Periodic Functions & Orthogonal Functions — Question 6

Question 6

On continuous real functions on $[0,1]$, define $$\langle f,g\rangle_w=\int_0^1x f(x)g(x)dx.$$ The weight vanishes at one endpoint. Seek a monic quadratic $q(x)=x^2+ax+b$ orthogonal to both $1$ and $x$ in this inner product.

Tasks

1. Prove positive definiteness despite the zero weight at $x=0$. Explain the role of continuity of the functions.

2. Determine $a,b$ from the two weighted moment equations. Verify both equations and prove uniqueness of the monic quadratic.

3. Compute $\|q\|_w^2$, give its unit-norm version with positive leading coefficient, and locate its roots. Sketch $q$ on $[0,1]$.

4. Test whether the same $q$ is orthogonal to $1$ for unweighted integration. Also explain why replacing the weight by $x-1/2$ would fail to define an inner product on this space.

Question 6 – Solution

Strategy. Include the weight in every moment, and distinguish an isolated zero of a nonnegative weight from a sign-changing weight.

Step 1: Establish positive definiteness. Symmetry and bilinearity follow from integration, and $\int_0^1xf^2dx\geq 0$. If a continuous $f$ is not identically zero, continuity supplies an interval of positive length inside $(0,1]$ on which $f^2>0$, even if the first known nonzero value is at zero. The weighted integral is then positive. Thus zero weighted norm forces $f=0$ everywhere.

Step 2: Solve the weighted moment equations. The two conditions are $$\frac 14+\frac a3+\frac b2=0,\qquad \frac 15+\frac a4+\frac b3=0.$$ Their coefficient determinant is $1/9-1/8=-1/72\ne 0$, so the solution is unique: $$\boxed{a=-6/5,\qquad b=3/10,\qquad q=x^2-6x/5+3/10.}$$ Substitution gives $1/4-2/5+3/20=0$ and $1/5-3/10+1/10=0$.

Step 3: Normalize and locate the sign changes. Expansion and integration give $$\int_0^1xq^2dx=\frac 16-\frac{12}{25}+\frac{51}{100} -\frac 6{25}+\frac 9{200}=\boxed{\frac 1{600}}.$$ Thus the unit-norm version is $\boxed{\sqrt{600}\,q}$. The roots are $\boxed{(6-\sqrt 6)/10,\ (6+\sqrt 6)/10}$, both strictly inside $(0,1)$. The quadratic is positive outside these roots and negative between them.

Step 4: Check two changes of weight. Without the weight, $\int_0^1q\,dx=1/3-3/5+3/10=\boxed{1/30\ne 0}$. Orthogonality therefore changes with the weight. For the proposed signed weight, the nonzero constant function one would have squared “norm” $\int_0^1(x-1/2)dx=0$. This violates positive definiteness, so that bilinear form is not an inner product on the stated space.