Convergence of Fourier Series — Question 8

Question 8 – Solution

Strategy. Use the weighted coefficient budget with two different dual estimates: one for a point value and one for energy.

Step 1: Prove uniform convergence from the budget. For any finite tail from $p$ to $q$, Cauchy–Schwarz gives $$\left|\sum_{n=p}^q(a_n\cos nx+b_n\sin nx)\right| \leq\left(\sum_{n=p}^q n^2(a_n^2+b_n^2)\right)^{1/2} \left(\sum_{n=p}^q\frac{\cos^2nx+\sin^2nx}{n^2}\right)^{1/2}.$$ It is at most $\sqrt M(\sum_{n\geq p}n^{-2})^{1/2}$, independently of $x$, and tends to zero as $p\to\infty$. The uniform Cauchy criterion applies. The limit of these continuous periodic partial sums is continuous and periodic.

Step 2: Bound the infinite tail. Passing to the uniform limit of the finite-tail inequality gives $$\boxed{\|f-S_N\|_\infty\leq\sqrt{MT_N}<\sqrt{M/N}\quad(N\geq 1).}$$ The weighted hypothesis controls the full collection of omitted modes, not just a finite list of observed coefficients.

Step 3: Attain the exact pointwise bound. Set every coefficient with $n\leq N$ to zero and, for $n>N$, choose $$\boxed{a_n=\sqrt{\frac M{T_N}}\frac{\cos(nx_0)}{n^2},\qquad b_n=\sqrt{\frac M{T_N}}\frac{\sin(nx_0)}{n^2}.}$$ The weighted sum equals $M$, and at $x_0$ the tail equals $\sqrt{MT_N}$. This proves sharpness of the first bound over all admissible sequences. The mean $m$ may be arbitrary because it cancels from the error.

Step 4: Optimize the integral error. Orthogonality, justified by the uniform convergence, gives $$\|f-S_N\|_2^2=\pi\sum_{n>N}(a_n^2+b_n^2) \leq\frac{\pi}{(N+1)^2}\sum_{n>N}n^2(a_n^2+b_n^2) \leq\boxed{\frac{\pi M}{(N+1)^2}}.$$ Equality is attained by $a_{N+1}=\sqrt M/(N+1)$ and all other coefficients zero. The integral optimizer concentrates its budget in the cheapest omitted mode; the pointwise optimizer distributes it across all omitted modes with phases aligned at the chosen point.