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I wonder if the following is true:

Conjecture: Let $I \subset \Bbb R$ be an open interval and $f, g: I \to \Bbb R$ be differentiable functions. Then the Wronskian $$ W(f,g) =\begin{vmatrix}f &g \\f' & g'\end{vmatrix} = f g' - f'g $$ is a Darboux function.

A Darboux function is a real-valued function $f$ which has the “intermediate value property”: for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$.

Motivation and thoughts:

  • In 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 $I\setminus A$? it what proved that

    If the Wronskian takes both positive and negative values on an interval, then it must be zero somewhere.

    and my conjecture would be a natural generalization.

    However, I do not yet see how the case of an arbitrary intermediate value $y$ can be reduced to the special case of $y = 0$ as the intermediate value.

  • The conjecture is (trivially) true if both $f$ and $g$ are continuously differentiable, since then $W(f, g)$ is continuous. So the interesting case is that $f$ and $g$ are just assumed to be differentiable.

  • Derivatives have the Darboux property, that covers the case that $f$ or $g$ is constant, e.g. $W(1, g) = g'$.

  • Sums and products of Darboux functions are not necessarily Darboux functions (see for example The sum of Darboux is a Darboux function?). So even if all terms in $f g' - f'g$ have the intermediate value property, there is no immediate way to conclude the conjecture.

A (failed) proof attempt:

Assume that $w = W(f, g)$ does not take a value $y \in \Bbb R$, and consider the sets $$ A = \{ x \in I \mid w(x) > y \} \, , \, B = \{ x \in I \mid w(x) < y \} \, . $$ If we can show that both $A$ and $B$ are open then one of them must be empty (since $I$ is connected), and we are done.

If $f(x_0) \ne 0$ then we can define $h(x) = y \int_{x_0}^x f(t)^{-2} dt$ in a neighborhood of $x_0$, and $$ W(f, g) -y = f^2 \left( \frac gf - h\right)' $$ shows that $W(f, g) -y$ does not change its sign near $x_0$, so that $x_0$ is an interior point of $A$ or of $B$.

A similar argument works if $g(x_0) \ne 0$. However, other than in my previous answer, one can not exclude the case $f(x_0) = g(x_0) = 0$. That is where my I am stuck in my current proof attempt.

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  • $\begingroup$ Why can’t the product rule be used on $-fg$? To show its a derivative. $\endgroup$ Jul 11, 2022 at 14:41
  • $\begingroup$ @VioletFlame: The derivative of $-fg$ is $-f'g - fg'$, and not equal to $W(f, g)$. $\endgroup$
    – Martin R
    Jul 11, 2022 at 14:42
  • $\begingroup$ Yes but it’s close enough. Maybe something can be done. $\endgroup$ Jul 11, 2022 at 14:46
  • $\begingroup$ They don't have to be open. Consider $f(x) = x^2 \sin(1/x)$, $g(x) = 1$, $y = 1/2$ - $0 \in B$, but it's not an inner point of $B$. $\endgroup$
    – mihaild
    Jul 11, 2022 at 22:07
  • $\begingroup$ @LostinSpace: No. As in math.stackexchange.com/a/4490486/42969, the idea is to prove the openness from the Darboux theorem, here applied to $\left( \frac gf - h\right)'$. That works as long as $f(x_0) \ne 0$ or $g(x_0) \ne 0$. $\endgroup$
    – Martin R
    Jul 12, 2022 at 8:44

1 Answer 1

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Wronskian of two differentiable functions may not have Darboux property. Below the conformation due to Józef Banaś; Wagdy Gomaa El-Sayed

Let $I=[0, 1]$ . Consider two functions $f, g $ defined by

$f(x) =\begin{cases}x^2 \sin(\frac{1}{x^4})&x\neq 0\\0 & \text{otherwise}\end{cases}$

$g(x) =\begin{cases}x^2 \cos(\frac{1}{x^4})&x\neq 0\\0 & \text{otherwise}\end{cases}$

For $x=0$ , $W(f, g) (0) =0$

For $x\neq 0$ , $\begin{align}W(f, g) (x) &=\begin{vmatrix} \sin(\frac{1}{x^4}) & \cos(\frac{1}{x^4})\\2x\sin(\frac{1}{x^4})-\frac{4}{x^3} \cos(\frac{1}{x^4})&2x\cos(\frac{1}{x^4})-\frac{4}{x^3} \sin(\frac{1}{x^4}) \end{vmatrix}\\&=\frac{4}{x}\end{align}$

Hence $x\to W(f, g) (x) $ doesn't have the Darboux property.


$\color{red}{\textbf{ Theorem}}:$

Let $f, g: I \to\Bbb{R}$ be functions differentiable on the interval $I$. Assume that the set $ Z_f $(or $Z_g$) has no accumulation points, and $Z_f \cap \overline{ Z_g} = \emptyset$ (or $ \overline{Z_f} \cap{Z_g} = \emptyset$). Then $W(f,g)$ has the Darboux property on $I$.

Source : BANAS, J.—EL-SAYED, W. G.: Darboux property of the Wronski determinant, Math. Slovaca $45 (1995), 57–61.$ (Online available here.)

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