I read the book "Finite-Dimensional Variational Inequalities and Complementarity Problems I", and try to solve Exercise 1.8.16.
1.8.16. Let K be a closed convex set in $\Re^n$ and $A$ be a symmetric positive definite matrix of order $n$. Use (1.5.11) to show that the skewed projector is nonexpansive under the $A$-norm; i.e., $\|P_{K,A}(u)-P_{K,A}(v)\|_A \leq \|u-v\|_A, \forall u,v\in\Re^n$, where $\|\cdot\|_A$ is a $A$-norm, e.g., $\|x\|_A=\sqrt{x^T Ax}$.
The inequality (1.5.11) is $(P_{K,A}(v)-P_{K,A}(u))^T A(v-u) \geq (P_{K,A}(v)-P_{K,A}(u))^T A (P_{K,A}(v)-P_{K,A}(u))$
It does not seem difficult to show but I cannot find any way to solve this.
If someone already knows how to solve it, could you answer this question? Thank you.
For vectors $x, y \in \Bbb{R}^n$, let $$\langle x, y \rangle = y^\top A x.$$ As you may or may not know, this forms an inner product on $\Bbb{R}^n$. Linearity in the first coordinate follows from basic properties of matrix multiplication. Similarly, symmetry follows from $A$ being symmetric, and basic properties of the transpose operator. Finally, positive-definiteness of the inner product corresponds precisely to the positive-definiteness of $A$.
Further, the canonical norm for this inner product, defined by $\|x\| = \sqrt{\langle x, x \rangle}$, is precisely the $A$-norm.
So, inequality 1.5.11 says $$\langle v - u, P_{K, A}(v) - P_{K, A}(u)\rangle \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A^2.$$ Cauchy-Schwarz inequality states, $$\|v - u\|_A \cdot \|P_{K, A}(v) - P_{K, A}(u)\|_A \ge \langle v - u, P_{K, A}(v) - P_{K, A}(u)\rangle.$$ Therefore, $$\|v - u\|_A \cdot \|P_{K, A}(v) - P_{K, A}(u)\|_A \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A^2.$$ If $\|P_{K, A}(v) - P_{K, A}(u)\|_A = 0$, then non-expansiveness holds trivially. If not, then $\|P_{K, A}(v) - P_{K, A}(u)\|_A > 0$, so we can divide both sides by it, to get $$\|v - u\|_A \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A,$$ as required.