Every continuous real valued function an be uniformly approximated by polynomial in these two functions Prove that every continuous real valued function on the square $[0,1] \times [0,1]$ can be uniformly approximated by polynomials in the functions $f(x,y)=x+e^y$ and $g(x,y)=x-e^y$.
To me, this looks like a Weierstrass approximation problem.
Let $f$ be a continuous real valued function on the square $[0,1] \times [0,1]$.
Basically we want to to find a polynomial in the functions $p(x)$ so that $|f(x)-p(x)|< \epsilon$.
But I'm not sure what it means by "polynomials in the functions".
Any help will be appreciated!
 A: As you write, we can apply the Stone-Weierstrass theorem. A polynomial in two variables $u,v$ has the form
$$p(u,v) = \sum_{i,j=0}^n a_{ij}u^iv^j$$
In your question we have to consider all polymials with $u = f(x,y)$ and $v = g(x,y)$, i.e. all functions in the two variables $x,y$ having the form
$$\phi(x,y) = p(f(x,y), g(x,y)) .$$
The set of these functions is clearly a subalgebra of $C([0,1] \times [0,1], \mathbb R)$. It contains all constant functions (take the constant polynomials $p(u,v) = a_{00}$). It is also separates points:
Let $(x,y), (x',y')$ be two distinct points of $[0,1]^2$. Then $(u,v) = (f(x,y), g(x,y))$ and $(u',v') =(f(x',y'), g(x',y'))$ are two distinct points of $\mathbb R^2$. In fact, $f(x,y) = f(x',y')$ and $g(x,y) = g(x',y')$ means that $x + e^y = x' + e^{y'}$ and $x - e^y = x' - e^{y'}$. Adding these equations gives $2x = 2x'$, i.e. $x = x'$. This implies $e^y = e^{y'}$, i.e. $y = y'$.
Therefore it remains to show that the polynomials $p(u,v)$ separate the points of $\mathbb R^2$. But if $(u,v) \ne (u',v')$, then $u \ne u'$ or $v \ne v'$. We only consider the case $u \ne u'$, the other case is treated similarly. Taking $p(u,v) = u$ gives $p(u,v) \ne p(u',v')$.
