The other day I was bounding some inequalities for my thesis. I had that some functions were Holder, but I haven’t the explicit derivatives of my functions.

I spent one hour or so thinking if I could bound correctly my expression without having the derivatives.

The answer is yes, and I will show how.

# Hölder condition

First, the definition according to Introduction to Nonparametric Estimation from A. Tsybakov for a Holder functional class is

\begin{equation}\label{eq:holder_fb}

\left| f^{(n)}(x) – f^{(n)}(x_0) \right| \leq C \left| x – x_0 \right|^{\beta-s},\ \forall x, x_0 \in D.

\end{equation}

However, the classical definition in the analysis textbooks, presents the last inequality as,

\left| f(x) – f(x_0) \right| \leq C \left| x – x_0 \right|^{\beta},\ \forall x, x_0 \in D.

\end{equation}

Silly or not, I couldn’t remember that those two inequalities were equivalents. %

This is one proof of its equivalence if you find it useful.

# Proof

I will prove that Equation \eqref{eq:holder_f} implies Equation\eqref{eq:holder_fb}. The inverse equivalence is similar and I will let it as an exercise.

First denote $n=\left\lfloor \beta\right\rfloor$ and $r=\beta – \left\lfloor \beta\right\rfloor$.

Next, remember that the derivative could be expressed as

\begin{equation} \label{eq:der}

f(x) = f(x_0)+f^\prime(x_0)(x-x_0) + o(x-x_0).

\end{equation}

Here the term $o(x-x_0)$ refers to the little-oh notation and means that

\begin{equation*}

\lim_{x\to x_0} \frac{f(x) – f(x_0)}{x-x_0} – f^\prime(x_0) = 0.

\end{equation*}

Thus, by Equation \eqref{eq:holder_f} and the identity \eqref{eq:der} we can prove that

\begin{equation}

\left\vert f^\prime(x_0) + o(1) \right\vert = \left\vert \frac{f(x) – f(x_0)}{x-x_0} \right\vert \leq C\vert x-x_0\vert^{n+r-1}.

\end{equation}

The points $x$ and $x_0$ belong to some interval $D$. Hence we can bound the difference between $(f(x) – f(x_0))/(x-x_0)$ and $f^\prime(x_0)$ implying that the term $o(1)$ is also controlled. It means we can always choose $C$ such as

\begin{equation}

\vert f^\prime (x_0)\vert \leq C \vert x-x_0\vert^{n+r-1}

\end{equation}

Therefore,

\begin{equation}

\vert f^\prime (x) – f^\prime (x_0)\vert \leq 2C \vert x-x_0\vert^{n+r-1}.

\end{equation}

We can apply the latter arguments iteratively $n$ times to Equation \eqref{eq:holder_f}. We notice in each step that subtracting $1$ from the exponent in the right side causes adding one derivative to left side.

Finally, we obtain our result

\begin{equation}

\left| f^{(n)}(x) – f^{(n)}(x_0)

\right| \leq C \left| x – x_0 \right|^{r},\ \forall x,

x_0 \in D.

\end{equation}

QED.