Find $a,b,c \in \mathbb{C}$ such that the integral $$ \int_{-1}^1 |x^3 - a -bx -cx^2|^2 dx$$ is minimized.
If we set a function $f: \mathbb{C}^3 \to \mathbb{R}$ equal to this integral and then evaluate it, we arrive at $$f(a,b,c) = 2a^2 + \frac{4ac}{3} + \frac{2b^2}{3}-\frac{4b}{5} +\frac{2c^2}{5} +\frac{2}{7}$$
My first thought was to evaluate each partial derivative and set them equal to zero and solve the system for a,b, and c. However, when I do this I end up getting an inconsistent system for a and c, which I know is incorrect. I am not quite sure how to continue from here.
NOTE: We have recently started working with Hilbert Spaces and I have the following theorems that I believe can be helpful, but I am not sure how to implement them:
If $\mathbb{H}$ is a Hilbert Space and $\mathbb{E}$ is a closed subspace of $\mathbb{H}$, then $\exists y \in \mathbb{E}$ such that $\forall x \in \mathbb{H}$ $||x-y|| = \inf\limits_{z \in \mathbb{E}}||x-z||$
If $\mathbb{H}$ is a Hilbert Space and $\mathbb{S}$ is a convex closed subset of $\mathbb{H}$, then $\exists ! y \in \mathbb{S}$ such that $\forall x \in \mathbb{H}$ $||x-y|| = \min\limits_{z \in \mathbb{S}}||x-z||$
Both of these say nearly the same thing and I don't see the difference in the result
Let $\mathbb{H} = \mathbb{C}_3[X]$ with the scalar product $$\langle P, Q\rangle = \int_{-1}^{1} P(x)\bar{Q}(x) \text{d}x$$ $\mathbb{H}$ is an hilbert space. Let $\mathbb{E} = \mathbb{C}_2[X]$ and $P(X) = X^3$. So we want to compute the orthogonal projection of $P$ onto $\mathbb{E}$. Let $a, b, c \in \mathbb{C}$ such that $Q(X) = a + bX + cX^2$ the projection of $P$ onto $\mathbb{E}$. Using $P -Q$ is orthogonal to $\mathbb{E}$. So $$\left\{\begin{array}{ccl}\int_{-1}^1 (x^3 - a - bx - cx^2)\text{d} x &=& 0\\ \int_{-1}^1 (x^3 - a - bx - cx^2)x\text{d} x &=& 0\\ \int_{-1}^1 (x^3 - a - bx - cx^2)x^2\text{d} x &=& 0\end{array}\right.$$
$$\left\{\begin{array}{ccl}a + \frac{1}{3}c &=& 0\\ \frac{1}{3}b &=& \frac{1}{5}\\ \frac{1}{3}a + \frac{1}{5}c &=& 0\end{array}\right.$$
So $a = 0, c = 0, b = \frac{3}{5}$