Implicit Function Thm: Let $f: \Bbb{R} \to \Bbb{R}$ be a non-constant $C^1$ function such that $f'(0) \neq 0$ and $f(x + y) = f(x)f(y)$.

Question

Let $f: \Bbb{R} \to \Bbb{R}$ be a non-constant $C^1$ function such that $f'(0) \neq 0$ and $f(x + y) = f(x)f(y)$. Define $F: \Bbb{R}^2 \to \Bbb{R}$ by $F(x, y) = f(x)f(y)$. Show that no other conditions are required to ensure that $y$ can be solved implicitly as a function of $x$ on the set {$(x, y) : F(x, y) = 1$}.

Don't really know where to start, maybe find the gradient of $F$? I don't really see where $f(x + y) = f(x)f(y)$ will come in play either. Any help appreciated.

Answer 1

We have $f(0)=f(0+0)=f(0)^2$ , hence $f(0)=0$ or $f(0)=1$. If $f(0)=0$, then $f(x)=f(x+0)=f(x)f(0)=0$ for all $x$, a contradiction. Therefore $f(0)=1$.

Now define $G(x,y)=F(x,y)-1$ and show:

$G(0,0)=0$ and $G_y(0,0) \ne 0$.

The implicit function theorem now says: there is a neigborhood $U$ of $0$ and exactly one function $g:U \to \mathbb R$ such that

$g(0)=0$ and $F(x,g(x))=1$ for all $x \in U$.