inner product of vector and its projection

Question

Let us consider Hilbert space $(\mathbb{R}^n, \langle \cdot ,\cdot\rangle_w)$, where $w\in\mathbb{R}^n$ and $w_i > 0$ and inner product $\langle x,y\rangle = \sum_{i=1}^n x_i y_i w_i$. Let $C$ be a closed convex cone in $\mathbb{R}^{n}$ and let $\Pi(x\mid C)$ be the projection of $x$ onto the cone $C$.

Is it always true that $\langle \Pi(x\mid C),x\rangle \geq 0$?

Answer 1

We have $\|x-\prod (x|C)\| \leq \|x\|$ since $0 \in C$. Squaring both sides and expanding we get $\|\prod (x\mid C)\|^2-2 \langle x, \prod (x\mid C) \rangle \leq 0$. Hence $\langle x, \prod (x\mid C) \rangle \geq 0$.

Answer 2

I think I have proved this fact.

Let $Z = \Pi(X\mid C)$, which means that $\displaystyle Z = \operatorname*{argmin}_{Y\in C}\langle(X-Y),(X-Y)\rangle$. Assume also that $\langle Z,X \rangle \,\, < \,\, 0$. Then $$ \langle (X-Y),(X-Y)\rangle = \langle X,X \rangle - 2 \langle Z,X \rangle + \langle Z,Z \rangle. $$ Next, let us take vector $Z_{1} = aZ \in C$, where $0<a<1$. Then $$ \langle(X-Z_{1}),(X-Z_{1}) \rangle = \\ \langle X,X \rangle - 2 \langle Z_{1},X \rangle + \langle Z_{1},Z_{1} \rangle =\\ \langle X,X \rangle - 2a \langle Z,X \rangle + a^{2} \langle Z,Z \rangle \,\, <\,\,\\ \langle X,X \rangle - 2 \langle Z,X \rangle + \langle Z,Z \rangle. $$ Therefore, $Z$ is not a projection.