# extending the implicit function theorem

Here's the problem:
Let $$f:\mathbb{R}^2 \to \mathbb{R}$$ be a $$C^1$$ map. Let $$c>0$$ be such that for all $$x,y_1,y_2 \in \mathbb{R}$$, we have
$$(f(x,y_1)-f(x,y_2))(y_1-y_2)\geq c \mid y_1-y_2\mid ^2$$

Prove that for any $$x\in \mathbb{R}$$, there is a unique $$y(x)\in\mathbb{R}$$ such that $$f(x,y(x))=0$$ and $$y:x\to y(x)$$ is a continuous differentiable map from $$\mathbb{R}$$ to $$\mathbb{R}$$.

Here's my idea: It seems that $$f(x,y_1)-f(x,y_2)$$ and $$(y_1-y_2)$$ are of the same direction, so $$g_x(y)=f(x,y)$$ is a monotonic function of $$y$$. Thus by the intermediate value theorem, we know the first part of the proof, i.e.for any $$x\in \mathbb{R}$$, there is a unique $$y(x)\in\mathbb{R}$$ such that $$f(x,y(x))=0$$.
We have to use the implicit function theorem in the second part. But the theorem only tells us that there are neighborhoods of $$x$$ and of $$y$$ where such function exists. How do we extend that to the whole $$\mathbb{R}$$?

When $$y(x)$$ is not unique, the implicit function theorem allows you to claim existence of a function within a neighborhood. However, if you know $$y(x)$$ is unique for all $$x$$, by definition $$y: x \rightarrow y(x)$$ is a function, without needing to use the implicit function theorem. You can verify its continuity and differentiability from that of $$f$$.
• I think I know how to show the continuity and differentiability of $y(x)$, but I don’t know how does the uniqueness extends the domain of the implicit function to $\mathbb{R}$ May 31, 2021 at 18:32