If a bounded $\mathcal{C}^n (\mathbb{R})$ function has a bounded n-th derivative, then all of its derivatives are bounded.

Question

Let $f$ be bounded $\mathcal{C}^n (\mathbb{R})$ function such as that $f^{(n)}$ is bounded.
Prove that all of its derivatives $f^{(k)}$ for $k \in$ $ \{ 1,..,n-1 \}$ are bounded.
So far I've been able to prove it for the case : n=2 using Taylor's inequality, however I've not been able to generalise this using induction, as I must prove that one of its derivatives other than $f^{(n)}$ is bounded.

Answer 1

For $f \in \mathcal{C}^n (\mathbb{R})$ and $0 < k < n$, Taylor's theorem can be used to show that there are coefficients $c_1, c_2, \ldots, c_{n-1}$ such that $$ \frac{f^{(k)}(x)}{k!} = \sum_{j=1}^{n-1} c_j \left( f(x+j) - f(x) - \int_x^{x + j} \frac{f^{(n)}(t)}{(n-1)!} (x+j - t)^{n-1} \, dt \right) $$ for all $x \in \Bbb R$, see for example If $f$ is Lebesgue integrable and $f''$ exists and is Lebesgue integrable, what can we say about the integrability of $f'$?.

In particular, the boundedness of $f$ and $f^{(n)}$ implies the boundedness of $f^{(k)}$ for $0 < k < n$.

See also Landau–Kolmogorov inequality for more estimates of this type.