# Are the roots of a smooth function, a smooth function?

Let $f(x,y)$ be a smooth function. It is given that for every $x$ there exists at least one $y$ such that $f(x,y)=0$. Is this possible to select one such $y$ for every $x$, such that the $y$'s are a smooth function of $x$?

I.e., is there a smooth function $Y(x)$, such that for every $x$: $f(x, Y(x))=0$?

Intuitively, if $f(x,y)$ describes the height of a certain landscape at coordinates $(x,y)$, then the zeros are the sea-level locations, and the question is: can one walk from west to east in a smooth path which is always on the sea-level?

• How about $f(x,y)=y^3-y-x$? Where $f=0$, you’re talking about $x=y^3-y$. Graph this. – Lubin Dec 14 '14 at 15:54
• @Lubin I see. You mean that $f=0$ will be a smooth curve, but not a function, since for some values of $x$ there is more than one value of $y$. So the answer to my question, as it is asked, is "no". – Erel Segal-Halevi Dec 15 '14 at 7:31
• Right. I guess the most precise way of posing your question is whether the zero-set of a smooth function contains the graph of a smooth function. – Lubin Dec 16 '14 at 0:13

Hint: Implicit function theorem.

• Thanks! This seems to solve the problem, although I am not sure about the pre-conditions. I have to study this theorem more deeply. – Erel Segal-Halevi Dec 14 '14 at 15:10