Showing a set is globally a graph

Question

I have a following problem: Let $B^2$ be an open unit ball in $\mathbb R^2$ and $$ S = \{(x, y, z) \in B^2 \times \mathbb R: \frac12(x^2 + y^2)z^3 + xyz^2 + z - 2 = 0\}$$

Show that $S$ is globally a graph of a $C^\infty(D)$ function $z = f(x, y)$.

I had no problem showing that it is true in a neighbourhood of every point using implicit function theorem, but how to show that this function is constructable globally? Does it require using something like the pasting lemma?

Answer 1

You have to show that for each $(x,y) \in B^2$ there exists a unique $z \in \mathbb R$ such that $$\phi(z) = \frac12(x^2 + y^2)z^3 + xyz^2 + z - 2 = 0 .\tag{1} $$ For $(x,y)= 0$ this is obvious. For $(x,y) \ne (0,0)$ write $(x,y) = r(\cos t, \sin t)$ with $0 < r < 1$. We get $$ \phi(z) = \frac12 r^2z^3 + \frac 12 \sin (2t)r^2z^2 + z - 2 \tag{2}.$$ Consider $$\phi(z) = \frac32 r^2z^2 + \sin (2t)r^2z + 1 . \tag{3}$$ We claim that $f'(z)$ does not have a zero. This implies that $\phi'(z) > 0$ for all $z$. Hence $\phi$ is strictly increasing and $\phi$ has a single (real) zero.

The disriminant of the quadratic polynomial in $(3)$ is $$\sin^2(2t)r^4 - 6r^2 \le r^4 - 6r^2 = r^2(r^2 - 6) < 1 \cdot (1-6) < 0$$ and this proves our claim.

As you mentioned in your question, the implicit function theorem allows to conclude that $f(x,y) =$ zero of $\phi$ is smooth.

Answer 2

Just a few thoughts.

The equation $1/2(x^2+y^2)z^3+xyz^2 + z-2 = 0$ defining $S$ is a polynomial equation in $z$, so finding an a function $f$ such that $z = f(x,y)$ comes down to solving the equation in $z$ for any given $x,y$. A cubic always at least one solution on $\mathbb R$ so we can always write $z = f(x,y)$ for some function $f$ (where $f$ maps x,y to a solution of the equation). Notice that this does not uniquely define a function $f$ as the equation could have several solutions. If you manage to prove that $1/2(x^2+y^2)z^3+xyz^2 + z-2 = 0$ always has a unique solution (if x,y are given) then you've found the map$f$ you're looking for. This shouldn't be too hard to show, indeed the map $z \mapsto 1/2(x^2+y^2)z^3+xyz^2 + z-2$ is strictly increasing (consider its derivative !).

I'm not sure how you can show that $f$ is smooth.