TITLE: A uniform bound of Hölder class-like densities.
QUESTION [2 upvotes]: Let $f(x)$ satisfies $f\geq 0$ and $\int_R f(x)dx = 1$, suppose it also in the so-called H$\ddot o$lder class, i.e. $\mid f(x)^{(l)}-f(y)^{(l)}\mid \leq L\mid x-y \mid^{\alpha}$ for all $x, y \in R$, where $f(x)^{(l)}$ is the $l-$order derivative of $f$, $l$ is a positive integer, $0<\alpha<1$ and $L$ is a positive number.
The question is to show that such class of $f$ is uniform bounded, i.e., $f(x)\leq M$ for all $x$ and all $f$ satisfies the conditions, where $M$ is a constant only depends on $l,\alpha, L$.
It seems like a mathematical analysis problem with a quite simple form. A proof or any references and directions are appreciated.

REPLY [3 votes]: All you need is the Holder condition on $f$, not on any of its derivatives. For an unbounded nonnegative function to have integral $1$, its graph would have to be extremely steep at points. The Holder condition prevents that.
To be precise, let $x_0 \in \mathbf R$ and let $M = f(x_0)$. For any point $x$ you have $$f(x_0) - f(x) \le L|x-x_0|^\alpha$$ and thus
$$f(x) \ge M - L|x-x_0|^\alpha.$$
Define $\epsilon = \left( \dfrac ML \right)^{1/\alpha}$. As long as $|x-x_0| < \epsilon$ you obtain a positive lower bound for $f$, and in particular
$$1 = \int_{\mathbf R} f(x) \, dx \ge \int_{x_0-\epsilon}^{x_0+\epsilon} f(x) \, dx \ge \int_{x_0-\epsilon}^{x_0+\epsilon} M-L|x-x_0|^\alpha \, dx$$ and a trivial calculation shows that
$$ \int_{x_0-\epsilon}^{x_0+\epsilon} M-L|x-x_0|^\alpha \, dx = 2M\epsilon - \frac{2L\epsilon^{\alpha+1}}{\alpha + 1} = 2\epsilon M \frac{\alpha}{\alpha + 1}.$$ You can conclude
$$2 \left( \frac ML \right)^{1/\alpha} M \frac{\alpha}{\alpha + 1} \le 1$$ which can be used to get an explicit upper bound of $M$ in terms of $L$ and $\alpha$.
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