No, that is the inner product in Euclide space, not Cartesian product.
the Norm is induced by the inner product as:
$\displaystyle |(r,f(x))|=\sqrt{|r|^2+\int_0^1|f(x)|^2 dx}$
The element of $\displaystyle \mathbb{C}\times L^2$ is not function, but a vector consisting of a complex number and a function. If you have problem about Cartesian Product, please turn to any function analysis book or set theory.
It is impossible to define a norm for function which has the required form and property.
I am sure I did not make any mistake, I've check it for several times.
Do notice that: the given basis is not a orthogonal basis of the closed subspace.
actually the answer is much simpler
i was expexting something like this
http://i45.tinypic.com/2jfhv9z.jpg
out of one of your posts
and as you see its nothing like your definition
do you know how to get to this formula from the given norm?
well, "two Defferent paths, the same destination"
Every route reaches Rome.
You can give you comment on these two method after you get the final result, and please let me know if you get the same result ,OKA?
I have been convinced for your solution ,I take back part of my words , finally thank you for your question and reply.
Think about it in a inverse direction, we can find some hint(or indication ) which lead to the given definition of the solution.
further, My method can be generalized to a more general case, It extends the dimension of the Hilbert Space and the space itself.
OK，I will do it in my way,then do it again in your way,and you will see that there is a clear corresponding between these two definition and method.
My way to solve it:
define the inner product as above mentioned In $\displaystyle \mathbb{C}\times L^2$, Then
The problem is equivalent to "Find the optimal solution and the distance between point $\displaystyle (1,3x^2)$ and the
closed subspace spanned by three vectors$\displaystyle (1,0),(1,1)$ and $\displaystyle (1,2x)$ in $\displaystyle \mathbb{C}\times L^{2}$."
By using Schimidt orthognalization, we get: $\displaystyle e_1=(1,0),e_2=(0,1)$ and
$\displaystyle e_3=\sqrt{3}(0,2x-1)$ are the orthogonal base in the colsed subspace spanned by those three vectors.
let $\displaystyle t=(1,3x^2)$, then the minimal point is
$\displaystyle <t,e_1>e_1+<t,e_2>e_2+<t,e_3>e_3$ that is,$\displaystyle (1,3x-\frac{1}{2})$ Thus we get:
$\displaystyle \alpha+\beta+\gamma=1$
$\displaystyle \beta+2\gamma x=3x-\frac{1}{2}$
which lead to $\displaystyle \alpha=0,\beta=-\frac{1}{2},\gamma=\frac{3}{2}$ .
Do it again in your way:
define the inner product as your definition.
The problem is equivalent to find the optimal solution and the distance between point $\displaystyle x^3$ and the closed subspace spanned by $\displaystyle 1(constant function),x,x^2$ in $\displaystyle L^2$.
Since $\displaystyle v_1=1,v_2=x-1,v_3=\sqrt{3}(x^2-x)$ are the orthogonal base in the closed subspace,(can you see the connection between these two basis,and the corresponding between these two methods),By using the Projection Theorem In Hilbert space, we get the same result: $\displaystyle \alpha=0,\beta=-\frac{1}{2},\gamma=\frac{3}{2}$ .
If you completely understand my way, you will see process from the norm to inner product, the $\displaystyle f(1)\overline{g(1)}$ part in the definition just simplify the process and computaion.
I thought it is impossible to define the inner product and norm in $\displaystyle L^2$ which has the required form and property, So I need to extend $\displaystyle L^2$ to a "bigger" space, and then define the inner product and norm. You can see your inner product is just a special case of my definition, Your Space is a closed subspace of the extended space.
$\displaystyle (1,3x^2) $ is the given point in the extended space, consider $\displaystyle \alpha,\beta,\gamma$ as the coefficients of the linear combination of the three vectors, thus 1 comes from here.
If you still have problems please review the Extended Space and its inner product definition(similar to $\displaystyle R^n$).