True/False: If the Wronskian of n functions vanishes at all points on the real line then these functions must be linearly dependent in R. I know that if a set of functions are linearly dependent, then its Wronskian = 0 at all values of t in the interval.
So can you conclude that if Wronskian = 0 for all values of t in the interval, then the functions must be dependent?
 A: The answer is no.  For instance, the functions $f_1(x) = x^2$ and $f_2(x) = x \cdot |x|$ are continuous with continuous derivatives, have a Wronskian that vanishes everywhere, but fail to be linearly dependent.
The Wronskian Wikipedia page has a good discussion about this. Note that if the set of functions considered is analytic, then their dependence over an interval is indeed equivalent to their having a Wronskian that is identically zero.  
A: The other direction does not hold. A counterexample is given by the functions $f_1,f_2:\Bbb R\to \Bbb R$ defined by
\begin{align*}
f_1(t)=\begin{cases}
0, &t\leq0\\
t^2, &t>0
\end{cases},\qquad f_2(t)=\begin{cases}
t^2, &t\leq0\\
0, &t>0
\end{cases}.
\end{align*}
The Wronkian is zero at every $t\in\Bbb R$ but the functions are linearly independent. Indeed, if $c_1f_1(t)+c_2f_2(t)=0$ for some constants $c_j\in\Bbb R$ and all $t\in\Bbb R$ we have
\begin{align*}
c_1&=c_1f_1(1)+c_2f_2(1)=0,\\
c_2&=c_1f_1(-1)+c_2f_2(-1)=0.
\end{align*}
A: Answer is no because for instance the functions 
\begin{equation}
\begin{aligned}
&\left\{\begin{array}{l}
f(x)=0 \text { for } x<0 \\
f(x)=x^{2} \text { for } x \geqslant 0
\end{array}\right.\\
&\left\{\begin{array}{l}
g(x)=0 \text { for } x>0 \\
g(x)=x^{2} \text { for } x \leq 0
\end{array}\right.
\end{aligned}
\end{equation}
Functions have 0 wronksian determinant but apparently they are linearly independent.
