The Gay Banach Fixed-Point Theorum

This system of equations.

Let $(X, d)$ be a non-empty complete metric space with a contraction mapping $T : X \to X$. Then $T$ admits a unique fixed-point $x^*$ in $X$ (i.e. $T(x^*) = x^*$). Furthermore, $x^*$ can be found as follows: start with an arbitrary element $x_0 \in X$ and define a sequence $(x_n)_{n\in\N}$ by $x_n = T(x_{n-1})$ for $n \ge 1$. Then $\lim_{x \to \infin} x_n = x^*$.

So, this is a pretty dense definition. If you are cut from a more mathematically inclined cloth you might understand what is being said here, but for the rest of you, I will break it down into simpler terms.

What is a non-empty complete metric space? Well a non-empty metric space is just some collection of things $X$ with a function $d(a,b)$ that gives you the distance between $a$ and $b$ where $a, b \in X$. (The $\in$ means that something is in something else, so here $a$ and $b$ are in $X$). For this just think of a 3D coordinate system. $X$ would just be the set of all points in 3D space, and we could just define $d(a,b)$ to just be the distance between those points using the Pythagorean theorem.

What does complete mean? Basically, it just means that if you have a sequence of points that are always getting closer to eachother in a specific way, there is a point they are always getting closer to. This way, specifically, is that you can choose some distance, call it $\epsilon$, and for however small of an $\epsilon$ you choose you know you can choose a point in that sequence where all the points after it are at most $\epsilon$ away from eachother.

This might seem stupid. “Of course if you can choose any arbitrarily small distance for points in a sequence to be from eachother they have to be getting closer to something!” you might think. But you would unfortunately be wrong!

Imagine if you took the 3D space you had before, and you have a spiral of points getting sucked to the origin. (In my mind there’s like a black hole there or something.) Now, in this imaginary space, lets say since the origin is a black hole, it isn’t actually part of our 3D space. Here, there is no point in our new space that the points are all getting closer to, because the origin no longer exits! Saying the space is complete means that this sort of thing is not allowed

So in that sense, another way of thinking about a complete space is that there are no spots you can get infinitely close to but never get to in it.

Luckly for us, our usual 3D space is complete, and since we already know it is a metric space and it has stuff in it, our usual 3D space is a non-empty complete metric space.

From now on, we are going to stick with $X$ as our usual 3D space with a regular (Cartesian) 3D coordiante system.

So then the next question is, “What the heck is a contraction mapping?” A contraction mapping is some function, in this case $T$, that takes an input from $X$ and gives an output in $X$. It also has the special property that if you apply $T$ to two different points in your space, it will bring them closer together.

In fancy math terms,

For all $a, b \in X$, $d(T(a), T(b)) \le k\cdot d(a,b)$ where $k$ is some constant number and $0 \le k \lt 1$.

A very simple example in our 3D space would be the function

$T((x,y,z)) = \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)$

Hopefully it is obvious that this brings points closer together. With a little work we can see that our $k = \frac{1}{2}$.

A more complex example would be

$T((x,y,z)) = \left(\frac{x + 1}{3}, 0.9z, y\right)$

See if you can figure out what $k$ is for this one.

The fruits of this labor

Now that we have both a non-empty complete metric space $X$ a contraction mapping $T$ on it, we can exercise the final part of the theorum.

… Then $T$ admits a unique fixed-point $x^*$ in $X$ (i.e. $T(x^*) = x^*$). Furthermore, $x^*$ can be found as follows: start with an arbitrary element $x_0 \in X$ and define a sequence $(x_n)_{n\in\N}$ by $x_n = T(x_{n-1})$ for $n \ge 1$. Then $\lim_{x \to \infin} x_n = x^*$.

The first part of that just means that there is some point in our 3D space $x^*$ where applying our function to it doesn’t change our point at all. For our first example of a contraction mapping that point is $x^* = (0,0,0)$ and for our second example it is $\left(\frac{1}{2}, 0,0\right)$. (Check to make sure these work yourself.)

The last part of that is the truly incredible part. It means that we can choose any point in our 3D space, and if we repeatedly apply $T$ to it, we will eventually find $x^*$.

Using this to calculate gayness

So, now if we look back at the first of our original equations

we can rewrite it by dividing by $C$ to be

For simplicity, lets say $\mu_a = \frac{C_a}{C}$, $\mu_{ab} = \frac{C_{ab}}{C}$ and so on. This is just to make our equations look nicer. Note, because $C_a + C_b + \cdots + C_{abc} = C$, this means that $\mu_a + \mu_b + \cdots + \mu_{abc} = 1$. Also, for all $\mu$, its true that $0 \le \mu \le 1$.

Now we have

we just say that $x = P_a$ to make this super clear, we can write

Now we are going to try to show that this function $f(x)$ is itself a contraction map (at least for $0 \le x \le 1$, since we know that already.)

Theres a few things we can say about this function pretty quickly. If we give it some input $a \in [0,1]$ ($a$ is in the range $[0,1]$) then we know that its output $f(a) \in [0,1]$.

Look at terms $2$ and $3$. Note that for all $x, b \ge 0$

$\frac{x}{x + b} \le 1$

because the denominator is always larger than the numerator.

If we look back at $f(x)$ we can reinterpret it as the weighted average of the terms

$\left[1, \frac{x}{x + P_b}, \frac{x}{x + P_c}, x, 0\right]$

with weights

$\left[\mu_a, \mu_{ab}, \mu_{ac}, \mu_{abc}, 1-(\mu_a + \mu_{ab} + \mu_{ac} + \mu_{abc})\right]$

Clearly the weights sum to 1, and all the weights must be non-negative, as we know all $\mu_?$ to be non-negative and we know the sum of all $\mu_?$ to be $1$, so a subsum of $\mu_?$ could not be greater than $1$.

A property of this is since we are averaging the terms, and we know that they are all clearly in $[0,1]$ that the average is in $[0,1]$ meaning $f(x) \in [0,1]$.

Now, take this theorum that I just happen to know to be true.

If you have a function $f(x)$ where $\left|f'(x)\right| < 1$ on some interval $I$, $f(x)$ is a contraction mapping on $I$.

Also,

If you have a function $g(x)$ where $g'(x) \ge 0$ on some interval $I$, that means it is an increasing function on that interval, which means for $a, b \in I$, if $a \le b$ then $f(a) \le f(b)$

Think about our $f(x)$ again. Its slope is clearly decreasing or constant, as $1$, $x$ and $0$ have constant slope while anything of form $\frac{x}{x + b}$ has decreasing slope. It is also clear that $f'(x)$ is never negative for similar reasons. Additionally, the maximum possible value for $f'(1)$ is $\frac{1}{4}$. I don’t really want to prove these from scratch so here are some graphs that show these facts pretty conclusively.

Lets do some small math real quick. Here $g(x)$ represents the amount a value $x$ would increase from calling $f$ on it.

The astounding result here is that either we are either

1. in a zone in which $f$ is a contraction mapping ($f'(x) < 1$) or
2. in a zone in which calling $f$ on $x$ will cause it to increase.

In case 2 since except for in very edge cases $f(x) \not = x$, its actually true that for $x \ge \epsilon$ where $\epsilon \gt 0$, there is some constant $\delta \gt 0$ where $g(x) \ge \delta$ meaning that repeatedly calling $f(x)$ on some value will eventually bring it into the zone where it is a contraction mapping.

Now here are some other things I may prove at some point.

We have now what I will call a safety zone for our $x$ AKA the zone on which it has a contraciton mapping. I will now start referring to $P_b$ as $y$ and $P_c$ as $z$. If $y$ or $z$ increases, this always increases the safety zone for our $x$. Now note, $\frac{x}{x + y} + \frac{y}{x + y} = 1$. From this we can actually develop something very interesting.

Note that the safety zone will only get narrower for $x$ if $y$ or $z$ decrease. $y$ or $z$ decreasing means inherently from this relationship (since these being low are the only things that will cause a $z$ decrease past linear causes) we know that they will then kick $x$ up enough to at least neglect the effect of the shrinking safety zone.

So, if you apply it to all of them at once what you see is that $x$ will still get to the safety zone in a finite amount of time. $x$ will stay in the zone with the contraction mapping. For reasons, that zone remains a contraction mapping in general.