If the Wronskian is zero at some point, does this imply linear dependency of functions? We know that

*

*If for functions $f$ and $g$, the Wronskian $W(f,g)(x_0)$ is nonzero for some $x_0$ in $[a,b]$ then $f$ and $g$ are linearly independent on $[a,b]$.

*If $f$ and $g$ are linearly dependent then the Wronskian is zero for all $x_0$ in $[a,b]$.

My doubt is : If for some $x$, $W(f,g)(x)$ is zero, can we conclude that Wronskian is identically zero as we know that wronskian is zero or never zero.
In one problem, Wronskian $W$ was coming as $-x^2$ on $(\infty,-\infty)$. Since $W$ is $0$ for $x=0$ can we say Wronskian is identically zero OR using point 1 we may conclude that we are getting more than one point where Wronskian is not zero and hence functions are linearly independent.
 A: Suppose $f, g$ two real valued differentiable functions defined on an interval $I$. $W(f, g) =\begin{vmatrix}f&g\\f'&g'\end{vmatrix}=fg'-gf'$
Supposed $f, g$ are Linearly dependent on  $I$.Then $\exists \lambda\in\Bbb{R}$ such that $f=\lambda g$ on $I$.
Then $W(f, g) =W(f, g) =\begin{vmatrix}\lambda g&g\\\lambda g'&g'\end{vmatrix}=0$
$\color{red}{\text{$\{f, g\}$ Linearly dependent on $I\implies W(f, g) =0$ on $I$}}$ $$[W(f, g) (x) =0\quad \forall x\in I]$$

But converse is not true in general. The celebrated example $f(x) =x^2 $ and $g(x) =x|x|$ on an interval $I$ containing $0$ is Linearly independent but $W(f, g) =0$ on $I$.

If $f, g$ are two solutions of $y"+p(x) y'+q(x) y=0\tag 1$  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 ${\displaystyle x} \in {\displaystyle I}$
$\color{red}{\text{ $f, g$ are solutions of $(1) $ then $W(f, g)(x) =0\quad \forall x\in I \iff \{f,g\}$ Linearly dependent}}$

But it's not difficult to find two functions $f, g$ such that $W(f, g)(x_0) =0$ for some $x_0\in I$ but $W(f, g) $ not identically $0$ on $I$.
Points to be considered :

*

*$W(f, g) (x_0) \neq 0\quad $ for some $x_0\in I$ implies Linearly independent.


*$f, g$ can't be solutions of $(1) $.
Example : $f=1$ and $g=x^2$ on $(-1, 1) $
Then $W(f, g) =2x $
It's clear that $W(f, g) (x_0) =0\iff x_0=0$

For further study (you might be interested in):

*

*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 $x$?


*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 $I\setminus A$?


*Does the Wronskian of three or more linearly independent functions change its sign?


*Do Wronskians have the intermediate value property?
A: "Identically zero" means "equal to zero for all values of $x$". 
The function $-x^2$ is not identically zero, because there are values of $x$ (such as $1,2,3,\dots$) for which it's nonzero. 
Since the Wronskian of linearly dependent functions is identically zero, the functions whose Wronskian is $-x^2$ are not linearly dependent. 

As an aside: there is a   scenario in which $W$ is either always zero or never zero: it happens when the two functions are solutions of the ODE of the form $y''+p(x)y'+q(x)y=0$. For such solutions, the Wronskian satisfies the identity $W(t)=W(s)\exp\left(-\int_s^t p(x)\,ds\right)$ which implies that if $W$ is zero at some point, it is zero everywhere.
