Convergence of Fourier Series — Question 4

Question 4 – Solution

Strategy. Cancellation controls a tail away from a jump, but the cancellation bound degenerates near the join.

Step 1: Bound finite oscillatory blocks. For $x\notin 2\pi\mathbb Z$, $$\left|\sum_{n=p}^q e^{inx}\right| =\left|\frac{e^{ipx}(1-e^{i(q-p+1)x})}{1-e^{ix}}\right| \leq\frac 1{|\sin(x/2)|}.$$ The same bound holds for the imaginary part. On the specified closed interval, it is at most $B=1/\sin(\delta/2)$.

Step 2: Sum by parts with a decreasing weight. Put $A_k=\sum_{n=N+1}^k\sin nx$, so $|A_k|\leq B$. Then $$\sum_{n=N+1}^M\frac{\sin nx}{n} =\frac{A_M}{M}+\sum_{k=N+1}^{M-1}A_k\left(\frac 1k-\frac 1{k+1}\right).$$ The absolute value is bounded by $B/(N+1)$. Passing to the convergent tail gives $$\boxed{\sup_{\delta\leq x\leq 2\pi-\delta}|G-P_N| \leq\frac 1{(N+1)\sin(\delta/2)}\longrightarrow 0.}$$

Step 3: Locate the global obstruction. For every fixed $N$, $P_N(x)\to 0$ as $x\downarrow 0$, while $G(x)\to\pi/2$. Therefore $\sup_{0<x<2\pi}|G-P_N|\geq\pi/2$ for every $N$. Excluding the endpoint itself does not exclude points arbitrarily close to it. At $\delta=0$ the geometric-sum bound has a zero denominator and gives no finite uniform control.

Step 4: Verify convergence in the integral norm. The sine functions have squared norm $\pi$ over this period. Hence $$\boxed{\|G-P_N\|_2^2=\pi\sum_{n>N}\frac 1{n^2}<\frac{\pi}{N}.}$$ The error tends to zero in mean square despite persistent near-jump suprema. Its assigned endpoint errors are zero, whereas its interior limits are $\pi/2$ at $0$ and $-\pi/2$ at $2\pi$.