Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have the following optimization problem:

$$\text{find } x= \min_{a} \max_{\lambda\in\Lambda} |R(h\lambda)|$$

where $\Lambda$ is some finite, fixed set of complex numbers and $R(z)$ is a polynomial of degree at most $s$ of the form

$$R(z) = 1 + z + \sum_{j=2}^{s}a_j z^j$$

where $h>0$ is fixed and the real coefficients $a_j$ are the decision variables. Can it be shown that $x$ is a differentiable function of $h$? Are there general tools that might be useful for proving this type of property?

share|improve this question
    
Your question isn't clear. Are you fixing $h$ and then minimizing over $a$, or are you actually minimzing over $h$ and $a$? If you're minimizing over $h$, then $x$ isn't a function of $h$. –  Brian Borchers Jun 29 '12 at 13:29
    
Thank you for catching the ambiguity -- I've corrected it ($h$ is fixed). –  David Ketcheson Jun 30 '12 at 20:50
    
Do you really want $x$ be to the value of $a$ that minimizes $\max_{\lambda \in \Lambda} | R(h\lambda) |$, or do you want $x$ to be the minimal objective value? –  Brian Borchers Jun 30 '12 at 21:06
    
No, $x$ is the minimum over $a$ of the maximum over $\lambda$ of the modulus of $R(h\lambda)$, as stated. In any case, $a$ is a vector. Is it not clear as written? –  David Ketcheson Jul 1 '12 at 9:35

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.