Finding the shortest vector in a subspace of an inner product space

Question

In the inner product space $L^2$[$0,1$], find a shortest vector to $f(x)=sin(πx)$ in the subspace $W=[a_0+a_1x+a_2x^2|a_0,a_1,a_2∈$R$]$. How does one prove that the vector found is a shortest one?

Answer 1

We know that the shifted Legendre polynomials $P_n(2x-1)$ give an orthogonal base of $L^2(0,1)$ with the usual inner product: $$ \int_{0}^{1}P_n(2x-1)P_m(2x-1)\,dx = \frac{\delta(n,m)}{2n+1} $$ so the projection of $\sin(\pi x)$ on the subspace generated by $\{1,x,x^2\}$ is given by $c_0 P_0(2x-1)+c_1 P_1(2x-1)+c_2 P_2(2x-1)$ where: $$ c_i = (2i+1)\int_{0}^{1}\sin(\pi x)P_i(2x-1)\,dx $$ so the projection is given by: $$ p(x)=\frac{2}{\pi}+\frac{10(\pi^2-12)}{\pi^3}P_2(2x-1) $$ and the squared distance between $\sin(\pi x)$ and $p(x)$ is given by: $$ \|\sin(\pi x)-p(x)\|_2^2 = \sum_{n>2} (2n+1) c_n^2 = \frac{1}{2}-\frac{2880}{\pi ^6}+\frac{480}{\pi ^4}-\frac{24}{\pi ^2}\approx 3\cdot 10^{-4}$$ that is minimal since $p(x)$ and $\sin(\pi x)-p(x)$ are orthogonal.