# How to prove Linear Independence of piecewise functions?

Suppose we have two functions: $f_1(x)=x^2,x\geq 0$ and $f_1(x)=0,x\leq 0$ and $f_2(x)=0,x\geq 0$ and $f_2(x)=x^2, x\leq0$. Show that these two functions $f_1(x)$ and $f_2(x)$ are linearly independent.

My attempt: Consider the interval $[0,\infty]$. Then $f_1(x)=x^2$ and $f_2(x)=0$. The Wronskian $W(f_1,f_2)=0$. Since the Wronskian is zero for every $x\in [0,\infty]$, we conclude that over this interval the functions are linearly dependent. Similar reasoning shows that for $x\in[-\infty,0]$ the functions are linearly dependent. Thus, one might conclude for all $x$ the functions are linearly dependent. However, this is not true. What is wrong with my argument?

I know that the Wronskian being equal to zero is not a sufficient condition of linear dependency, but then several solutions make use of this condition directly to establish linear independency.

• Why not assume that $af_1+bf_2=0$ and show that $a=b=0$? – Did Jul 3 '16 at 9:40
• How should I go about doing that? They are piece wise functions, I've no idea how to add them. – nls Jul 3 '16 at 9:47
• Piecewise or not, the function $g=af_1+bf_2$ is defined by $g(x)=af_1(x)+bf_2(x)$ for every $x$. Thus $g=0$ if and only if $af_1(x)+bf_2(x)=0$ for every $x$, hence... – Did Jul 3 '16 at 9:56

remember that a zero vector can never be a vector in a free system; what you have to demonstrate is that the restrictions of $f_1$ and of $f_2$ to every interval are linearly dependent and not the functions themselves
to be clear in my answer, let $I=]-\infty,0]$ and $J=[0;\infty[$ for $i=1,2$ let $f_{i,I}$ and $f_{i,J}$ the restriction of $f_i$ respectively to $I$ and $J$. I said in my answer above that for showing that the system $\{f_{1,I},f_{2,I}\}$ (resp. $\{f_{1,J},f_{2,J}\}$) is linearly dependant, you need not in this case to use the Wronksian of $(f_{1,I},f_{2,I})$ (resp.$(f_{1,J},f_{2,J})$ because we know that a zero vector can not be in a linearly independent system.
in the seconde affirmation that if two maps $f$ and $g$ him restriction $f_{A},g_{A}$ to A (resp. $f_{B},g_{B}$ to B) are linearly dependent, where $A\cup B=D$ is the domain on which $f$ and $g$ are defined, the system $\{f,g\}$ can be not linearly dependant, as shown your example, note in your example you can not us the Wronksian for $f$ and $g$ because $f$ and $g$ are not derivable functions on $\Bbb{R}$, but you can use the definition directly to show that are linearly independents.