Continuity of solutions to convex optimization problems

Question

Let $x_A$ solve $$ \min J(x) \quad \text{subject to} \quad Ax=b $$ and $x_B$ solve $$ \min J(x) \quad \text{subject to} \quad Bx=b $$ given that $\|A-B\|_\text{operator} \leq \epsilon$ and that $J$ is convex (though not necessarily differentiable) , what can I say about $\| x_A - x_B \|_2$ ?

Answer 1

Let $J(x) = x_1^2+(x_2-10)^2$. Let $b=0$, $A_\epsilon=\begin{bmatrix}0 & \epsilon\end{bmatrix}$, $B_\epsilon=\begin{bmatrix}\epsilon & 0\end{bmatrix}$. Then if $\epsilon>0$, $x_{A_\epsilon}=\binom{0}{0}$, $x_{B_\epsilon} = \binom{0}{10}$. The norm $\|A_\epsilon-B_\epsilon\|$ can be made as small as you want, but $\|x_{A_\epsilon}-x_{B_\epsilon}\|_\infty = 10$.