## Friday, December 27, 2013

### A New Generalized Function

There is a sort of mathematical function (technically a distribution) called a Dirac delta function, symbolized as $\delta(x)$. This function has the property of being identically zero everywhere, except at the origin, at which point it is positive infinity, but in a specific sense. It goes to infinity in such a way such that the integral of the function over any interval containing the origin (but not as an endpoint), will be unity. That is, ${\int_{-\infty }^{+\infty}\delta(x)dx}=1$
An additional important property of the Dirac delta function is that, for any function $f(x)$, $f(x)\delta(x)=f(0)\delta(x)$. This is called the sifting property of the delta function.

A typical way to define the delta function involves the rectangular pulse function, defined as $\mathrm{rect}(x) = \begin{cases} 0 & \mbox{if } |x| > \frac{1}{2} \\ \frac{1}{2} & \mbox{if } |x| = \frac{1}{2} \\ 1 & \mbox{if } |x| < \frac{1}{2} \end{cases}$We then have $\delta (x)=\lim_{a\rightarrow \infty} a \, \mathrm{rect}(a\,x)$Notice that this would indeed have the required properties.
For fun, let us introduce two modified delta functions: a right handed and left handed one. We define these using the right and left handed rectangular pulse functions, defined respectively as $\mathrm{rect}_R(x) = \mathrm{rect}(x-\frac{1}{2})\,,\,\mathrm{rect}_L(x) = \mathrm{rect}(x+\frac{1}{2})$. We would then have $\delta_{R} (x)=\lim_{a\rightarrow \infty} a \, \mathrm{rect}_{R}(a\,x)\,,\,\delta_{L} (x)=\lim_{a\rightarrow \infty} a \, \mathrm{rect}_{L}(a\,x)$ These would have the same properties as the centered delta function, except that the sifting property becomes, respectively, $f(x)\delta_{R}(x)=f(0^{+})\delta_{R}(x) \, , \, f(x)\delta_{L}(x)=f(0^{-})\delta_{L}(x)$. In addition, we can now better formulate the integral properties of the various delta functions. Let $A<0$ and $B>0$. We then have: ${\int_{A }^{B}\delta(x)dx}=1,\, {\int_{0 }^{B}\delta(x)dx}=\frac{1}{2},\, {\int_{A }^{0}\delta(x)dx}=\frac{1}{2}\\ {\int_{A }^{B}\delta_{R}(x)dx}=1,\, {\int_{0 }^{B}\delta_{R}(x)dx}=1,\, {\int_{A }^{0}\delta_{R}(x)dx}=0\\ {\int_{A }^{B}\delta_{L}(x)dx}=1,\, {\int_{0 }^{B}\delta_{L}(x)dx}=0,\, {\int_{A }^{0}\delta_{L}(x)dx}=1$
Notice that, in every case, ${\int_{A }^{B}}={\int_{A }^{0}}+{\int_{0 }^{B}}$, as must be the case.

Suppose now we consider the opposite case: we begin with the rectangular pulse function, but instead of making it narrower and narrower and taller and taller, we make it wider and wider and shorter and shorter. That is, suppose we define a new function $\eta(x)$ that also has unit total area, defined as $\eta (x)=\lim_{a\rightarrow \infty} a^{-1} \, \mathrm{rect}\left(\frac{x}{a}\right)$ This function is the same as the uniform distribution function over an infinite interval. Oddly enough, this means that, for $-\infty<A<0$ and $0<B<+\infty$, ${\int_{A }^{B}\eta(x)dx}=0,\, {\int_{-\infty }^{A}\eta(x)dx}={\int_{-\infty }^{B}\eta(x)dx}=\frac{1}{2},\, {\int_{B }^{+\infty}\eta(x)dx}={\int_{A }^{+\infty}\eta(x)dx}=\frac{1}{2}$ This suggests some funny business going on at infinity: it seems the cumulative function is discontinuous at $\pm\infty$.

The function also has some other interesting properties:

• It seems to be a counterexample to the claim "for a function $f$ identically zero everywhere, $\int_{-\infty}^{+\infty}f(x)dx=0$", as $\eta(x)$ seems to be identically zero everywhere and yet has a nonzero improper integral. Furthermore, it seems that we cannot interpret its improper integral in the usual way, e.g. as the Cauchy principal value.
• It is invariant under a finite translation, that is, $\eta(x-a_{0})=\eta(x)$. This would mean that convolving a function with it would merely involve taking the inner product, that is, $\eta(x)\ast f(x)=\int_{-\infty}^{+\infty}\eta(x)f(x)dx$.
• Convolving it with another function (if the convolution exists) gives a constant function which with the value of the global average of the convolvend. That is, $f(x)\ast\eta(x)=\lim_{L\rightarrow \infty}\frac{\int_{-L/2}^{+L/2}f(x)dx}{L}$
The resultant function is just the constant component of the input function. For instance, $e^{-x^2}\ast\eta(x)=0\,,\, \cos^{2}(x) \ast\eta(x)=\frac{1}{2}$
• It almost seems like the function could be represented by a pair of delta functions at $\pm\infty$. Namely, it seems almost that $\eta(x)=\frac{\delta(x+\infty)+\delta(x-\infty)}{2}$. However, this is not quite true.
• The Fourier transform of the eta function can be determined by considering the limit of the transform of the rectangular pulse $\hat{\eta}(\omega)=\int_{-\infty}^{\infty}\eta(x) e^{-2\pi ix\omega}\,dx=\lim_{a\rightarrow \infty} a^{-1}\int_{-\infty}^{\infty} \mathrm{rect}\left(\frac{x}{a}\right)e^{-2\pi ix\omega}\,dx\\ \hat{\eta}(\omega)=\lim_{a\rightarrow \infty}\frac{\sin(\pi\omega a)}{\pi\omega a}= \begin{cases} 1 & \mbox{if } \omega=0 \\ 0 & \mbox{if } \omega\neq0 \end{cases}$

It is also possible to modify the eta function by making it right- or left-sided (as with the delta function).$\eta_{R} (x)=\lim_{a\rightarrow \infty} a^{-1} \, \mathrm{rect}_{R}\left(\frac{x}{a}\right)\,,\, \eta_{L} (x)=\lim_{a\rightarrow \infty} a^{-1} \, \mathrm{rect}_{L}\left(\frac{x}{a}\right)$ The properties are mostly similar (including translation-invariance, zero finite integrals, etc.). While the symmetric eta function takes the global average, the right-handed one takes the average over positive values, and the left-handed one takes the average over negative values. For instance, we would have $\int_{-\infty}^{+\infty}\eta_{R}(x)\left(\cos^{2}(x+|x|) \right)dx=\frac{1}{2},\\ \int_{-\infty}^{+\infty}\eta_{L}(x)\left(\cos^{2}(x+|x|) \right)dx=1$I'm sure there is more to say on this, and I will continue to look into the properties of this new, strange function.