I'm on trouble proving the next analysis proposition:
Let $(H,||\cdot||_q)$ be a Hilbert Space (over $\mathbb{R}$). Let $w_1\in W<H$ ($W$ is a subspace of $H$) and $v\in H$, such that $v-w_1\in W^{\perp}$. Prove that $w_1$ is the best approximation of $W$ to $v$ , i.e., prove that $||w_1-v||_q\leq ||w-v||_q \quad \forall w\in W$
My attempt.
I really tried to bring on all the learned machinery. For example, the Pythagorean theorem:
$||w_1-v+w||_q^2 =||w_1-v||_q^2 +||w||_q^2 \quad$ for all $w\in W$.
Then using triangle inequality over the term in the left (arriving to nothing, of course):
$||w_1-v+w||_q^2=||w_1+w-v||_q^2\leq ||w_1||_q^2 +2||w_1||_q||w-v||_q +||w-v||_q^2$
I also found certain equalities, for example, since $v-w_1\in W^{\perp}$, in particular for $w_1\in W$ we have $v-w_1\perp w_1$, i.e. $q(v-w_1,w_1)=0 \quad$ so $\quad q(v,w_1)=q(w_1,w_1)=||w_1||_q^2$.
But I'm really failing in the process of putting all the pieces together, i don't know how to keep incorporating this information. So any advice would be really appreciated. Thank you.
Choose $w \in W$, then $w-w_1 \bot v-w_1$ so by Pythagoras we have $\|v-w\|^2 = \|v-w_1 - (w-w_1)\|^2 = \|w-w_1\|^2 + \|v-w_1\|^2$.
In particular, $\|v-w\|^2 \ge \|v-w_1\|^2$ for all $w \in W$.