prove that exist $c$ at $[a,b]$ ,so that for every $\varepsilon>0$ there is $x$ that $0<|x-c|<\varepsilon$ and $f(x)=0$

Question

Let $f(x)$ Differentiable at $[a,b]$ and have infinity numbers of zero at $[a,b]$.

• prove that exist $c$ at $[a,b]$ ,so that for every $\varepsilon>0$ there is $x$ that $0<|x-c|<\varepsilon$ and $f(x)=0$

• $f^{(i)}(c)=0$ for every $0\le i$

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my answer for the first one :

Let $x_n$ sequence with different elements at $[a,b]$ that $f(x_n)=0$.

then i tried to use Bolzano–Weierstrass theorem .

so we get that for $x_n$ there is convergent subsequence $a_k\le x_{n_k} \le b_k$ , that convergent to $c$ when $k\to \infty$.

is that enough to prove that first one ? if not how to continue?

for the second :

i tried to use Rolle's theorem $f$ Derivative at $[a,b]$ and $f(a)=f(b)=0$ then there is $c$ such that $f'(c)=0$ but how to prove it when $f^{(i)}(c)=0$ ?

thanks a lot .

Answer 1

Your proof is good to me for the first point.

(Watch out that the $c$ in the second point is the same in the first one)

For the second point (assuming $f$ is smooth), it is a simple consequence of Rolle's theorem that if $f$ has $n$ zeros $x_1, \dots, x_n$, then $f'$ has $n-1$ zeros $x'_1, \dots ,x'_{n-1}$ (with $x_1\leq x'_1 \leq\cdots \leq x'_{n-1} \leq x_n$), then $f''$ has $n-2$ zeros (between $x'_1, \dots ,x'_{n-1}$), and so on...

So, with this remark in head it is quite straightforward that (for a fixed $i\geq 0$), using the first point, $$ \forall \varepsilon'>0, \quad \exists x \ \text{such that} \ 0<|x-c|<\varepsilon' \ \text{and} \ f^{(i)}(x)=0. $$

So, you can make a sequel $(y_n)$ such that $y_n \rightarrow c$ and $f^{(i)}(y_n)=0$ ; but since $f^{(i)}$ is continuous, $$0=f^{(i)}(y_n) \rightarrow f^{(i)}(c)$$ so $$f^{(i)}(c)=0.$$