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?

  • 1
    $\begingroup$ 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 $\endgroup$ – Jakobian Apr 17 at 17:12
  • 1
    $\begingroup$ 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. $\endgroup$ – Jakobian Apr 17 at 17:20

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