Let $f \in C^k(0,1)$ and assume that the $k$-th derivative is $\alpha$-Hölder continuous. Assume that $f(x) = 0$ in a fixed interval $(a,b) \subset (0,1)$. Can we characterize (or at least find some examples of) non-constant functions $f$ as above such that
$$|f^{(k)}|\_{C^{0,\alpha}(0,1)} \le \Vert f \Vert\_{L^1(0,1)},$$
where $|g|\_{C^{0,\alpha}} = \sup\_{x,y \in (0,1)} \frac{|g(x)-g(y)|}{|x-y|^\alpha}$?



Functions for which $|f^{(k)}|_{C^{0,\alpha}(0,1)} \le \Vert f \Vert_{L^1(0,1)}$
The answer is no.


Indeed, let $t:=\alpha\in(0,1)$ and $c:=\|f\|\_1:=\|f\|\_{L^1(0,1)}\in(0,\infty)$. Suppose that $f\in C^k(0,1)$ and $f=0$ on $(a,b)$, with $0\le a<b\le1$. Suppose that the inequality in question holds.


Then for $x\in[b,1]$ we have $|f^{(k)}(x)|=|f^{(k)}(x)-f^{(k)}(b)|\le c(x-b)^t\le c$, $|f^{(k-1)}(x)|\le c(x-b)\le c$, $\ldots$, $|f(x)|\le c$, $\int\_b^1|f|\le c(1-b)$.


Similarly, $\int\_0^a|f|\le ca$. So,
$$c=\|f\|\_1=\int\_0^a|f|+\int\_b^1|f|\le ca+c(1-b)<c,$$
a contradiction. $\Box$


