Sensitivity analysis in minimum norm problems under a linear constraint

Question

Suppose $\Delta$ is some nice topological space, say compact, and Hausdorff. Let $A:\Delta \rightarrow \mathbb{R}^{m\times n}$ be a continuous $m\times n$ matrix valued map. Let $b\in \mathbb{R}^{m}$ be a fixed vector that belongs to the range of $A(\delta)$ for all $\delta\in \Delta$. Then we know that there is a unique solution $x_*(\delta)$ to the optimization problem $$ \left\{ \begin{array}{ll} \textrm{minimize}& \frac{1}{2}\|x\|^2\\ \textrm{subject to}& A(\delta) x=b. \end{array}\right. $$ Question: Is the map $\delta\mapsto x(\delta):\Delta \rightarrow \mathbb{R}^n$ continuous?

Answer 1

For a counterexample with $m=n=2$, consider the case $$ A(\delta) = \pmatrix{\delta - 1 & 1\cr -1 & 1\cr},\ b = \pmatrix{1\cr 1\cr}$$ For $\delta \ne 0$, $Ax = b$ has the unique solution $x(\delta) = \pmatrix{0\cr 1\cr}$. For $\delta = 0$, the solutions are $\pmatrix{t\cr t+1}$, whose norm is minimized at $t = -1/2$, so $x(0) = \pmatrix{-1/2\cr 1/2\cr}$.

Answer 2

Yes if all matrices have the same rank. Your $x_*(\delta)$ is $A(\delta)^+b$ where $A^+$ is the Moore-Penrose pseudoinverse of $A$, which depends continuously on $A$ if (and only if) restricted to matrices of the same rank.

For the more general case, I don't know.