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

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    $\begingroup$ It is even in the sense of distribution theory, yes. You shouldn't think of it as being even in the pointwise sense. $\endgroup$
    – Ian
    May 9 '16 at 21:36
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    $\begingroup$ yes it is even. define evenness of a distribution $S$ by $\langle S,\varphi \rangle = \langle S,T \varphi \rangle$ for every test function $\varphi$, with $T$ the time reversal operator $T \varphi(x) = \varphi(-x)$. and use the definition of the distributional derivative : $\langle S',\varphi \rangle = - \langle S, \varphi' \rangle$, hence if $S$ is even then $\langle S'',T \varphi \rangle = - \langle S',(T \varphi)' \rangle = \langle S,(T \varphi)'' \rangle = \langle S,T \varphi'' \rangle = \langle S,\varphi'' \rangle = -\langle S',\varphi' \rangle = \langle S'',\varphi \rangle$ $\endgroup$
    – reuns
    May 9 '16 at 22:15
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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]

enter image description here

$f_s$ ($\color{green}{green \ curve}$) 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)

enter image description here

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).

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  • $\begingroup$ I am deceived by the downvote. I would have liked the person who has downvoted to say what is "choking" in my vision of the "evenness" of the second derivative of $\delta$. $\endgroup$
    – Jean Marie
    May 10 '16 at 0:46
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    $\begingroup$ I'm not the downvoter, but one problem is that it is not entirely obvious that evenness of an approximating sequence implies evenness of the limit, or even really what evenness of the limit necessarily means. It takes some distribution theory to even really formulate this question; talking about approximating sequences mostly serves to give intuition. $\endgroup$
    – Ian
    May 10 '16 at 0:57
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    $\begingroup$ Thanks for comment @Ian The way I have written my answer was clear : it intended to give an intuitional grasp on the issue. I am aware of subtilities of distributions and I know the rigorous definition of an even distribution (as used by user1952009). Besides, in the second part of my answer, it seems to me that I have brought some supplementary information (also on the side of understanding). It is not absurd to recall that one of the main uses of $\delta''$ is through the convolution operation $\delta'' * f = f''$ and its connection with image/signal processing through the mask [1,-2,1]... $\endgroup$
    – Jean Marie
    May 10 '16 at 1:10
  • $\begingroup$ I have improved my text and added a figure of the difference of gaussians. $\endgroup$
    – Jean Marie
    Aug 31 '16 at 15:28
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    $\begingroup$ @JeanMarie if you want to make things rigorous, you can say that a distribution $S$ is even if and only if for every even test function $\varphi$ : $\ S \ast \varphi$ is even. $\endgroup$
    – reuns
    Aug 31 '16 at 15:43

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