How can I compute the infimum of the following non linear functional 
I was trying to solve this problem from previous exam of functional analysis and I am stuck
Clealy $ \inf_{f \in \mathcal{M}}\phi(f) \leq 0$.(I can choose f=0) If I compute the infimum over the bigger set $L_{\mathbb{R}}^2[-1,1]$, the infimum should be $-2/5$. This follows from completing the square as follows
$\phi(f)=\int_{-1}^{1}((f(x)-x^2)^2 -x^4)dx$ and since $x^2 \in L_{\mathbb{R}}^2[-1,1]$
So I am certain that $\inf_{f \in \mathcal{M}} \phi(f) \in [-2/5,0]$
I am having a hard time narrowing it down to a number. The problem I am facing is that the set $\mathcal{M}$ clearly contains functions which are not necessarily odd. If the functions were odd I could easily conclude that the infimum was 0.
Can someone help me? Any hints would be appreciated. 
 A: Since 
$$
\phi(f) = \int_{-1}^1 (f(x)-x^2)^2 - x^4 \, dx = \int_{-1}^1 (f(x) - x^2)^2 \, dx - \int_{-1}^1 x^4 \, dx, 
$$
you essentially want to minimize the first integral. But the first integral is the $L^2$ norm of $\|f-x^2\|$ squared. We have a nice Hilbert basis for $L^2_{\mathbb R}([-1,1])$, and the subspace of functions with zero integral perhaps decomposes nicely with respect to this basis... I haven't worked out the details, feel free to ask questions. I just thought this was too big for a comment.
Hope that helps,
A: Let g be a continuous linear functional on a Banach space B.Let the norm of g be G and let the null space of g be M. If g is not zero then G has co-dimension 1 in B. From this,and from the definition of G, the absolute value of g(x) is G.d(f,M) for any f in B, where d(f,M) is the distance from f to M. So if g(f) is the integral of f from -1 to 1, then G = sqrt(2) .So for the function f(x)=x2 , we have d(f,M) = 2/(5.sqrt(2)). Combine this with Patrick Da Silva's work ,and you see that the inf in the question is -1/5.
