I am given the following: Let $f$ be a real function, which itself and all its derivatives at $0$ are $0$.
Assume there exists $b>0$ such that for all real $x$ and all natural $n$: $|f^{n}(x)|\leq n!\cdot b^n$
- Prove that $f$ is the zero function for the interval $(-1/b, 1/b)$.
- Prove that it is the zero functions for all $x$.
Regarding the closed interval:
- I know that as $b$ goes to infinity, the interval closes-in on $0$, and then we were given that $f(0)=0$.
- As $b$ goes to 0, $f$ is bounded to be $0$ by the condition above.
But for an arbitrary $b$, I have no solution. I tried considering the fact that it is enough to find that the condition holds for $f'(x)$, because if $f'(x)=0$ for all $x$ and $f(0)=0$, then it is the zero function. I also tried considering FTC and claim something about $$f'(b)=\int_0^b f''(x) dx.$$ Last thought was that if $f$ is not the zero function, by taylor polynomial there must exist a point which $$f'\left( \frac{1}{b} \right)=f\left( \frac{1}{b} \right) b, \quad f''\left( \frac{1}{b} \right)= f\left( \frac{1}{b} \right) 2b^2,$$ and so forth- $$f^{(n)}\left( \frac{1}{b} \right) = f \left( \frac{1}{b} \right) n! \cdot b^n.$$ It gets me close but not enough, and I'm not even sure it is true.
It seems the following.
Let $x\in (-1/b,1/b)$ be an arbitrary point and $n$ be a natural number. Taylor expansion of $f(x)$ with the remainder in Lagrange form gives
$$\left|f(x)\right|=$$ $$\left|f(0)+\frac{f’(0)}{1!}(x-0)+\frac{f’’(0)}{2!}(x-0)^2+\dots+\frac{f^n(0)}{n!}(x-0)^n+\frac{f^{n+1}(c)}{(n+1)!}(x-0)^{n+1}\right|=$$ $$\left|\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}\right|\le \left|\frac{n!b^n}{(n+1)!}x^{n+1}\right|\le \frac{1}{(n+1)b},$$
(where $c$ is a point from a segment with the ends $0$ and $x$). Taking the limit when $n$ goes to infinity we obtain that $f(x)=0$.
Since the function $f$ is constant zero on the interval $(-1/b,1/b)$, all its derivatives at $x$ are $0$ for any point $x_0\in (-1/b,1/b)$. Applying the already proved results to a function $g(x)=f(x-x_0)$ we obtain that the function $f$ is constant zero on the interval $(x_0-1/b,x_0+1/b)$. Proceeding in such a way, we obtain that the function $f$ is constant zero on the whole real line.