Invariance of the Sign of the First Derivative

According to one of the properties established before for the first-order filter [5], if $f(\theta)$ is continuous in the domain where the filter is applied, then $f_{\epsilon}(\theta)$ is differentiable, and its derivative is given by

$$\frac{df_{\epsilon}(\theta)}{d\theta} = \frac { f\!\le... ...on\right) - f\!\left(\theta-\epsilon\right) } {2\epsilon}.$$ (8)

This is valid so long as the support interval of the filter fits completely inside the region where $f(\theta)$ is continuous. This immediately implies that, in a region where $f(\theta)$ increases monotonically we have

$$\begin{eqnarray*} f\!\left(\theta+\epsilon\right) & \geq & f\!\left(\theta-\e... ...tarrow \\ \frac{df_{\epsilon}(\theta)}{d\theta} & \geq & 0. \end{eqnarray*}$$

We may therefore conclude that $f_{\epsilon}(\theta)$ also increases monotonically within the sub-region where the support interval of the filter fits inside the region in which $f(\theta)$ is continuous. In the same way, in a region where $f(\theta)$ decreases monotonically we have

$$\begin{eqnarray*} f\!\left(\theta+\epsilon\right) & \leq & f\!\left(\theta-\e... ...tarrow \\ \frac{df_{\epsilon}(\theta)}{d\theta} & \leq & 0. \end{eqnarray*}$$

We may therefore conclude that $f_{\epsilon}(\theta)$ also decreases monotonically within that same sub-region. In other words, the monotonic character of the variation of a function is invariant by the action of the filter. In particular, at points where $f(\theta)$ is differentiable the sign of its derivative is invariant by the action of the filter.