Do Wronskians have the intermediate value property?

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:

• 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.

• Why can’t the product rule be used on $-fg$? To show its a derivative. Jul 11, 2022 at 14:41
• @VioletFlame: The derivative of $-fg$ is $-f'g - fg'$, and not equal to $W(f, g)$. Jul 11, 2022 at 14:42
• Yes but it’s close enough. Maybe something can be done. Jul 11, 2022 at 14:46
• 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 $B$. Jul 11, 2022 at 22:07
• @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 $g(x_0) \ne 0$. Jul 12, 2022 at 8:44

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.)