Let $\left \langle .,.\right\rangle$ be an inner product of $\Re^n$ and {$p_i$|$i=0,1,..., n-1$} be a basis of $\Re^n$ satisfying $\left \langle p_i,p_j\right\rangle=0$ if $i\not=j$. For any $i=0,1,..., n-1$, let $$K_i=span(p_0, p_1,...,p_{i-1})$$
A) Then for any $x\in K_i$ and $x^*\in\Re^n$ $$\left \langle x-x^*,x-x^*\right\rangle\ge \sum_0^{n-1} |a_k|^2,$$ where $$a_k=\frac{\left \langle x^*,p_k\right\rangle}{\left \langle p_k,p_k\right\rangle}$$
B) The vector $$x_i=\sum_0^{i-1}a_kp_k$$ minimises the finction $f(x)=\left \langle x-x^*,x-x^*\right\rangle$ defined for $x\in K_i$
For part A, i let $$x=\sum_0^{i-1}b_jp_j$$ and $$x^*=\sum_0^{n-1}a_jp_j$$ thus $\left \langle x-x^*,x-x^*\right\rangle=\sum_0^{i-1}{(b_j-a_j)^2\left \langle p_j,p_j\right\rangle}+\sum_i^{n-1}a_j^2\left \langle p_j,p_j\right\rangle$. But I am stuck after this. Am I on the correct path?
Also I do not know what I need to do for B)
Your help is greatly appreciated
There's a typo at the indices of the summation: we are about to prove $$\langle x-x^*, \, x-x^*\rangle \ \ge\ \sum_{k={\bf i}}^{n-1}a_k^2 $$ And, you just proved it, as each $(a_k-b_k)^2\ge0$, for $k< i$.
What is not written yet, is that the $a_k$'s as defined in the question, coincide with the coordinates of $x^*$.
Indeed, suppose $x^*=\sum_kc_kp_k$. Then $$\langle x^*, p_k\rangle\ =\ c_k\langle p_k, p_k\rangle\,. $$
For B, we consider $x^*=\sum_ka_kp_k$ fixed. Just observe that $f(x_i)=\sum_{k\ge i}a_k^2$, which is the minimal possible value for $f(x)$ by A.
To visualize it, let $p_1,p_2,p_3$ be arbitrarily long (but nonzero) vectors in the space of directions along the $x, y, z$ axes. Then $K_2$ is just the $x, y$ plane, and for any vector $x^*=(a,b, c) $, its orthogonal projection to $K_2$ is just $(a, b, 0)$ which is the closest point to $x^*$ within the plane..