A polynomial in $g$ approximates every $f$ iff $g$ is injective

Question

Prove or disprove the following statement:

There exists a continuous function $g$ defined on $[a,b]$ with $g(x)\neq x $ for at least one $x\in[a,b]$ such that for every continuous function $f$ defined on $[a,b]$, given any $\varepsilon>0$, there exist constants $a_0,a_1,...,a_n$ depending only on $g,f,\varepsilon$ such that $$|f(x)-\sum_{k=0}^na_k(g(x))^k|<\varepsilon$$ for all $x\in[a,b]$ IF AND ONLY IF $g$ is injective.

After zhw's comment, the simpler version of this question stands as:

Prove or disprove that if $g$ is a continuous function on $[a,b]$ then polynomials in $g$ are dense in $C[a,b]$ iff $g$ is injective.

I think that the statement is true. I attempted to prove it as follows:

First the IF part. If $g$ is given to be injective, then I let $g(x)=\dfrac{x}{2}$ for all $x\in[a,b]$. Then $g(x)\neq x$ for at least one $x\in[a,b]$ but I can use a polynomial in $g$ to approximate any continuous function on $[a,b]$ since I will basically be using a polynomial of the form $\sum_{k=0}^n\dfrac{a_k}{2^k}x^k=\sum_{k=0}^nb_kx^k$ where $b_k$ will be obtained from the approximating polynomials of Weierstrass Approximation Theorem and hence I will get $a_k=2^kb_k$ which depend only on $f,\varepsilon$ as $b_k$ depend only on $f,\varepsilon$. Dependency on $g$ is removed as we have boiled down to a polynomial only.

Now the ONLY IF part. Suppose $g$ is not injective, then there exist $x_1\neq x_2\in[a,b]$ such that $g(x_1)=g(x_2)$. Now suppose this $g$ satisfies all the conditions mentioned, then we can write, for every continuous function $f$ defined on $[a,b]$, given any $\varepsilon>0$, that there exist constants $a_0,a_1,...,a_k$ depending only on $f,g,\varepsilon$ such that $$|f(x)-\sum_{k=0}^na_k(g(x))^k|<\varepsilon$$ for all $x\in[a,b]$. In particular, take $x=x_1$ and $x=x_2$ which will give,$$|f(x_1)-\sum_{k=0}^na_k(g(x_1))^k|<\varepsilon$$ and $$|f(x_2)-\sum_{k=0}^na_k(g(x_2))^k|<\varepsilon$$ Now, $$|f(x_1)-f(x_2)=|f(x_1)-\sum_{k=0}^na_kg(x_1)^k+\sum_{k=0}^na_kg(x)^k-f(x_2)|$$$$\leq|f(x_1)-\sum_{k=0}^na_kg(x_1)^k|+|\sum_{k=0}^na_kg(x)^k-f(x_2)|<2\varepsilon$$ This implies that every continuous function $f$ defined on $[a,b]$ must have $f(x_1)=f(x_2)$ but this is not true as in particular, taking $f(x)=x$ which is bijective contradicts this. Hence $g$ must be injective.

Is my proof correct?

Answer 1

A continuous injective function $g$ on $[a,b]$ maps $[a,b]$ bijectively onto some $[c,d],$ and its inverse $g^{-1}$ has the same property going backwards. Let $f\in C([a,b]).$ Then $f\circ g^{-1}\in C([c,d]).$ Hence there is a sequence of polynomials $p_n\to f\circ g^{-1}$ uniformly on $[c,d].$ This implies $p_n\circ g \to (f\circ g^{-1})\circ g = f$ uniformly on $[a,b].$

Answer 2

If you go to Principles of Mathematical Analysis (W. Rudin) and read Stone's-Weirstrass Theorem Concerning algebras that separate points and their densities in C[a,b] then the result you are proving is a consequence of that: Just take the smallest algebra containing g.

Answer 3

The Stone-Weierstrass theorem applies here: the algebra generated by $g$ is dense in $C([a,b])$ if and only if it separates points, and this is true if and only if $g$ is injective.