Let $y_1, y_2, . . .$ be a sequence in a Hilbert space.
Let $V_n$ be the linear span of $\{y_1, y_2, . . . , y_n\}$.
Assume that for $n \ge 1, ∥y_{n+1}∥ \le ∥y − y_{n+1}∥$ for each $y \in V_n$. Show that $\langle y_i, y_j\rangle = 0 \forall i \neq j.$oblem
Please give some hints on how to start the problem.I feel totally clueless
W.l.o.g I will assume $\|y_i\|=1$ for any $i$.
By induction, we only have to show that $\langle y_{n+1}, y_i \rangle = 0$ for any $i = 1, \dotsc, n$.
Consider $y = \sum_{i=1}^n \langle y_{n+1},y_i\rangle y_i$ and $z = y_{n+1}-y$. By assumption we have $\| y_{n+1} \|^2 \leq \|z\|^2$.
Computing $\|z\|^2=\langle z,z \rangle$ we get
$$\langle z,z \rangle = \|y_{n+1}\|^2-\sum_{i=1}^n \langle y_{n+1},y_i \rangle^2,$$
hence the inequality can only hold if $\langle y_{n+1},y_i \rangle=0$ for any $1 \leq i \leq n$.
This has the following geometric interpretation:
$\inf_{y \in V_n }\|y-y_{n+1}\|$ is the smallest distance from $y_{n+1}$ to $V_n$. The infimum is well known to be taken on, when $y$ is the orthogonal projection of $v_{n+1}$ on $V_n$, i.e. the assumption of your exercise states that the orthogonal projection of $y_{n+1}$ onto $V_n$ is zero, which means that $v_{n+1}$ is orthogonal to $V_n$.