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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 $\| x_A - x_B \|_2$ ?

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Sadly, you can say nothing. All you know is that the optimization occurs on one of the vertex points, so changing the slope every so slightly, could send you to the next vertex point, which is very far away. – Calvin Lin Jan 6 '13 at 0:57
Ok thanks. Do you know if there are some additional constraints I can place on A or B that would allow me to say something? – dranxo Jan 7 '13 at 3:35
up vote 3 down vote accepted

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 $\|x_{A_\epsilon}-x_{B_\epsilon}\|_\infty = 10$.

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