I was doing some exercises of linear functionals, then I spend hours on one item of one of those exercises. Well, the first item I did it. I'm struggling on the second item. Anyway, i'll post them here.
On the three dimensional vector space of real quadratic polynomials in x, define the linear functional $F(f) = \int_0^1 f(x)dx$. Suppose that $1, x, x^2$ are an orthonormal basis, then what vector A represents this functional F, so that $A\cdot v=F(f)$, where the vector v means a quadratic polynomial.
In the preceding example, take the scalar product to be $(f,g)=\int_{-1}^1f(x)g(x)dx$ and find the vector that represents the functional in this case.
Well, I was doing this exercise replacing one of the functions $f$ or $g$ with the polynomial basis, i.e., $$(x,g)=\int_{-1}^1x(a+bx+cx^2)dx=(2/3)b=1.$$ So I was able to find $b$ but I could not find $a$ and $c$.
In the first case, you have $$ F(a+bx+cx^2)=a+\frac b2+\frac c3=(a+bx+cx^2)\cdot\left(1+\frac x2+\frac {x^2}3\right). $$ In the second one, you have $$\tag{1} 1=F(1)=\int_{-1}^1(a+bx+cx^2)=2a+\frac{2c}3, $$ $$\tag{2} \frac12=F(x)=\int_{-1}^1(ax+bx^2+cx^3)=\frac {2b}3, $$ $$\tag{3} \frac13=F(x^2)=\int_{-1}^1(ax^2+bx^3+cx^4)=\frac {2a}3+\frac{2c}5. $$ From $(2)$ we obtain $b=3/4$. From $(1)$ and $(3)$ we get the system $$ 6a+2c=3,\ \ \ 10a+6c=5, $$ with solution $a=1/2$, $c=0$. That is, $$ A=\frac12+\frac{3x}4. $$