# Excercise of Convex Analysis

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

1. $x_1$ is the projection of $x$ to $K$ and $x_2$ is the projection of $x$ to $K^*$, where $K^*$ is the polar cone of $K$.
2. $x = x_1 + x_2$, 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 $\mathbb{R}^n$.

• Which definition of $K^*$ ? – MathOverview Oct 22 '13 at 19:04

(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 $K^*$.