Solving non-linear (convex) systems of equations

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I have a system of non-linear equations that takes the following form

\begin{align} \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots\\ y_n \end{array} \right] = \left[ \begin{array}{c} f_1({\bf x}) \\ f_2({\bf x}) \\ \vdots\\ f_n({\bf x}) \end{array} \right] \end{align} where $y_1,y_2,\ldots,y_n$ are given, ${\bf x}\in\mathbb{R}^{n-1}$ is the unknown variable, and each $f_i:\mathbb{R}^{n-1}\to\mathbb{R}$ is convex, with at least one $f_i$ strictly convex. I am attempting to determine a method for obtaining ${\bf x}$ given $y_1,y_2,\ldots,y_n$.


Attempt at a solution:

Newton's method applied to problems with more variables than unknowns requires us to form the generalized inverse of the non-square Jacobian matrix, denoted $J^+ = (J^\top J)^{-1}J^\top$. Finding the solution of a function is equivalent to minimizing the sum of squares of residuals \begin{align*} S({\bf \hat x}) = \sum_{i=1}^n r_i({\bf \hat x})^2 \end{align*} where $r_i({\bf \hat x}) = y_i - f_i({\bf \hat x})$. A solution is obtained iteratively by ${\bf \hat x}^{t+1} = {\bf \hat x}^{t} - J^+{\bf r}({\bf \hat x}^t)$ from initial condition ${\bf \hat x}^0$ (the generalized inverse $J^+$ may depend on ${\bf \hat x}^t$). The convergent point ${\bf \hat x}^*$ is a solution to the non-linear system of equations, if it exists, or a least-squares solution, if a solution does not exist.


Questions:

Does the convexity of the functions $f_i$ (with at least one being strictly convex) allow us to claim existence or uniqueness of a solution to the above system? Will the outlined Newton's method, above, be guaranteed to converge to it? Or is there a better method for finding ${\bf x}$?

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Does the convexity of the functions $f_i$ (with at least one being strictly convex) allow us to claim existence or uniqueness of a solution to the above system?

Not necessarily, consider the function $f(x,y)=(e^x,y^2)$, then $f=(f_1,f_2)$ is such that $f_1$ is strictly convex and $f_2$ is convex. Observe that $f(x,y)=(-1,-1)$ has no solution and $f(0,1)=f(0,-1)$.

Will the outlined Newton's method, above, be guaranteed to converge to it?

Let us suppose that there is a solution to the system. If the $f_i$ are smooth enough (say $f_i \in C^2$), theoretical convergence can be guaranteed. But there are pathological cases where in practice convergence is extremely slow. Note that thanks to convexity of $f_i$, you can use the generalized version of the Newton algorithm with subgradient, in case $f_i$ would not be differentiable.