A weird theorem of existence

Question

Im struggling with a problem which seems to be an application of implicit theorem function. It's really hard for me and I would love some help.

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Let $F: \mathbb{R}^2 \to \mathbb{R}$ a $C^1$ function. Suppose that :

$\bullet$ $\forall x \in \mathbb{R}$, $\lim\limits_{u \rightarrow -\infty} F(x,u) = -\infty$.

$\bullet$ $\forall x \in \mathbb{R}$, $\lim\limits_{u \rightarrow +\infty} F(x,u) = +\infty$.

$\bullet$ $\forall K \in \mathbb{R}$, $\exists C \in \mathbb{R}$, $\forall (x,u) \in \mathbb{R}^2$, $-K \leq F(x,u) \leq K$ $\implies$ $-C \leq u \leq C$.

$\bullet$ $\exists M \in \mathbb{R}$, $\forall (x,u) \in \mathbb{R}^2$, $F(x,u)=M$ $\implies$ $\partial_uF(x,u) \neq 0$.

Then exists $\nu : \mathbb{R} \to \mathbb{R}$ a $C^1$ function such that : $\forall x \in \mathbb{R}$, $F(x,\nu(x))=M$.

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In my opinion we have to apply implicit theorem function to show existence, and compacity to show definition (by hypothesis 3 we can have totally bounded, and completeness leads to compactness), but i have absolutely no clue how to write a proof.

Answer 1

Let us call $(I,f)$ a solution if $I$ is a non-empty real interval and $f:I\to\Bbb R$ is $C^1$ on $I$ and satisfies $F(x,f(x))=M.$

The set of solutions is non-empty: by your 1st and 2nd hypothesis, there exist pairs $(a,b)$ such that $F(a,b)=M,$ i.e. $(\{a\},a\mapsto b)$ is a solution (in such a case, the $C^1$ property is true by convention).

It is inductively ordered by : $(I,f)\le(J,g)\iff I\subset J\text{ and }f$ is the restriction of $g$ to $I.$

Let $(I,\nu)$ be a maximal solution. Then $I$ is non-empty and:

• open: if $x\in I$ then, by the local existence and uniqueness in the implicit function theorem, $\nu$ "extends" to a solution on a "larger" interval which is a neighborhood of $x,$ but by maximality, this interval is $I$ itself;
• closed: if $x\in\bar I,$ admit for the moment (lemma below) that $\nu$ admits at $x$ a finite limit, and call $\bar\nu$ the continuous extension of $\nu$ to the interval $I\cup\{x\}.$ By continuity, $F(x,\bar\nu(x))=M$ so by your 4th hypothesis, $\partial_uF(x,\bar\nu(x))\ne0.$ Since $F$ is $C^1,$ this proves that the function $t\mapsto\nu'(t)=-\frac{\partial_xF}{\partial_uF}(t,\nu(t))$ admits a limit at $x,$ hence $\bar\nu$ is $C^1,$ so by maximality $x\in I.$

By connexity, $I=\Bbb R.$

Now the lemma: By the 3rd hypothesis, let $C\in\Bbb R$ be such that $\forall(t,u)\in\Bbb R^2\;(F(t,u)=M\implies |u|\le C).$ On $[x-1,x+1]\times[-C,C],$ $\frac{\partial_xF}{\partial_uF}$ is bounded hence on $[x-1,x+1]\cap I,$ $\nu'$ is bounded, so that $\nu$ is Lipschitz continuous hence uniformly continuous. By the Cauchy criterion for functions, this ensures that $\nu$ admits at $x$ a finite limit.