# Trigo system completeness..

This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $$C(-\pi,\pi)$$ instead of $$C(0,2\pi)$$, but this makes no difference of course.

Suppose we had trigonometric polynomials $$Q_k \geq 0$$ such that

• $$\frac{1}{2\pi}\int_{-\pi}^{\pi} \! Q_k(t) \, dt = 1$$
• $$Q_k \to 0$$ uniformly on $$[-\pi, -\delta] \cup [\delta, \pi]$$ for every $$\delta >0$$

(i.e. the $$Q_k$$ are a summation kernel). Associate to each continuous $$f$$ the function $$P_k(t) = f \ast Q_k(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! f(t-s)Q_k(s) \, ds.$$ Substitution shows that also $$P_k(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! f(s)Q_k(t-s) \, ds$$ so that each $$P_k$$ is a trigonometric polynomial. Now let $$\epsilon > 0$$. Since $$f$$ is uniformly continuous, there exists $$\delta > 0$$ such that $$|f(t) - f(s)| < \epsilon$$ whenever $$|t-s| < \delta$$. Since each $$Q_k$$ has average equal to one, we see that $$P_k(t) - f(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! (f(t-s) - f(s)) Q_k(s) \, ds$$ and the positivity of $$Q_k$$ implies $$|P_k(t) - f(t)| \le \frac{1}{2\pi}\int_{-\pi}^\pi \! |f(t-s) - f(s)|Q_k(s) \, ds$$ $$= \frac{1}{2\pi}\int_{|s| \le \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds + \frac{1}{2\pi}\int_{|s|\geq \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds$$ In the first term, the uniform continuity of $$f$$ shows that the integrand is smaller than $$\epsilon Q_k(s)$$ and the assumption on the average of $$Q_k$$ makes this term small. For the second the integrand converges uniformly to $$0$$ as $$k \to \infty$$ (since $$f$$ is bounded). Since the estimates are independent of $$t$$ we have shown that $$\|P_k - f\|_\infty < \epsilon$$ for $$k$$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $$Q_k$$ with the desired properties. To do this we let $$Q_k(t) = c_k \left(\frac{1 + \cos t}{2}\right)^k$$ where the $$c_k$$ is chosen in such a way as to satisfy the assumption on the averages. Since the $$Q_k$$ are clearly positive, we only need to show that $$Q_k \to 0$$ uniformly away from $$0$$. But $$Q_k$$ is even (on $$(-\pi,\pi)$$) and decreasing on $$[0,\pi]$$. Thus for any $$\delta > 0$$ we have $$|Q_k(t)| \le Q_k(\delta) \le \frac{\pi(k+1)}{2}\left(\frac{1 + \cos \delta}{2} \right)^k \to 0$$ independently of $$t$$ as $$k \to \infty$$ since $$\frac{1 + \cos \delta}{2} < 1.$$

My questions are as follows:

Substitution shows that also $$P_k(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! f(s)Q_k(t-s) \, ds$$ so that each $$P_k$$ is a trigonometric polynomial.(?)

we see that $$P_k(t) - f(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! (f(t-s) - f(s)) Q_k(s) \, ds$$ and the positivity of $$Q_k$$ implies $$|P_k(t) - f(t)| \le \frac{1}{2\pi}\int_{-\pi}^\pi \! |f(t-s) - f(s)|Q_k(s) \, ds$$ $$= \frac{1}{2\pi}\int_{|s| \le \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds + \frac{1}{2\pi}\int_{|s|\geq \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds$$.(?) I think there are many details omitted in the proof. Thanks for the help in advance.