Show that projections on a convex closed cone and its polar cone are orthogonal Let be $ K \subset \mathbb{R}^n $ a convex and closed cone, $ x \in \mathbb{R}^n $. Show that the following asserts are equivalent:


*

*$x_1$ is the projection of $x$ to $ K $ and $x_2$ is the projection of $x$ to $ K^* $, where $ K^* $ is the polar cone of $ K $.

*$ x = x_1 + x_2 $, where  $\: x_1 \in K, \: x_2 \in K^*$ and $\: \langle x_1, x_2 \rangle = 0 $


Note: $ \langle , \rangle $ is the inner product of $ \mathbb{R}^n $.
 A: (1) $\Rightarrow$ (2): It is known that $0 \in K$ since $K$ is a closed cone and $0 \in K^*$.
If $x_1$ is the projection of $x$ to $K$ then
$ \langle x-x_1 , x_1-y \rangle \geq  0$, $\: \forall \: y \in K$.
In particular, for $y=0$:  $\: \langle x-x_1 , x_1 \rangle \geq  0$. 
And for $y=2x_1$: $\: \langle x-x_1,-x_1 \rangle \geq  0 \:$ then $\: \langle x-x_1 , x_1 \rangle \leq  0$ 
Thus  $\: \langle x-x_1 , x_1 \rangle =  0$
Similarly, if $x_2$ is the projection of $x$ to $K^*$ then
$ \langle x-x_2 , x_2-z \rangle \geq  0$, $\forall \: z \in K^*$.
In particular, for $z=0$:  $\: \langle x-x_2 , x_2 \rangle \geq  0$.
And for $z=2x_2$: $\: \langle x-x_2,-x_2 \rangle \geq  0 \:$ then $\: \langle x-x_2 , x_2 \rangle \leq  0$ 
Thus  $\: \langle x-x_2 , x_2 \rangle =  0$
Affirmation: $x-x_1=x_2$
Indeed, let be $y \: \in K^*$ then $\: \langle x-(x-x_1), (x-x_1)-y \rangle = \langle x_1, (x-x_1)-y \rangle$ 
$\langle x-(x-x_1), (x-x_1)-y \rangle = \langle x_1,x-x_1 \rangle + \langle x_1,-y \rangle = \langle x_1,-y \rangle \geq  0$.
Ergo, $\: \langle x-(x-x_1), (x-x_1)-y \rangle \geq  0$, $\: \forall \: y \in K$ $\:\: \Rightarrow \: x-x_1=x_2$
i.e. $\: x=x_1+x_2$
Finally: $\: \langle x_1,x_2 \rangle = \langle x-x_2,x_2 \rangle = 0$.
(2) $\Rightarrow$ (1): Let be $y \: \in K$ then $\: \langle x-x_1,x_1-y \rangle = \: \langle x_2,x_1-y \rangle = \langle x_2,x_1 \rangle + \langle x_2,-y \rangle = \langle x_2,-y \rangle \geq  0$.
Namely, $\: \langle x-x_1,x_1-y \rangle \geq  0$, $\: \forall \: y \in K$.
From this it is concluded that $x_1$ is the projection of $x$ to $K$.
Similary, let be $z \: \in K^*$ then $\: \langle x-x_2,x_2-w \rangle = \: \langle x_1,x_2-w \rangle = \langle x_1,x_2 \rangle + \langle x_1,-w \rangle = \langle x_1,-w \rangle \geq  0$.
Namely, $\: \langle x-x_2,x_2-w \rangle \geq  0$, $\: \forall \: w \in K^*$.
From this it is concluded that $x_2$ is the projection of $x$ to $K^*$. 
