Let $W$ be a subespace of $V$, the best approximation for $v\in V$ for vectors of $W$ is a vector $w\in W$ that satisfies $||v-w||\leq||v-u||$ for every $u\in W$.
Prove:
$w\in W$ is the best approximation for $v\in V$ for vectors in $W$ iff $v-w\in W^{\perp}$
I was able to show the $\Leftarrow $ direction of the sentence, but I'm having problems using $||v-w||\leq||v-u||$ to show that $v-w\in W^{\perp}$.
I tried to write $||v-w||\leq||v-u||$ in terms of inner products using that $||v-w||^2\leq||v-u||^2$:
$<v-w,v-w>\leq <v-u,v-u>$
$<v,v-w>-<w,v-w>\leq <v,v-u>-<u,v-u>$
$\overline{<v,v>}-\overline{<w,v>}-\overline{<v,w>}+\overline{<w,w>}\leq \overline{<v,v>}-\overline{<u,v>}-\overline{<v,u>}+\overline{<u,u>}$
$||v||+||w||-\overline{<w,v>}-\overline{<v,w>}\leq ||v||+||u||-\overline{<u,v>}-\overline{<v,u>}$
$||w||-<v,w>-<w,v>\leq ||u||-<v,u>-<u,v>$
I don't know how to continue, some hint? Thanks!
Suppose that $w$ is the best approximation for $v$ in the subspace $W$, i.e. $||w-v||\leq ||u-v|| \quad \forall u\in W$. Suppose $w-v\notin W^\perp$, let $P$ be the orthogonal projector onto $W$, then you may write $v=P(v)+(v-P(v))$ with $P(v)\in W$, $v-P(v)\in W^\perp$. We get
$$||w-v||^2=||w-P(v)||^2+||P(v)-v||^2\geq ||P(v)-v||^2 \tag{1}\label{eq1}$$
because $w-P(v)\in W$ and $P(v)-v\in W^\perp$. So
$$||P(v)-v||\leq ||w-v||\leq \inf_{u\in W}||u-v||,$$
that means that $P(v)$ is an approximation for $v$ in $W$ at least as good as $w$, since the inequality in \eqref{eq1} is strict if $w\neq P(v)$ we get $w=P(v)$, finally $w-v=P(v)-v\in W^\perp$.