# Proving that $f$ functions of compact support are linearly independant.

I am reading some notes, along with that I am trying to follow the mathematics behind.

I have a set of functions that are the derivatives of the head functions. Regardless from where they come, they are defined as

$f_i(x)= \begin{cases} 4 & x_{i_1}\leq x < x_i \\ -4 & x_i \leq x\leq x_{i+1} \\ 0 & \text{otherwise} \end{cases}.$

So the text gives the example of $f_1,f_2,f_3,f_4,f_5$ on an interval $[0,1]$ where we make the step size between the $x_i$ and $x_{i+1}$ to be $1/4$. It also claims that they are linearly dependent, but $f_2,f_3,f_4$ are independent. My thinking first was that this makes sense because

for example $$f_1(x)+f_2(x)+f_3(x)+f_4(x)+f_5(x) =0,$$

where I just pick my $a_i=0$ for all $i$ and $x=0.2$. This shows that the $f_i$ are dependent. But my problem is why are $f_2,f_3,f_4$ independent?

Can't we do the same thing again $f_2(0.6)+f_3(0.6)+f_4(0.6)=-4+4+0=0$

Can someone help me understand this better?

It's not clear to me why $f_1,\dots, f_5$ are dependent; that seems to depend on which $x_i$ were chosen.
In any case, the core confusion seems to be this: in order to say that $f_2,f_3,f_4$ are linearly dependent functions, we must be able to say something like $f_2(x) + f_3(x) + f_4(x)=0$ for every $x$. That is, some (non-trivial) linear combination of $f_2,f_3,f_4$ must produce the zero-function.