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We know that the inverse function of

$ y=\log(x) $ is $ y =\exp(x) $.

However, what would be the inverse of

$ y=\log(x)+ \sum_{n=1}^{\infty}\delta (x-n) $?

I have tried with Mathematica, and whenever there is a Dirac Delta, for example at the point $ x=1 $, at this point the computer gives a 'null' value, there is a white point on the screen at this point $ x=1 $. Very curious.

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migrated from Jan 26 '13 at 3:02

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Phew...I believe it is ill-defined even if you approximate the delta function in the usual ways (a peak and then take the limit as it gets infinitely sharp at constant unit area) because that function is not monotonic. – dmckee Jan 25 '13 at 21:46
i guess the inverse function of $ log(x) +\sum_{n=1}^{\infty} \delta (x-n) $ is just $ y=exp(x) $ except for the points $ n=1,2,3,4,5,6,.... $ where is ill-defined. – Jose Garcia Jan 25 '13 at 22:24
Jose, I put some questions on your deleted question "Fourier transform (inverse) of a Heavside function" at the meta site and I hope you don't mind. Cheers... – draks ... Jan 27 '13 at 16:35
it does not matter draks :) regards.. – Jose Garcia Jan 27 '13 at 20:11
up vote 5 down vote accepted

The delta function, viewed as a distribution, takes a (sufficiently well-behaved) "test" function as its input, and spits out the value of the function at a certain point as its output. For example, let $\delta_a$ be the delta distribution centered at $a$, then for any real-valued test function $f$ on the real line, one obtains $\delta_a(f) = f(a)$. Notice that if $g\neq f$ is any other function that has the same value as $f$ at $a$, then we also have $\delta_a(g) = g(a)$ so that


This means that $\delta_a$ is not one-to-one and therefore it is not invertible. For the function you gave, it is essentially the same story.

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Why can't you find the inverse for a function that's not one-to-one? For example, $f(x)=x^2$ is not one-to-one, but the inverse is: $f^{-1}(x)= \pm\sqrt{x}$ which is a multiple-valued function. – Garrett Feb 16 '14 at 22:35
@Garrett See…, for example. Notice also that saying "the" function $f(x) = x^2$ is misleading. One needs to both specify the domain of the function, and what it does to each element of the domain. If we define $f$ as the function whose domain is $[0,\infty)$ and for which $f(x) = x^2$ for all $x\in [0,\infty)$, then $f$ is invertible. But if we extend its domain to $(-\infty, \infty)$, then it is not invertible. – joshphysics Feb 17 '14 at 3:47
This leads to another question if there exists a set of distributions $\{f_s\}_{s\in S}$ such that for any test function $\phi $ the set $\{f_s(\phi)\}_{s\in S}$ completely describes $\phi$. – TZakrevskiy Mar 8 '14 at 19:20
@TZakrevskiy Could you clarify what "completely describes" means? What if you were to take $\{\delta_x\}_{x\in\mathbb R}$? – joshphysics Mar 8 '14 at 19:24
@joshphysics well, means, allows to obtain $\phi$. Your example is quite illustrative, I think, to answer my question. – TZakrevskiy Mar 8 '14 at 19:55

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