# Functions for which $|f^{(k)}|_{C^{0,\alpha}(0,1)} \le \Vert f \Vert_{L^1(0,1)}$

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 $$|g|_{C^{0,\alpha}} = \sup_{x,y \in (0,1)} \frac{|g(x)-g(y)|}{|x-y|^\alpha}$$?

• Take any smooth $f$ and add a sufficiently large constant to it. Or am I missing something? Jan 24 at 23:00
• @MateuszKwaśnicki Sorry, I forgot to add the assumption that $f(x) = 0$ ona given interval $(a,b) \subset (0,1)$ (so we cannot just add a constant to make its $L^1$ norm larger without also making the $|\cdot|_{C^{0,\alpha}}$ seminorm larger)
– Hiro
Jan 24 at 23:07

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 $$0\le a. Suppose that the inequality in question holds.
Then for $$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) a contradiction. $$\Box$$