Specific proof technique of the complex Stone-Weierstrass theorem

The question is as follows:

If $$f:\mathbb{T}\rightarrow\mathbb{C}$$ is continuous, prove that there is a sequence of polynomials $$p_n(z,\bar{z})$$ such that $$p_n\rightarrow f$$ uniformly for every $$z\in\mathbb{T}$$.

(Note: $$\mathbb{T}$$ denotes the unit circle.) I've seen proofs of the more general statement of the complex version (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Stone%E2%80%93Weierstrass_theorem,_complex_version), but this is asked in the context of a first course in complex analysis, so we have not developed the foundation to even understand the more general statement. We are given the following hint, however.

Let $$g(re^{i\theta})=P_r(f)$$ and show that for each $$r<1$$ there is a sequence of polynomials $$p_n(z,\bar{z})$$ such that $$p_n$$ converge uniformly for every $$z\in\mathbb{T}$$.

(Note: $$P_r(f)$$ denotes the Poisson kernel.) Here are my specific questions:

1) Can we prove this by simply writing $$f$$ as $$f=u+iv$$ for some real-valued, continuous functions $$u$$ and $$v$$ and then applying the real version of Stone-Weierstrass? I.e. approximating $$u$$ and $$v$$ with polynomials of real variables and claiming that the supremum norm of $$f$$ minus the sum of these polynomials is arbitrarily small? (Applying the fact that polynomials in 2 real variables can be transformed into polynomials in complex conjugates of 1 variable.)

2) If the above is an invalid approach, how does introducing the Poisson kernel fix the logical error (as what I'm proposing is a similar idea to the hint)?

It is quite possible that I just have a fundamental misunderstanding of the Poisson kernel. Maybe my claim in 1) that polynomials in 2 real variables can be transformed into polynomials in complex conjugates of 1 variable is dependent on the Poisson kernel?

The purpose of this post is to request assistance in interpreting this problem (and required tools to prove it), not to ask for a solution.

• $p_n\rightarrow f$ uniformly for every $z\in\mathbb{T}$. that doesn't really make sense. – zhw. Nov 25 '18 at 1:13
• @zhw. ...why not? – Atsina Nov 25 '18 at 1:15
• @Atsina what is your definition of convergence in $\mathbb{C}?$ – Idonknow Nov 25 '18 at 3:43
• @Idonknow A sequence of complex numbers $\{z_1,z_2,\cdots\}$ converges to $w\in\mathbb{C}$ if $\lim_{n\rightarrow\infty}|z_n-w|=0$? I mean, there are plenty of equivalent definitions and theorems involving them, but this is probably the simplest definition of convergence in $\mathbb{C}$ – Atsina Nov 25 '18 at 3:51