# Existence of unique root of mixed linear/logarithmic equations

I have functions $$f: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ and $$g$$ that look like this: \begin{aligned} f(x) &= Ax + b + g(x) \\ g(x_i) &= -c \log(1/x_i - 1), \quad x_i\in(0,1) \end{aligned}

where $$A$$ is square, $$x_i$$ are the components of $$x\in\mathbb{R}^n$$ and $$g(x)$$ is intended to indicate the vectorization of $$g(x_i)$$ (abuse of notation).

For $$A$$ full rank, I want to know if the equation $$f(x) = 0$$ has a unique solution $$x^\star\in(0,1)^n$$ (note that $$g$$ is undefined for $$x_i\in(-\infty,0] \cup [1,\infty)$$, so this specification is not particularly constraining). Namely, if there is a unique $$x^\star$$ which satisfies $$x^\star = A^{-1}(-g(x^\star) - b)$$.

Also, note that $$\dfrac{\partial g}{\partial x_i} \geq 4c > 0$$.

1) Does this even seem to be true or should I expect multiple roots?

2) How to prove it?

My approach so far has been via global versions of the Implicit Function Theorem, e.g. let $$\tilde{f}(x,y) = Ax + b + y = 0$$, where $$y = g(x)$$. That doesn't seem to be particularly useful here because it's approaching the problem from the opposite direction -- I've already defined the global implicit function for $$y$$ in terms of $$x$$. What I need to know is if the function $$f$$ has a (unique?) root.

• What can you say when $n=1$? – Michael Burr Oct 5 '18 at 23:22
• @MichaelBurr Ah! This helps me a bit - for $n=1$, we have $df/dx = a + c/(x-x^2)$. A sufficient condition for uniqueness would be $df/dx > 0$, i.e. $(x-x^2)^{-1} > -a/c$. Since $(x-x^2)^{-1}$ achieves a minimum value of $4$ on $x\in(0,1)$, we have $4c + a > 0$. If this isn't the case, $b$ can be arbitrary to imply multiple roots. In the case of my problem (in an engineering application), I can design $A$ and $c$. So I'm wondering if it's sufficient to say $4cI + A \succ 0$. I'm not sure how to proceed here in the multivariate case, since it's not as straightforward as requiring $df/dx > 0$. – Tor Anderson Oct 5 '18 at 23:47
• I suppose we could compute the whole Jacobian of $f$ and then justify that if it is elementwise positive, a unique root exists. This seems like it's unnecessarily conservative (sufficient, but not necessary). Is my intuition correct here? – Tor Anderson Oct 6 '18 at 0:01
• Perhaps it is enough for $A$ to be positive definite. – Michael Burr Oct 6 '18 at 2:23