Convergence of Fourier Series — Question 7

Question 7 – Solution

Strategy. The signed Fourier kernel controls integral projection, while its absolute integral controls worst-case pointwise error.

Step 1: Bound the projection in the supremum norm. Taking absolute values in the given integral gives $$|S_1f(x)|\leq\frac{\|f\|_\infty}{2\pi} \int_{-\pi}^{\pi}|1+2\cos t|\,dt.$$ Null-set changes do not affect this estimate or the projected polynomial. Taking the supremum over $x$ proves the claimed inequality.

Step 2: Integrate the absolute kernel. The kernel is nonnegative for $|t|\leq 2\pi/3$ and negative outside. Its full signed integral is $2\pi$. Its integral on the two negative intervals is $$2\int_{2\pi/3}^{\pi}(1+2\cos t)\,dt=\frac{2\pi}{3}-2\sqrt 3.$$ Changing the sign of that contribution gives $$\boxed{\Lambda=\frac 13+\frac{2\sqrt 3}{\pi}>1.}$$

Step 3: Attain the bound with an explicit input. Choose $f(t)=\operatorname{sgn}(1+2\cos t)$, with value zero at its two zeros. It is even and piecewise constant with essential supremum one. Its positive region has length $4\pi/3$, so its mean is $1/3$. Direct integration gives $$a_1=\frac 2\pi\left(\int_0^{2\pi/3}\cos t\,dt -\int_{2\pi/3}^{\pi}\cos t\,dt\right)=\frac{2\sqrt 3}{\pi},\qquad b_1=0.$$ Hence $S_1f(0)=1/3+2\sqrt 3/\pi=\Lambda$, proving sharpness.

Step 4: Interpret the stability statement. For input error $e$ with $\|e\|_\infty\leq\varepsilon$, linearity gives $\|S_1(f+e)-S_1f\|_\infty\leq\Lambda\varepsilon$. The choice $e=\varepsilon\operatorname{sgn}(1+2\cos t)$ attains this bound. Orthogonal projection still satisfies $\|S_1e\|_2\leq\|e\|_2$ by the Pythagorean identity. That statement uses an integral norm, and does not prevent a larger pointwise peak. Contraction is a property of a specified norm.