Convergence of Fourier Series — Question 6

Question 6 – Solution

Strategy. Point evaluation on a finite-dimensional space has a norm that grows with the number of available modes.

Step 1: Apply Cauchy–Schwarz in coefficient space. An orthonormal basis is $1/\sqrt{2\pi}$, $\cos nx/\sqrt\pi$, $\sin nx/\sqrt\pi$. The sum of their squared values at any point is $$\frac 1{2\pi}+\frac 1\pi\sum_{n=1}^N(\cos^2nx+\sin^2nx) =\frac{2N+1}{2\pi}.$$ Cauchy–Schwarz against the orthonormal coefficient vector proves $$\boxed{\|p\|_\infty\leq\sqrt{\frac{2N+1}{2\pi}}\,\|p\|_2.}$$

Step 2: Identify equality cases. Equality at $x_0$ holds exactly when the coefficient vector is a real nonzero multiple of the basis-value vector at $x_0$. Equivalently, $$\boxed{p(x)=C D_N(x-x_0),\qquad C\in\mathbb R\setminus\{0\}.}$$ This exhibits the sharp constant, not merely an upper estimate.

Step 3: Build a vanishing-energy sequence with a fixed peak. The triangle inequality and the value at zero give $\|D_N\|_\infty=2N+1$. Orthogonality gives $\|D_N\|_2^2=2\pi(2N+1)$. Consequently $$\boxed{\|p_N\|_\infty=1,\qquad \|p_N\|_2^2=\frac{2\pi}{2N+1}\to 0.}$$ At $0$ modulo $2\pi$, $p_N=1$. At every other fixed point it tends to zero by the quotient below. Point evaluation is therefore not continuous in the $L^2$ norm on the union of these polynomial spaces.

Step 4: Explain the oscillatory peak. Summing $\sum_{n=-N}^N e^{inx}$ yields $$\boxed{D_N(x)=\frac{\sin((N+1/2)x)}{\sin(x/2)} \quad(x\notin 2\pi\mathbb Z).}$$ The removable value at zero is $2N+1$. The normalized peak narrows with $N$. These $p_N$ cannot be successive partial sums of one fixed function: their constant coefficients $1/(2N+1)$ change with $N$, whereas truncation preserves every already-included Fourier coefficient.