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I have the following parametric constrained quadratic optimization problem:

$\min_{x\in\mathbb{R}^n} \dfrac{1}{2}x^TK(\alpha)x+c(\alpha)^Tx$ subject to $Ax\preceq b$, $Cx=d$

where $\alpha\in[M_1,M_2]$ is the parameter with $0<M_1<M_2<\infty$ and $K(\alpha)\succeq 0$.

This is a convex problem and suppose for some $\alpha_0\in[M_1,M_2]$, the optimal solution to this problem is $x_{\alpha_0}$. Informally, my goal is to show that if I perturb $\alpha_0$ to some $\alpha_1\in[M_1,M_2]$ in this problem then $\|x_{\alpha_1}-x_{\alpha_0}\|\le \beta|\alpha_1-\alpha_0|$ where $\beta>0$ is a constant independent of $\alpha$.

I would really appreciate any hint or references which may be useful to get some sufficient conditions that guarantee this sort of a result. Thanks!

Ergodic
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  • You can get an upper bound on the distance by dropping the constraints. At that point, the distance you want is equal to $|K(\alpha_0)^{-1}c(\alpha_0) - K(\alpha_1)^{-1}c(\alpha_1)|$. You should be able to bound from there given Lipschitz assumptions on $K$ and $c$. – cdipaolo Sep 05 '19 at 04:10
  • Note that the first sentence above applies because you can rewrite a quadratic objective function as a distance from the variable to some fixed point in an inner product metric, so adding constraints just gives a projection onto a convex set, which is a (weak) contraction. – cdipaolo Sep 05 '19 at 04:12
  • @cdipaolo: Thanks! The $K(\alpha)$ here positive semi-definite and so the inverse will not exist in this case. – Ergodic Sep 05 '19 at 04:16

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