Consider the following maximization problem.

$$ \begin{aligned} &\max_{x \in \mathbb{R}^n}x^TPx\\ &a_i \leq x_i \leq b_i,~ \forall i \in \mathbb{Z}_{[1,n]}\,, \end{aligned} $$ where $x_i$ is the $i$-th scalar component of vector $x$ and $P>0$ is positive definite. As the objective is a convex function, the maximizer should be where all constraints are active, i.e. at an extreme point of the domain of the form $x^\star_i=a_i$ or $x^\star_i=b_i$. Is the global maximizer the one of such points such that its norm ${x^\star}^Tx^\star$ is maximal?


1 Answer 1


Maximizing $\vec x^T\vec x$ is not generally equivalent to maximizing $\vec x^TP\vec x$. As a simple counterexample, take this choice of parameters for $\vec x\in\mathbb R^2$.

$$P=\begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}$$

$$-1\le x_1\le 2$$

$$-1\le x_2\le 2$$

We see that $\vec x^T\vec x$ is maximized at $\vec x=\begin{bmatrix}2\\2\end{bmatrix}$, while $\vec x^TP\vec x$ is maximized at $\vec x=\begin{bmatrix}2\\-1\end{bmatrix},\begin{bmatrix}-1\\2\end{bmatrix}$.


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