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


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