Prove that $f$ is bounded when higher derivative is bounded

Let $$f:(-2019,2020) \rightarrow \mathbb R$$ is $$1000$$ times differentiable and $$f^{(1000)}$$ is bounded then also $$f$$ is bounded. If $$f:\mathbb R \rightarrow \mathbb R$$ it is also right?

My try:

$$f^{(1000)}$$ bounded $$\Rightarrow$$ $$f^{(999)}$$ has Lipschitz continuity $$\Rightarrow$$ $$|\frac{f(x)-f(y)}{x-y}|\le L$$

Then I tried to use this for $$f^{(998)}, f^{(997)},\dots,f^{(0)}=f$$. However I don't know how I can do it.

Can you help me?

• Just prove that if $f$ is differentiable then if $f'$ is bounded, then $f$ is bounded. Now, by induction, if $f$ is $n$ times differentiable, then $f^{(n)}$ is bounded implies $f$ is bounded – Jakobian Apr 17 at 17:12
• Now, if we have function from $R$ to $R$, then for example for $f(x) = x$, $f'(x) = 1$ is bounded, but $f$ isn't. Now we see that if we just take $f(x) = x^{1000}$, it's a counterexample. – Jakobian Apr 17 at 17:20