Let $K$ be a closed convex set in $\mathbb{R}^n$, $K^*$ be the dual cone of $K$, and $\prod_K(x)$ denote the Euclidean projection onto $K$.
Show that $\prod_K(x)+\prod_{-K^*}(x)=x$ $\forall x \in \mathbb{R}^n$.
This decomposition is called Moreau decomposition. Intuitive example is projection of a point onto the first and third quadrants because the first quadrant $K$ is a cone and its dual $K^*$ is itself so $-K^*$ is the third quadrant.
The following from characterization of projection holds for every $p \in K$
$$ \langle x-\Pi_K(x),p - \Pi_K(x) \rangle \leq 0 $$
$p=0 \in K$ so $ \langle x-\Pi_K(x), - \Pi_K(x) \rangle \leq 0 $
$p=\Pi_K(x) \in K$ then $2\Pi_K(x) \in K$ so VI $ \langle x-\Pi_K(x), \Pi_K(x) \rangle \leq 0 $
Hence, $$ \langle x-\Pi_K(x), \Pi_K(x) \rangle = 0 $$
what is left is to show $\Pi_{-K^*}=x-\Pi_K(x)$.
The point is ${-K^*}={K^{\circ}}$.
Claim: $z=x-\Pi_K(x)$ is in $K^{\circ}$.
$$ \langle x-\Pi_K(x),p - \Pi_K(x) \rangle \leq 0 \rightarrow \langle z,p - \Pi_K(x) \rangle = \langle z,p \rangle + \langle z,\Pi_K(x) \rangle = \langle z,p \rangle \leq 0 $$ $\Box$
For any $q$ in $K^{\circ}$ we have $\langle q, \Pi_K(x) \rangle \leq 0$. Because $ \langle z, \Pi_K(x) \rangle = 0 $, we have $\langle q, \Pi_K(x) \rangle - \langle z, \Pi_K(x) \rangle \leq 0$
Since $\Pi_K(x)=x-z$,
$$ \langle q- z, x-z \rangle \leq 0$$
The above inequality means $q=\Pi_{K^{\circ}}(x)$.