I would like to prove Weierstrass' second theorem (trigonometric polynomials are dense in the space of continuous $2\pi$-periodic functions for the uniform norm) using the following version of Korovkin's Lemma:
Let $C_{2\pi}$ be the space of continuous $2\pi$-periodic functions and let $H_1, H_2, \ldots$ be a sequence of positive linear operators on $C_{2\pi}$ such that $\displaystyle \lim_{n \rightarrow \infty} ||H_n(f) - f||_{\infty} = 0$ for $f = 1, f = \cos$ and $f = sin$. Then $\displaystyle \lim_{n \rightarrow \infty} ||H_n(f) - f||_{\infty} = 0$ holds for all $f \in C_{2 \pi}$.
First of all I would like to prove the above lemma, and I don't know how.
Secondly, define $\displaystyle F_n(t) = \frac{1}{2} + \sum_{k=1}^{n-1} (1 - \frac{k}{n}) \cos(kt)$ and define the the so-called Fejér operator $H_n$ by $\displaystyle H_n(f, x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) F_n(t - x) dt$. Then I've been told that we can use this Fejér operator and the above lemma to conclude Weierstrass' second theorem.
In other words, we have to show that $H_n(f, x)$ is a trigonometric polynomial (this is done here) and that $\displaystyle \lim_{n \rightarrow \infty} ||H_n(f) - f||_{\infty} = 0$ holds for $f = 1, f = \cos$ and $f = sin$. For $f = 1$ this is easy to see, as $H_n(f, x) = 1$ for all $n$ and all $x$. For $f = cos$ or $f = sin$ I however do not see it. Some help would be much appreciated.