Applying Bishop’s Theorem to $\langle xy;x^2y\rangle$

Question

I am using Bishop’s Theorem in the version given by Wikipedia¹:

Let $\mathfrak{A}$ be a closed subalgebra of the Banach space $C(X,ℂ)$ of continuous complex-valued functions on a compact Hausdorff space $X$. Suppose that $f∈C(X,ℂ)$ has the following property:

• $f|_S∈\mathfrak{A}_S$ for every maximal set $S⊂X$ such that all real functions of $\mathfrak{A}_S$ are constant.

Then $f∈\mathfrak{A}$.

I apply this to $X$ being some neighbourhood of the origin and: $$ \mathfrak{A} = \big\langle ~ (x,y)↦xy; ~ (x,y)↦x^2y ~ \big\rangle,$$ where $⟨⟩$ denotes the generated set.

Then $S^*=\left\{ (x,y) \,\middle |\, x=0 \lor y=0 \right\}$ is the only maximal subset on which all real functions of $\mathfrak{A}_{S^*}$ are constant and that has more than one point².

My problem is that, according to Bishop’s theorem, the following functions must be in $\mathfrak{A}$, as they are in $\mathfrak{A}_{S^*}$ (being constantly $0$ on $S^*$):

• $(x,y) ↦ x y^2$: My problem with this is that I only get a higher power of $y$ by multiplying two generating functions with each other, which in turn yield a higher power of $x$ (than $1$).
• $(x,y) ↦ x^3 y$: My problem with this is that I only get an $x^3$ in the term by multiplying two generating functions, which in turn yield a higher power of $y$.

My questions are:

• Did I make a mistake somewhere and the above functions are not in $\mathfrak{A}$?

• If not, how can I explicitly see that they are in $\mathfrak{A}$? What sequence in $\mathfrak{A}$ is converging to the functions in question?

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¹ According to some sources, there is an additional requirement on $\mathfrak{A}$ to contain constant functions. But this does not solve my problem.

² If we had another such subset, there would have to be $(x_1,y_1), (x_2,y_2) \notin S^*$, such that $(x_1,y_1) ≠ (x_2,y_2)$ and $x_1 y_1 = x_2 y_2$ and $x_1^2 y_1 = x_2^2 y_2$. However, this implies: $$x_1 = \frac{x_1^2 y_1}{x_1 y_1} = \frac{x_2^2 y_2}{x_2 y_2} = x_2,$$ and from that $y_1 = \frac{y_1 x_1}{x_1} = \frac{y_2 x_2}{x_2} = y_2$, which contradicts with the requirements.

Answer 1

I see no problem in your argumentation. But remember that, by Stone-Weierstrass, the function $t\mapsto \sqrt{t}$ is a limit of polynomials in $t$, so the obtaining a "lower power" than expected is not a real problem.

Here is a sketch on how to construct a sequence in $\mathbb{R}[xy,x^2y]$ converging to $xy^2$ uniformly on a compact neighborhood $X$ of $0$. (the other one can probably be dealt with a similar construction). The idea is to create a contracting operator preserving the algebra $\mathbb{R}[xy,x^2y]$ (not strictly contracting, unfortunately), with fixed point the required function $xy^2$.

Pick $\alpha>0$ sufficently small so that $\alpha x^4y^2<1$ on the compact neighborhhod $X$ of the origin (so $0<1-\alpha x^4y^2<1$ on $X-S^*$). Look at the map on $C^0(X,\mathbb{R})$:

$$\Phi(f)=f+\alpha x^2y \left( (xy)^3-x^2y f \right)=(1-\alpha x^4y^2)f+\alpha x^5y^4,$$

We have $\Phi(xy^2)=xy^2$, i.e. $f_0=xy^2$ is a fixed point of this map. Define a sequence $P_0=0, \; P_{n+1}=\Phi(P_n)$, it is a sequence in $\mathbb{R}[xy,x^2y]$ because of the expression of $\Phi$. On the other hand $$(P_{n+1}-f_0)(x,y)=(\Phi(P_n)-\Phi(f_0))(x,y)=(1-\alpha x^4y^2)(P_n-f_0)(x,y), $$ i.e outside $S^*$, the sequence $(P_n-f_0)(x,y)$ converge to zero exponentialy fast. I skip some details: one has to be a little careful to deal with the uniform convergence on neighborhoods of $S^*$, but since both $P_n$ and $f_0$ vanish on $S^*$, it gives a uniform convergence of $P_n$ toward $f_0$ on $X$, as required.