$f:R→R$ is twice differentiable. $f$ and $f''$ are bounded. How to prove with Taylor-theorem that $f'$ is bounded as well?
How to prove with Taylor-theorem that if $f$ and $f''$ are bounded, than $f'$ is bounded, too?
1
$\begingroup$
functions
derivatives
taylor-expansion
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0What's restricted? – 2017-01-28
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0What does "resctricted function" mean here? – 2017-01-28
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0Do you mean "bounded"? – 2017-01-28
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0yeah, sorry, I mean bounded. – 2017-01-28
1 Answers
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For any $x$, by Taylors theorem, $$f(x+1) = f(x) + f^\prime (x) + \frac{1}{2}f^{\prime\prime}(\xi)$$
for some $\xi\in (x, x+1)$. Now solve for $f^\prime(x)$.
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0smart solution! :) – 2017-01-28