Real Polynomials on Compact sets of Complex numbers

Question

Setting: $\mathbb{R}[x]$ is the set of polynomials with real coefficients. All $f\in \mathbb{R}[x]$ has domain $\mathbb{C}$. $K$ is a compact subset of $\mathbb{C}$. $\mathbb{R}[x]|_{K}$ is the set of restrictions of functions in $\mathbb{R}[x]$. $\mathcal{C}(K)$ is the set of continuous complex-valued functions on $K$. By a set $F$ of complex-valued functions on some set $R$ being self-adjoint, we mean for all $f\in F,$ there exists some $\overline{f}\in F$ such that $\overline{f}(x)=\overline{f(x)}$ for all $x\in R$.

Questions:

1. Is $\mathbb{R}[x]$ self-adjoint?
2. Is $\mathbb{R}[x]|_{K}$ dense in $\mathcal{C}(K)$?
3. If $\mathbb{R}[x]|_K$ is not dense in $\mathcal{C}(K)$, is there a continuous function $f:K\rightarrow \mathbb{C}$ such that $f$ is not in the uniform closure of $\mathbb{R}[x]|_{K}$ and can be explicitly written down?

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Motivation (for me to ask this question): I am currently studying the section of Stone-Weierstrass in Baby Rudin, and it seems that Rudin doesn't answer the questions I asked above.

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Attempts:

1. I think $\mathbb{R}[x]$ is not self-adjoint. I have played around with some real polynomials of low degrees and tried some inputs like $1+i,1+2i$, but not much concrete results came out with a proof.

2. Another reason I think $\mathbb{R}[x]$ is not self-adjoint is that if it is, then $\mathbb{R}[x]|_{[a,b]}$ will be dense in $\mathcal{C}(K)$, which implies each continuous complex-valued function on $[a,b]$ can be approximated by real polynomial. If this is true, I don't think Rudin will say at the end of the statement of Theorem 7.26 (Weierstrass approximation Theorem for $f:[a,b] \rightarrow \mathbb{R}$, as his proof goes): If $f$ is real, then $P_n$ may be taken real.

3. For the same reason above, I think $\mathbb{R}[x]|_K$ is not generally dense in $\mathcal{C}(K)$, but I can't come up with an example of a continuous complex-valued function that can't be approximated by real polynomials, since I haven't learned much about complex functions.

Thanks for any help in all forms. Truly appreciated.

Answer 1

Consider the polynomial $f(x) = x$, which is certainly a real polynomial. Is there a real polynomial $\bar{f}$ such that $\bar{f}(x) = \bar{x}$? More generally, if both $f$ and $\bar{f}$ satisfying $\bar{f}(x) = \overline{f(x)}$ also individually satisfy the Cauchy-Riemann equations then they have to be constants.

In fact, if $f$ is a real polynomial then it satisfies $\overline{f(x)} = f(\bar{x})$, which is a closed (check!) condition on $\mathcal{C}(K)$, and is non-trivial (by above) if $K$ is infinite (and therefore has a limit point). If $K$ is finite then $\mathcal{C}(K)$ is the set of all functions $K \to \mathbb{C}$, so the same condition is again non-trivial iff $K$ contains some pair of conjugate points or some real point, ie some $x \in K$ with $\bar{x} \in K$. If $K$ does not contain such an $x$ then for any function $f : K \to \mathbb{C}$, the unique complex polynomial of degree $\le 2|K|$ taking the value $f(x)$ at $x \in K$ and $\overline{f(\bar{x})}$ at $x$ with $\bar{x} \in K$ will be real.

If you want a set of functions that is self-adjoint, consider real polynomials in the two real variables $(\Re(x), \Im(x))$.

Answer 2

Question 1

Let $f(x)=ax+b$, $\overline{f}(x)=a\overline{x}+b, z=r_1+ir_2\in K\subseteq\mathbb{C}$ ,where $a,b\in\mathbb{R}$ is fixed and $r_2\neq 0$. Assume $\overline{f}(x)=g(x)$, we note that $\deg(g)=\deg(\overline{f})=1$ since fundamental theorem of algebra. Suppose $g(x)=a_1x+b_1$ ,where $a_1,b_1\in\mathbb{R}$. You can easy to see (by calculus) that

\begin{align} a_1&=-a\\ b_1&=b+2ar_1. \end{align}

Since $r_1$ is arbitrary, $g(x)$ doesn't exist.

Note that we can assume that there is $z\in K$ such that $z=r_1+ir_2$,$r_2\neq 0$. Otherwise $K$ will be a compact set of $\mathbb{R}$. Then that will go back to the Stone-Weierstrass Theorem.

Question 2

I think you can get answer from exercise 21 of Rudin "Principles of Mathematical Analysis" 2-edtion. It implies that:

Let $K=\{z\in\mathbb{C} : |z|=1\}$ and $h(z)=1/z$. Then $h$ is not in the uniform closure of $\mathbb{R}[x]|_K$.