$ a contradiction which should be enough but it doesn't feel right if I'm not using parts of the question. Also, I'm not sure on how to formally express the counter example, should i use the $$\epsilon - \delta$$ ? Heine with two sequences ?

• I don't think this can be done. There are many smooth functions with compact support, what's to keep you from inserting one of these functions between $x=1/2$ and $x=1$? Commented Jun 2, 2014 at 15:06
• @PA6OTA I don't know what compact support means but how can a function be uniformly continuous and have 'jumps' ? Commented Jun 2, 2014 at 15:13
• What about $$f(x) = e^{-1/x^2}\sin\frac{\pi}{x}$$ ? Commented Jun 2, 2014 at 18:54
• Oh, I see. The derivatives are not bounded by the same constant. Ok. Commented Jun 2, 2014 at 18:56
• @JackD'Aurizio they are bound by the same $M$. Commented Jun 2, 2014 at 19:03

Let $$R_n (x,0) =f(x)-\left( f(0) +\sum_{k=1}^n \frac{f^{(k)} (0) x^k}{k!} \right) ,$$ then we have $$|R_n (x,0)|\leq \frac{M}{(n+1)!}\longrightarrow 0$$ hence $f$ is analytic and therefore $f\equiv 0.$

• The reminder is bounded by that because each derivative is bounded by $M$, I see. I'm not familiar with analytic functions, does it mean that there's no error in the Taylor expansion ? Why does that means that $f=0$ ? Commented Jun 2, 2014 at 18:45
• Yes, since $R_n (x,0)\to 0$ we have $$f(x)=\sum_{k=0}^{\infty} \frac{f^{(k)} (0) x^k }{k!}$$ on $[-1,1]$ and hence $f$ is analytic. But zeros of analytic function $f\neq 0$ are isolated.
– user110661
Commented Jun 2, 2014 at 19:03
• What do you mean by the zeros of an analytic function are isolated ? Commented Jun 2, 2014 at 19:09
• – user110661
Commented Jun 2, 2014 at 19:16