Proving the uniqueness of a fixed point on $(0,1]^N$

Question

I'll preface this by saying I am not very experienced with formal proofs, so that's mainly why I'm asking for help here. I found several other threads on this topic here on Math SE, but most of them either speak in the abstract about various theorems or, when a specific example is provided, a commenter will say "just use [x] theorem", which is not helpful to someone of my level.

So, I have a system of equations of the form:

$$X_p = F_p(\mathbf{X}) := \frac{1}{1 + \sum\limits_q^N A_{qp} X_q}$$

where:

• $\mathbf{X} \in \mathbb{R}^{N}$ is the vector/1D array of independent variables.
• $F(\mathbf{X}) : \mathcal{D} \to \mathcal{D}$ is a map (don't know how else to describe this, sorry; maybe a "self-map"?).
• $\mathcal{D} := (0,1]^N$ is a domain of interest.
• $\mathbf{A} \in \mathbb{R}^{N \times N}_{\ge 0}$ is a nonzero symmetric matrix/2D array of nonnegative (constant) weight parameters. By symmetry, $A_{qp} = A_{pq}$. By nonzero, at least one pair of elements $A_{qp} = A_{pq}$ in $\mathbf{A}$ will be nonzero.
• $p,q \in \{1, ..., N\}$ are indices.
• $N$ is the number of equations/variables.

I want to prove that this system of equations has exactly one unique solution on $\mathcal{D}$.

To be clear, this system will have other solutions, at either $X_p < 0$ or $X_p > 1$, but always one and only one in $\mathcal{D}$.

One additional assumption we can make about the system is that all variables $X_p$ will be "active", that is, no variable $X_p$, due to zero-valued weights, will have a value of $X_p = \frac{1}{1 + \sum\limits_q (0) X_q} = 1$. I don't know how exactly to notate this symbolically, but it guarantees that more elements of $\mathbf{A}$ will be nonzero. Outside the scope of this post, in my actual unsimplified problem, this can happen (hence the closed upper bound "$]$" on the domain), but this need not be the case for this proof.

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I have already proven that $F$ maps $\mathcal{D}$ to itself, and, via Brouwer's Fixed-Point Theorem, have proven that there exists at least one solution in $\mathcal{D}$. My issue is with proving the uniqueness of this solution.

I have seen two approaches to proving this:

1. Banach Fixed-Point Theorem, with contraction mappings
2. Uniqueness via monotonicity

I don't really understand the first option, because I don't know how to prove that $F$ is a contraction mapping; I don't know how to choose/find a metric for this problem.

The second approach I admittedly originally got from an LLM, but I checked it manually, and everything seems fine except for one part, which is either false or I just can't understand why it is true. I will include that proof at the bottom of this post, but if anyone can already point me to a simple way to prove the uniqueness of a fixed point of $F$ on $\mathcal{D}$, then your help is much appreciated.

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(Supposed) Uniqueness Proof via Monotonicity of $F$

Let $\mathbf{X},\mathbf{Y} \in \mathcal{D}$ be two fixed points: $F(\mathbf{X}) = \mathbf{X}, F(\mathbf{Y}) = \mathbf{Y}$.

Assume for contradiction that $\mathbf{X} \neq \mathbf{Y}$.

Note that $F$ is monotone decreasing in all $X_q$ for $A_{qp} \ge 0$, and strictly so if $A_{qp} > 0$.

Consider the difference of reciprocals of $X_p$ and $Y_p$:

$$\frac{1}{X_p} - \frac{1}{Y_p} = \sum\limits_q A_{qp} (X_q - Y_q)$$

• If $\mathbf{X} > \mathbf{Y}$ componentwise, then because $X \mapsto \frac{1}{X}$ is strictly decreasing, the LHS: $\frac{1}{X_p} - \frac{1}{Y_p} < 0$ for all $p$. However, because $X_q > Y_q$ for all $q$, the RHS sum is strictly positive. But this contradicts the fact that the LHS is strictly negative. [CONTRADICTION]

• If $\mathbf{X} < \mathbf{Y}$ componentwise, then the same logic applies. LHS: $\frac{1}{X_p} - \frac{1}{Y_p} > 0$ for all $p$. However, because $X_q < Y_q$ for all $q$, the RHS sum is strictly negative. This contradicts the fact that the LHS is strictly positive. [CONTRADICTION]

• If we can only say in general that $\mathbf{X} \neq \mathbf{Y}$, i.e., some $X_p < Y_p$ and other $X_p > Y_p$, then I feel like this proof fails here, because the product of weights and differences $\sum\limits_q A_{qp} (X_q - Y_q)$ can be positive, negative, or zero, depending on the exact values. [???]

Because I don't know what I'm doing here, I asked the LLM which claimed that the last case cannot happen if $\mathbf{X}, \mathbf{Y}$ are fixed points and $F$ is strictly decreasing in all variables, but I never got a convincing explanation as to why this would be the case, and I am also not sure if $F$ indeed is strictly decreasing in all variables because some weights $A_{qp}$ can be zero.

As I said above, any help with this problem is greatly appreciated, either in patching the hole in the above proof, or with a different proof entirely. Thank you in advance!

Answer 1

Assume that $X$, $Y$ are different fixed points in $(0,1]^N$. Set $$ \alpha:= \min\{\beta \ge 0 \mid \forall q=1,\dots,N: ~ (Y_q \le \beta X_q) \wedge (X_q \le \beta Y_q)\}. $$ Then $Y_q \le \alpha X_q$ and $X_q \le \alpha Y_q$ $(q=1,\dots,N)$, and $\alpha >1$ since $X \not= Y$. Now for each $p\in \{1,\dots,N\}$ we have $$ Y_p = \frac{1}{1+\sum_{q=1}^N A_{qp}Y_q} \le \frac{1}{1+\frac{1}{\alpha}\sum_{q=1}^N A_{qp}X_q} = \frac{\alpha}{\alpha+\sum_{q=1}^N A_{qp}X_q} < \frac{\alpha}{1+\sum_{q=1}^N A_{qp}X_q} = \alpha X_p, $$ and vice versa $X_p < \alpha Y_p$. Thus we have strict inequalities in all components, and therefore there is some $0\le \beta < \alpha$ such that $(Y_p \le \beta X_p) \wedge (X_p \le \beta Y_p)$ for all $p\in \{1,\dots,N\}$, in contradiction to the minimality of $\alpha$.