A real subalgebra of $C(X)$

Question

Let $X$ be a compact Hausdorff space and $C(X)$ be the space of all complex-valued continuous functions on $X$. Let $\tau:X\rightarrow X$ be a homeomorphism such that $\tau^2=\tau\circ\tau=identity$. I want to show that the subalgebra $$C(X,\tau)=\{f\in C(X):f(\tau(x))=\overline{f}(x)\}$$ is a real subalgebra of $C(X)$ and not a complex subalgebra. Please help.

Answer 1

Take a scalar $\alpha$ (member of the complex field) which has a non zero imaginary part and try and multiply a member $f$ of your subset. You will find that you would need to see what happens to $\alpha . f \left(\tau(x)\right)$. This unfortunately will not end up as $\overline {\left(\alpha . f \right)(x)}$ unless the imaginary part of $\alpha$ was zero.

Answer 2

If $f\in C(X,\tau)$ then $i\ f\notin C(X,\tau)$ and hence $C(X,\tau)$ cannot be a complex subalgebra.