Does there exists two differentiable functions $f, g$ on $I$ such that $W(f, g) (x) >0$ on $A$ and $W(f, g) <0$ on $I\setminus A$?

Question

Let $I=(0, 1) $ and $A=\mathcal{C}\cap (0, 1) $ where $\mathcal{C}$ denote Cantor set.

$\color{red}{Question}$ : Does there exists two differentiable functions $f, g$ on $I$ such that $W(f, g) (x) >0$ on $A$ and $W(f, g) <0$ on $I\setminus A$ ?

Where $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}$ denote the Wronskian of $f, g$ at $x\in I$

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Previous Question :

Does there exists two functions $f, g\in C^1(I)$ for which $W(f, g) (x) >0$ for some $x$ and $W(f, g) (x) <0$ for some $x$?

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Let me summarize

$W(f, g) (x) \neq 0$ for some $x\in I$ implies $\{f, g\}$ linearly independent.

If two functions are solutions of a differential equation $y"+p(x) y'+q(x) y=0$ on $I$ where $p, q\in C(I) $ then by Abel's identity we have

$$W(f, g) (x) =W(f, g) (x_o) e^{-\int_{x_0}^{x} p(t) dt}$$

Then $W(f, g) (x_0) \neq 0$ for some $x_0\in I$ implies $W(f, g) \neq 0$ on $I$

Moreover $W(f,g)$ different from zero with the same sign at every point ${\displaystyle x} \in {\displaystyle I}$

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If $f, g \in C^1(I) $ then

$w(x) =W(f, g) (x) =f(x) g'(x) -f'(x) g(x) $ is a continuous map on $I$ .

Then $S=\{x\in I : W(f, g) (x) >0\}$ is an open set.

Further if $W(f, g) $ attains both positive and negative values then by Darboux property $W(f, g) (x) =0$ for some $x\in I$ .

Hence if $f, g$ are solution of the same differential equation and $f, g$ has continuous derivative. Then $ W(f, g) (x) =0$ on $I$ .

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Hence to get two functions $f, g$ with the mentioned properties, we need to consider the following :

1. $f, g$ can't be the solution of a Linear Homogenous Differential Equation.

2. $f, g$ can't be continuously differentiable on $I$.

3. $f, g$ can't be linearly dependent on $I$.

Does there exists two differentiable functions $f, g$ on $I$ such that $W(f, g) (x) >0$ on $A$ and $W(f, g) <0$ on $I\setminus A$ ?

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Answer 1

We can prove the following:

Let $I$ be an interval and $f, g: I \to \Bbb R$ be differentiable functions on $I$. If the Wronskian $w = W(f, g)$ has no zeros on $I$ then it does not change its sign on $I$, i.e. it is either positive everywhere or negative everywhere.

This answers your question in the negative: It is not possible that the Wronskian takes both positive and negative values on an interval, without being zero somewhere. The nature of the set $A$ as a Cantor set is irrelevant.

Remark: If $f, g$ are continuously differentiable then $w$ is continuous, and the claim follows immediately from the intermediate value theorem. The following proof does not require that $f'$ or $g'$ are continuous.

Proof: It suffices to prove the claim for a compact interval $I$.

Let $x_0 \in I$. $f$ and $g$ have no common zeros, so $f(x_0) \ne 0$ or $g(x_0) \ne 0$ must hold. In the first case is $$ w(x) = W(f, g)(x) = f(x)^2 \left( \frac{g(x)}{f(x)}\right)' $$ in some neighborhood $U(x_0)$, and it follows from Darboux's theorem applied to $(g/f)'$ that $w(x)$ does not change its sign on $U(x_0)$. Similarly, if $g(x_0) \ne 0$, $$ w(x) = W(f, g)(x) = -g(x)^2 \left( \frac{f(x)}{g(x)}\right)' $$ does not change its sign on some neighborhood $U(x_0)$.

So every $x_0 \in I$ has a neighborhood $U(x_0)$ on which the Wronskian $w(x)$ does not change its sign. $I$ is compact, therefore it can be covered with finitely many such neighborhoods. It follows that $w$ does not change its sign on $I$. This finishes the proof.

Alternative proof: As above, show that every $x_0 \in I$ has a neighborhood $U(x_0)$ on which the Wronskian $w(x)$ does not change its sign.

It follows that both sets $$ A = \{ x \in I \mid w(x) > 0 \} \, , \, B = \{ x \in I \mid w(x) < 0 \} \, . $$ are relatively open in $I$. Since the interval $I$ is connected, it can not be the union of two non-empty disjoint relatively open sets, so one of $A$ and $B$ must be empty.