Periodic Functions & Orthogonal Functions — Question 8

Question 8 – Solution

Strategy. Use an orthogonal basis for the integral objective, then test the distinct uniform-error objective separately.

Step 1: Compute the projection. The needed moments are $\langle f,1\rangle=1$ and $\langle f,p_2\rangle=1/2-1/3=1/6$. Thus $$g_*=\frac 12+\frac{1/6}{8/45}(x^2-1/3) =\boxed{\frac 3{16}+\frac{15}{16}x^2.}$$

Step 2: Verify residual orthogonality and uniqueness. For $r=f-g_*$, $$\int_{-1}^1r\,dx=1-\frac 38-\frac 58=0,\qquad \int_{-1}^1rx^2dx=\frac 12-\frac 18-\frac 38=0.$$ For any other $g=a+bx^2$, the cross term vanishes, so $\|f-g\|_2^2=\|r\|_2^2+\|g_*-g\|_2^2$. The second term is zero only when the two continuous polynomials agree, proving both optimality and uniqueness.

Step 3: Evaluate the two errors for the optimum. Subtracting the squared projection norm from $\|f\|_2^2=2/3$ gives $$\boxed{\|r\|_2^2=\frac 23-\frac 12-\frac{(1/6)^2}{8/45}=\frac 1{96}.}$$ On $[0,1]$, $r=x-3/16-15x^2/16$ has maximum $19/240$ at $x=8/15$, and endpoint values $-3/16,-1/8$. By evenness, the largest absolute error is $\boxed{\|r\|_\infty=3/16}$, attained at zero.

Step 4: Disprove uniform optimality. For $h=1/8+x^2$, put $t=|x|\in[0,1]$. Then $f-h=t(1-t)-1/8\in[-1/8,1/8]$, with both extremes attained. Hence $\boxed{\|f-h\|_\infty=1/8<3/16}$. This competitor proves that the unique least-squares solution is not a uniform best approximation. Different error objectives need not select the same function.