I'm reading this theorem in my lecture notes
My questions:
If $x^\star = 0$ then $f(x^\star) = \langle x^\star, x^\star \rangle = \langle 0, 0\rangle = 0$. How can I infer that $x^\star$, the projection of the origin $0$ on $A$, is not $0$?
Since $f(x) := \langle x^\star, x \rangle$. Should it be $f(c) := \langle x^\star, c \rangle$ rather than $f(c) := \langle c, x^\star \rangle$?
Thank you for your help!
1) $C$ and $D$ are disjoint so $0$ does not belong to $A$. Hence the projection on $A$ cannot be $0$.
2) There is no difference between $\langle a, b \rangle$ and $\langle b, a \rangle$.