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 $I\setminus A$? 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 $x\in I$

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 $x$?

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 ${\displaystyle x} \in {\displaystyle I}$

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 $I$ .

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

*

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


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


*$f, g$ 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 $I\setminus A$ ?


 A: 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.
