Convergence of Fourier Series — Question 5

Question 5 – Solution

Strategy. Check a separate convergence condition for each differentiation; one valid differentiation does not authorize the next.

Step 1: Justify the first differentiation. Every summand is $C^1$. The original series converges at $0$ (all terms are zero), and the derivative terms satisfy $|\cos nx/n^2|\leq 1/n^2$. The Weierstrass test gives uniform convergence of the derivative series. The stated theorem therefore gives $$\boxed{H'(x)=\sum_{n=1}^{\infty}\frac{\cos nx}{n^2}.}$$ This includes one-sided endpoint derivatives on the closed interval.

Step 2: Verify the explicit function. Integrating the provided formula and using $H(0)=0$ gives $$\boxed{H(x)=\frac{\pi^2x}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}, \qquad 0\leq x\leq 2\pi.}$$ Both endpoint values are zero. Both endpoint slopes equal $\pi^2/6$, so the periodic extension is $C^1$ across the join.

Step 3: Identify the second derivative obstruction. Inside the period, $H''(x)=-\pi/2+x/2$. Its right limit at zero is $-\pi/2$, whereas its left periodic limit is $\pi/2$. Thus the periodic function is not twice differentiable at the join and is not $C^2$. The formal second derivative series is $-\sum\sin(nx)/n$. The piecewise smooth Fourier theorem, or direct coefficients of the displayed linear function, gives $H''$ on $0<x<2\pi$. At a join the series equals zero, the average of the limits, where the classical periodic second derivative does not exist.

Step 4: Rule out uniform second differentiation. Every finite sum $-\sum_{n=1}^N\sin(nx)/n$ tends to zero as $x\downarrow 0$, but the interior target tends to $-\pi/2$. Thus its error supremum on the full open period is at least $\pi/2$ for every $N$. The derivative series for the first differentiation had summable bounds $n^{-2}$; the next series has no such global uniform convergence. There is no contradiction between a valid first differentiation and failure of the stronger second-derivative conclusion.