## Tuesday, June 26, 2012

### I am an Analytic Bastard and I will fight Delusional Geometers to death

This post is dedicated to AMG: friend and enemy, mentor and destroyer, wise and fool.

AMG is obsessed to equate generalized functions (as in Swartz distribution theory) to probability distributions (as in measure theory). According to AMG I am an Analytic Bastard, I agree. And this was the least thing I could take from a Delusional Geometer. Therefore I left AMG.

Schwarz distributions are NOT probability distributions

I can say it louder but not clearer.

It doesn't matter that they are both called distributions, sometimes it does happen in mathematics that two different things are similarly called. It doesn't matter how hard you try to make them the same, it doesn't matter how proud you are and how little you think of the people that surround you and that are not Field medalists.

The fact that Strichartz's book shows a bell-like $C^{\infty}$, compactly supported function does not imply it is a distribution. What is more, this bell-like $C^{\infty}$, compactly supported function is clearly stated to belong to the set $\mathcal{D}$, the set of test functions. Therefore it is not a distribution itself, but the objects to which the linear functional (the Swartz distribution) is applied to. If $\varphi \in \mathcal{D}$ then there is a test function. It looks similar to a DENSITY function (the Gaussian) but the density is not the distribution, nor the test function is the (other kind of) distribution.

Furthermore, forcing my brains so as to accept $\varphi \in \mathcal{D}'$ and call it a (Swartz) distribution, then you can't write $\varphi(x)$ outside the integral symbol. It is a functional, which means that it is applied to some $\phi \in \mathcal{D}$, so what makes sense is $\varphi(\phi)$, don't get angry with me because of this, this is a fact. If $\varphi(x)$ were a linear functional, it could be written as $\int \varphi(x) \phi(x) dx$, why on Earth do you say $\varphi(x)$ anyway?

At this point, why the hell do you use $\varphi(x)$ to name a "bell-like" function with $x= \arg \max_y \varphi(y)$?

Then it remains going full retarded and try to apply density estimation methods to image analysis following this logic:
1. David Mumford develops an axiomatic theory that describes images as generalized functions (check)
2. We have methods that work in the density estimation field fairly well (check)
3. Since YOU (and only you) say generalized functions = probability distributions, then our methods must be very powerful in image analysis (FAIL)
FAIL! Because the only supporting argument you have is your pride.

So, let me out!