$f \in C^{\infty}(A)$, $A$ compact then $f^{(k)} \in L_p$

Question

Let $f \in C^{\infty}(A)$, $A \subset \mathbb{R}$ compact. Is it true that $|| f^{(k)} ||_p < \infty$, for each naturals pair $p,k$?

My attempt. Since $A$ is compact and $f \in C^{\infty}(A)$ then for each $k$ natural $f^{(k)}(A)$ is compact as well, and it is bounded specifically. So is it is $|f^{(k)}|$, and same as $|f^{(k)}|^p$, so for each pair $(k,p)$ the function $g(x;p,k) = |f^{(k)}(x)|^p$ is Reinmann integrable, and therefore is Lebesgue integrable. And the integral

$$I_{p,k} = \int_{A} g(x;p,k) dx$$

Since $I_{p,k} = ||f^{(k)}||_p$ this completes the proof.

Assuming is correct, can the argument be generalized for any measure $\mu$? If not where's my mistake?

(I'm a little bit rusty in these things).

Answer 1

Since $f^{(k)}$ is bounded, i.e. $|f^k(x)|\leq M_k $, for any finite measure $\mu$ on $A$, i.e. with $\mu(A)<\infty$, you have $$ \int|f^{(k)}(x)|^pd\mu\leq M^p_k\mu(A)<\infty, $$ i.e. $f^{(k)}\in L^p(A,d\mu)$.

Answer 2

Your topological approach is good.

What you hideously use, is that a bounded function over a compact set has finite integral, which holds only for a certain type of measures.

It is quite clear that the condition on the measure is that it is finite on compact sets to have the same result. By taking the constant function $1$ you see that this is actually an iff condition.