Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\| x_A - x_B \|_2$ ?

share|improve this question
    
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.