0
$\begingroup$

I have functions $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $g$ that look like this: \begin{equation} \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} \end{equation}

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.

$\endgroup$
  • 1
    $\begingroup$ What can you say when $n=1$? $\endgroup$ – Michael Burr Oct 5 '18 at 23:22
  • $\begingroup$ @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 $df/dx > 0$. $\endgroup$ – Tor Anderson Oct 5 '18 at 23:47
  • $\begingroup$ 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? $\endgroup$ – Tor Anderson Oct 6 '18 at 0:01
  • 1
    $\begingroup$ Perhaps it is enough for $A$ to be positive definite. $\endgroup$ – Michael Burr Oct 6 '18 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.