# Real Polynomials on Compact sets of Complex numbers

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?

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.

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.

• You say you know the answers. Maybe write in what you think the answers are and why. That might be the reason for the down vote and close vote. The MathSE community tends to be happier when question askers put in more details on their attempts at a solution. I think this is an interesting question which is worthy of a response, personally. Commented Mar 3 at 10:42

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 $$(\Re(x), \Im(x))$$.

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 $$\mathbb{R}[x]|_K$$.

• Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. Commented Mar 3 at 13:34