Fourier Cosine Series — Question 7

Question 7 – Solution

Strategy. Sum the geometric coefficients exactly, then locate the mass accumulating near the endpoint.

Step 1: Derive the uniformly convergent expansion. The geometric identity gives $1+2\operatorname{Re}[\rho e^{ix}/(1-\rho e^{ix})]=P_\rho(x)$. Since $\sum\rho^n<\infty$, its cosine series converges absolutely and uniformly: $$\boxed{P_\rho=1+2\sum_{n=1}^\infty\rho^n\cos nx,\qquad a_0=2,\quad a_n=2\rho^n.}$$

Step 2: Compute mass, energy and extrema. Uniform integration gives $\boxed{\int_0^\pi P_\rho=\pi}$. Parseval and the geometric sum give $$\boxed{\int_0^\pi P_\rho^2=\pi\frac{1+\rho^2}{1-\rho^2}.}$$ The denominator increases with $x$. The maximum at zero is $(1+\rho)/(1-\rho)$; the minimum at $\pi$ is $(1-\rho)/(1+\rho)$.

Step 3: Explain the apparent loss of mass. For fixed $x>0$, the numerator tends to zero and the denominator to $2(1-\cos x)>0$, so $P_\rho(x)\to 0$. At zero it tends to infinity. The almost-everywhere limit is zero, yet the integral stays $\pi$. Uniform convergence and interchange of the integral with this limit are not available; the exact squared norm also diverges.

Step 4: Prove endpoint concentration. Given $\eta>0$, choose $\delta>0$ so $|\varphi(x)-\varphi(0)|<\eta$ on $[0,\delta]$. That part of $\int P_\rho(\varphi-\varphi(0))$ is at most $\pi\eta$ in magnitude. On $[\delta,\pi]$, $P_\rho\leq(1-\rho^2)/[2\rho(1-\cos\delta)]\to 0$ uniformly, so the remaining part tends to zero. Letting $\eta\downarrow 0$ proves the claim. The limit is mass $\pi$ concentrated at zero, not an ordinary integrable profile.