C[0,2]\\

f:[0,2]->c\\

i dont know how to solve the complement.

i tried to prove there easier version without the complement first.

<f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\

A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g( 1)+f(2)g(2)=<f,g>

B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+x f(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg>

C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y( 2))g(2)

step d:

<x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2)

i dont know how to prove that

<x,x> greater or equal 0

i dont know the values of x

??