let $W$ be an inner product space, $\dim W = n$, and let $ψ: W → \mathbb{R}$ be linear.
Then there exists a unique vector $w_0 ∈ W$ such that $ψ (w) = (w, w_0)$ for all $w ∈ W$,
While $w_0 = \sum_{i=0}^n ψ (u_i)u_i$, where {$u_1, . . . , u_n$} is an orthonormal basis for $W$.
Here is my way of proof (I wrote $T$ as $\psi$ for readability):
Let $w \in W$.
$(w,w_0)=(w,T(u_1)u_1+...+T(u_n)u_n)=(w,T(u_1)u_1)+...+(w,T(u_n)u_n)=$
$T(u_j) \in \mathbb{R}$ so:
$=(w,u_1)T(u_1)+...+(w,u_n)T(u_n)=T((w,u_1)u_1+...+(w,u_n)u_n)=T(w)$
I will appreciate if someone can make this statment more clear for me to understand.
Thanks a lot!