Points of Non-Differentiability

Let us show that in the $\varepsilon\to 0$ limit the derivative of the function $f_{\varepsilon}(x)$ essentially reproduces the derivative of the original function $f(x)$. Stating it more precisely, we will show that, if the function $f(x)$ is continuous but has an isolated point of non-differentiability at $x_{0}$, then in the $\varepsilon\to 0$ limit the derivative of $f_{\varepsilon}(x_{0})$ tends to the average of the two lateral limits of the derivative of $f(x)$ to the point $x_{0}$, that is,

$$\lim_{\varepsilon\to 0}\frac{df_{\varepsilon}}{dx}(x_{0}) = \frac{1}{2}\left({\cal L'}_{+}+{\cal L'}_{-}\right),$$

where

$${\cal L'}_{\pm} = \lim_{x\to x_{0\pm}}\frac{df}{dx}(x),$$

regardless of any value that may be artificially given to the derivative of $f(x)$ at $x_{0}$. In particular, if $f(x)$ is differentiable at $x_{0}$, then ${\cal L'}_{+}$ and ${\cal L'}_{-}$ are both equal to the derivative of $f(x)$ at $x_{0}$, and hence the derivative of $f_{\varepsilon}(x)$ tends to the derivative of $f(x)$ at $x_{0}$ in the limit, thus reproducing the derivative of the original function at that point.

Here is the proof: if $f(x)$ is continuous at $x_{0}$ and has an isolated point of discontinuity there, then there are two neighborhoods of $x_{0}$, one to the left and another one to the right, where $f(x)$ is continuous and differentiable. According to the results of Section A.3 of this Appendix, since $f(x)$ is continuous at and around $x_{0}$, $f_{\varepsilon}(x)$ is differentiable and its derivative at $x_{0}$ is given by

$$\frac{df_{\varepsilon}}{dx}(x_{0}) = \frac { f(x_{0}+\varepsilon) - f(x_{0}-\varepsilon) } {2\varepsilon}.$$

However, since $f(x)$ is not differentiable at $x_{0}$, the $\varepsilon\to 0$ limit of the right-hand side of this equation does not give us any definite results. We may however separate this expression in two, each one making reference to only one of the two neighborhoods,

$$\begin{eqnarray*} \frac{df_{\varepsilon}}{dx}(x_{0}) & = & \frac { f(x_{0}... ... \frac { f(x_{0}) - f(x_{0}-\varepsilon) } {\varepsilon}. \end{eqnarray*}$$

It is now clear that, since $f(x)$ is differentiable in the two lateral neighborhoods, in the $\varepsilon\to 0$ limit the two terms in the right-hand side of this equation converge respectively to the right and left derivatives of $f(x)$ at $x_{0}$. We therefore have

$$\lim_{\varepsilon\to 0}\frac{df_{\varepsilon}}{dx}(x_{0}) = \frac{1}{2}\left({\cal L'}_{+}+{\cal L'}_{-}\right),$$

where

$${\cal L'}_{\pm} = \lim_{x\to x_{0\pm}}\frac{df}{dx}(x).$$

This establishes the result. In particular, if $f(x)$ is differentiable at $x_{0}$, then ${\cal L'}_{+}$ and ${\cal L'}_{-}$ are both equal to the derivative of $f(x)$ at $x_{0}$, and therefore we have for the derivative of $f_{\varepsilon}(x)$ at $x_{0}$

$$\lim_{\varepsilon\to 0} \frac{df_{\varepsilon}}{dx}(x_{0}) = \frac{df}{dx}(x_{0}),$$

thus reproducing the derivative of the original function at that point.