inner product space and polynomial Let $V = \mathrm{span}\{1,x,x^2,x^3\}$ be a real inner product space with the
inner product defined by 
$$
\langle f,g\rangle =\int\limits_{-1}^{1} fg 
$$
Check that $T(f) = f(0)$ is a linear functional on this space and find the element of $V$ which represents it (corresponds to).
So how should I do it? From where to begin?
I am guessing the first part is easy because I have to prove that 
$$T(f+g)=(f+g)(0)=f(0)+g(0)=T(f)+T(g)$$
and that
$$T(af)=af(0)=aT(f)$$
but what about this vector that I need to find?
I know that it has something to do with Gram-Schmidt but I do not know how.
 A: As you said, in order to prove that $T[f] = f(0)$ is a linear functional you just need to check the two conditions of linearity, ie.
$$T[f + g] = (f + g)(0) = f(0) + g(0) = T[f] + T[g]$$ and
$$T[af] = af(0) = aT[f]\qquad \forall a \in K = \mathbb{R}\ \text{or}\ \mathbb{C}.$$
Now you are asked to find an element representing a linear functional, therefore I suppose you are asked to use the Riesz Representation Theorem. This is what I am talking about: http://en.wikipedia.org/wiki/Riesz_representation_theorem.
Roughly speaking, you are asked to look for the vector $g$ satisfying the equality
$$Tf = (f,g).$$
Before doing this just a couple of observations:
$1)$ I suppressed the notation [] as is usually done with linear operators.
$2)$ I am assuming that the field is $\mathbb{R}$ only because there is no conjugacy in the integral defining the inner product.
$3)$ The existence of such a $g$ is given by the above Theorem by Riesz. (It is also unique!)
Finally the problem is just a matter of calculations: let $$g = a x^3 + bx^2 + cx + d.$$ It is determined if we determine its coefficients. To do this just remember that the integral of an odd function over an interval symmetric with respect to the origin equals $0$ and it is two times the integral on the positive subinterval if the function is even. In formulas, if $f(x) = -f(-x)$ then $\int_{-1}^1f(x)dx = 0$, while if $f(x) = f(-x)$ we have $\int_{-1}^1f(x)dx = 2\int_0^1f(x)dx$.
Now let's compute $\int_{-1}^1f(x)g(x)dx$ for some (non-random) $f$.
If f$(x) = 1$ then we impose
$$1 = f(0) = Tf = \int_{-1}^1(a x^3 + bx^2 + cx + d)dx = 2\int_0^1(bx^2 + d)dx = \frac{2}{3}b + 2d.$$
If $f(x) = x^2$ then we impose
$$0 = f(0) = Tf = \int_{-1}^1(ax^5 + bx^4 + cx^3 + dx^2)dx = 2\int_0^1(bx^4 + dx^2)dx = \frac{2}{5}b + \frac{2}{3}d$$
If can now find $b$, $d$ by solving the system
$$1 = \frac{2}{3}b + 2d$$
$$0 = \frac{2}{5}b + \frac{2}{3}d.$$
You now you should be able to continue by yourself, and you should also have an idea of what could be nice choises for $f$ to find $a$ and $c$.
I hope this helps!
A: Hint:
Find an orthonormal basis for $\,V\,$ using Gram-Schmidt, say the basis $\,\{u_1,...,u_4\}\,$ you get by applying G-S to the given basis of $\,V\,$ . From here, you get that any element $\,w\in V\,$ has a representation (I'm copying (one of) the proof(s) of Riesz Representation Theorem):
$$w=\sum_{k=1}^4\langle w,u_i\rangle u_i\implies Tw=\sum_{k=1}^4\langle w,u_1\rangle Tu_i=\sum_{k=1}^4\langle w,T(u_i)\,u_i\rangle u_i$$
Well, define now
$$u:=\sum_{k=1}^4 T(u_i)\,u_i\;\ldots\ldots$$
