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 $\mathbb{R}$?


1 Answer 1


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 $f$.

  • $\begingroup$ 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 $\mathbb{R}$ $\endgroup$
    – WinnieXi
    May 31, 2021 at 18:32
  • $\begingroup$ I edited my answer. I hope I understood you correctly. $\endgroup$
    – imkevinkuo
    May 31, 2021 at 18:34

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