How get the upper bound (maximum) of a convex function with inequality constraints?

Question

Condition: $h,f\in \mathbb{C}^{N\times1}, \text{where}f =\hat{f} + e \text{ and } e^H e \leq 1,\ \ \ Q=h^Hff^Hh$. The function $ Q$ is convex.

Now I want to get the maximum (not minimum), i.e., the upper bound of $Q$ over the variable $e$. How to get it? Will the Lagrangian function of $Q$ be useful in this case?

The Lagrangian function of $Q$ is $\mathcal{L} = h^H(\hat{f} + e)(\hat{f} + e)^Hh + \lambda (e^H e - 1) $ which is convex, where the $\lambda \geq0$.

Thanks!

Didn't asked this question yesterday?: http://math.stackexchange.com/questions/511146/how-to-get-the-minimum-and-maximum-of-one-convex-function — user91011, Oct 02 '13 at 15:48
yes, I did ask.Because I did not give a clear expression about the question yesterday. so I did not get the answer which I want. — begforopt, Oct 02 '13 at 15:52
If the KKT conditions are used, How to understand the $\lambda \leq 0$? — begforopt, Oct 02 '13 at 15:57

score 1 · Accepted Answer · answered Oct 02 '13 at 16:12

The problem can be written as $\mu = \max_{\|e\| \le 1} | \langle h, \hat{f}+e \rangle |= \max_{\|e\| \le 1} | \alpha+\langle h, e \rangle |$, where $\alpha = \langle h, \hat{f}\rangle$.

It should be clear that $\{\langle h, e \rangle \}_{\|e\| \le 1} = \{z \in \mathbb{C} | |z| \le \|h\| \}$.

The problem reduces to $\max_{|z| \le \|h\|} |\alpha + z|$, and a maximizer is easily seen to be $z = \frac{\alpha}{|\alpha|} \|h\|$, if $\alpha \neq 0$, and $z=\|h\|$, otherwise.

Hence $\mu = |\alpha|+\|h\| = | \langle h, \hat{f}\rangle | + \|h\|$.

Thanks for your help, copper.hat! – begforopt Oct 03 '13 at 01:24 — begforopt, Oct 03 '13 at 01:24

How get the upper bound (maximum) of a convex function with inequality constraints?

1 Answers1