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Spring 2020 - Problem 5

Hilbert spaces

Rigorously determine the infimum of

∫−11∣P(x)−∣x∣∣2 dx\int_{-1}^1 \abs{P\p{x} - \abs{x}}^2 \,\diff{x}

over all choices of polynomials P∈R[x]P \in \R\br{x} of degree not exceeding three.
Solution.

Viewing this problem in L2([−1,1])L^2\p{\br{-1,1}}, we see that we need to orthogonally project ∣x∣\abs{x} onto the subspace spanned by {1,x,x2,x3}\set{1, x, x^2, x^3}. By Gram-Schmidt, we see that

e0(x)=1e1(x)=xe2(x)=x2−13e3(x)=x3−35x\begin{aligned} e_0\p{x} &= 1 \\ e_1\p{x} &= x \\ e_2\p{x} &= x^2 - \frac{1}{3} \\ e_3\p{x} &= x^3 - \frac{3}{5}x \end{aligned}

are orthogonal. By calculation, we also have

∥ek∥L22=22k+1.\norm{e_k}_{L^2}^2 = \frac{2}{2k + 1}.

Thus, the orthogonal projection is given by

P(x)=∑k=03⟨∣x∣,ek⟩ek.P\p{x} = \sum_{k=0}^3 \inner{\abs{x}, e_k}e_k.