Second derivative of the delta function

Question

Is the second derivative of the delta 'function' even? My intuition tells me yes, and my calculation relies on delta''(-x) = delta''(x).

Answer 1

The first objective of this answer is to provide an intuitional perspective, that I have found useful when working/teaching in image/signal processing to various audiences, some of them with very little mathematical background. I am aware that a theoretical approach (that I have taught as well) is at some distance from this "raw" presentation.

A fruitful way to consider (and sometimes to work on) $\delta$ distribution is as the limit of (gaussian density) functions $f_s(x)=\frac{1}{s \sqrt{2 \pi}}e^{-x^2/(2s^2)}$ when parameter $s \rightarrow 0$ (note that $s$ is the standard deviation). Thus $\delta'$ and $\delta''$ can be considered as resp. limit of functions $f'_s$ and $f''_s$ resp.

In order to answer your question [see figure below]

$f_s$ ($\color{green}{green \ curve}$) is even , $f'_s$ ($\color{red}{red \ curve}$) is odd and $f''_s$ ($\color{blue}{in \ blue}$) even. We assume that these "even" and "odd" properties are preserved when taking (distributional) limits.

These functions correspond to a same value of $s$, with a certain rescaling. The larger $s$, the closer to the $y$-axis are these curves.

A last point: It is well known that $\delta'$ modelizes the "doublet" or dipole in physics (two particles with opposite charges "very close one to the other"). See (https://www.quora.com/What-is-the-physical-significance-of-a-unit-doublet-as-the-derivative-of-dirac-delta-function).

Sometimes, one finds in the literature that this derivation is assimilated to "DoG"s (Difference of Gaussians). Indeed there is a striking similarity of the curve of

$$\tag{0}y=g(x+1)-g(x-1) \ \ \text{with} \ \ g(x)=e^{-x^2/2}$$

(see below)

with the curve of $f'_s$ displayed above; in fact, convolution of a function $f$ by $\delta'$ amounts to take the first derivative. Its discrete counterpart is covolution with mask [1,-1], and this is equivalent to expression (1).

The discrete equivalent of the second derivative $f''$ is known to be equivalent to convolution with mask $[1,-2,1]$. If this mask is used simultaneously on the horizontal direction and on the vertical direction, one generates the so-called "laplacian mask":

$$\begin{array}{|r|r|r|}\hline 0&1&0\\ \hline 1&-4&1\\ \hline 0&1&0\\ \hline \end{array}$$

with many applications like this one (https://angeljohnsy.blogspot.com/2013/07/image-sharpening-using-second-order.html).