Linear independence of a set of mappings $Map(\mathbb{R},\mathbb{R}):=$ The set of all mappings from $\mathbb{R} \rightarrow \mathbb{R}$
For every $a \in \mathbb{R}$ there is a Funktion $f_{a}:\mathbb{R} \rightarrow \mathbb{R}$ with:
$$
f_{a}(x) = \left\{ 
  \begin{array}{l l}
    (a-x)^{3} & \quad \text{if $x \le a$}\\
    0 & \quad \text{else}
  \end{array} \right.
$$
Show, that the Set $\{f_{a}|a \in \mathbb{R}\} $ in the real vector space $Map(\mathbb{R},\mathbb{R})$ is linear independent.
I wonder how to determine that there is no linear combination that equals zero, other than with all koefficients being zero. Any hints how to start on this?

You have to show that if you have a finite linear combination of functions of the form $f_a$, which is identically zero, then all coefficients have to be zero. So let $\lambda_1f_{a_1}+\ldots+\lambda_nf_{a_n}=0$, that is, $\lambda_1f_{a_1}(x)+\ldots+\lambda_nf_{a_n}(x)=0$ for each $x\in\mathbb R$. We want to show that $\lambda_1=\ldots=\lambda_n=0$. Now, use that for each $x\leq a$ the function $f_a$ is a polynomial of degree 3. Then your linear combination is a polynomial also of degree 3 for $x\leq\min\{a_1,\ldots,a_n\}$, which is zero. What do you know about such polynomials?
