Prove the original Riesz Representation using the bilinear form

Question

In the linear space $R^n$ over $R$, we have an inner product which is the usual one. The problem is to use the following theorem to prove the Riesz Representation Theorem:

For any bilinear form $B(x,y)$ in $R^n$, there exists a linear operator $T$ such that $B(x,y) = <x,Ty>$.

Here is my effort:

Given a linear functional $S$, let $\epsilon_i=(0,0,0,...,1,...)$ whose $i$-th coordinarte is $1$ and others being $0$, then $\epsilon_i's$ form a basis. Let $y=\sum_{i=1}^{n}S(\epsilon_i)\epsilon_i$, then $S(x)=<x,y>$ for any $x$.

I am wondering how to use theorem to prove the Riesz Representation Theorem.

Any suggestion would be appreciated!

Answer 1

Given a linear functional $S$ define $B(x,y)=(Sx)(Sy)$. This is a bilnear functional. Hence there exists a linear operator $T$ such that $B(x,y)=\langle x, Ty \rangle$. Now, if $S$ is not the zero functional we can choose $y$ such that $Sy=1$. We then get $Sx=(Sx)(Sy)=B(x,y)=\langle x, Ty \rangle$ and this proves Riesz Theorem. [ If $S=0$ then $Sx=\langle x, 0 \rangle$].