$\inf\int_0^1|f(x)-x|^2dx$

Question

So this is an old prelim problem. It asks me to find $\inf\int_0^1|f(x)-x|^2dx$ where $f\in L^2([0,1])$ are the functions that satisfy $\int_0^1 f(x)(x^2-1)dx=1.$ Now I tried manipulating the expression $\int_0^1|f(x)-x|^2dx$ to incorporate our conditions on $f$, but I got nowhere. Are there some hints you guys have for how to proceed with this problem?

Answer 1

This is like the classical case of finding the distance from a point $(u_0,v_0,w_0)$ to the plane $a u + b v + c w + d = 0$, that is $(a,b,c)(u,v,w) + d = 0$. The distance there is $$\left| \frac{ a u_0 + bv_0 + c w_0 + d}{\sqrt{ a^2 + b^2 + c^2}}\right |$$ In your case the distance is $$\left|\frac{\langle x, x^2 -1\rangle -1}{\|x^2 -1\|}\right|$$ The square of that is what you need, it's even simpler.

$\bf{Added:}$ Let's prove the result: the distance from the vector $v$ to the hyperplane $H_{u, \alpha}\colon =\{ w, \langle w, u \rangle = \alpha\}$ equals $$\left | \frac{ \langle v, u\rangle - \alpha }{\|u\|}\right |$$

Indeed, consider $v_0$ the closest point in $H$ to $v$. The vector $v-v_0$ is perpendicular to $H$, so has the same direction as $u$. Therefore $$d(v, H) = \|v-v_0\| = \left |\langle v-v_0, \frac{u}{\|u\|}\rangle \right|= \left|\frac{\langle v- v_0, u\rangle }{\|u\|} \right|= \left | \frac{ \langle v, u \rangle - \alpha }{\|u\|}\right |$$