I'm struggling to show this:
For all vector fields $ Y\in C^\infty_0([a,b],\mathbb{R}^n) $ along a parametrized curve $ \gamma_0:\ [a,b]\rightarrow \mathbb{R}^n $ there is a variation $ \gamma $ with variational vectorfield $ Y $.
$ C^\infty_0([a,b],\mathbb{R}^n):=\{Y:\ [a,b]\rightarrow \mathbb{R}^n \text{smooth}|\exists \ \delta > 0 : Y_{|[a,a+\delta]\cup[b-\delta,b]}\equiv 0\} $
My idea: Let $ Y\in C^\infty_0([a,b],\mathbb{R}^n) $ ba an arbitrary vector field along an arclenght-parametrizied curve $ \gamma_0:\ [0,L]\rightarrow \mathbb{R}^n $. This means that for this vector field $ Y:\ [0,L]\rightarrow \mathbb{R}^n $ there is a $ \delta>0 $ with the property $ Y_{|[0,\delta]\cup[L-\delta,L]}\equiv 0\in \mathbb{R}^n $. Define a curve $ \gamma: \ [0,L]\times ]-\varepsilon, \varepsilon[\rightarrow \mathbb{R}^n $ such that $ \gamma(x,t)=\gamma_0(x) $ for all $ x\in [0,\delta]\cup[L-\delta,L] $. Then $ \gamma $ is a compactly supported variation of $ \gamma_0: \ x\mapsto \gamma(x,0) $.
From here I have no idea how to proceed. My first concept would be trying to extend my compactly supported variation to a variation of $ \gamma_0 $.
EDIT: A $C^\infty$-Map $ \gamma: \ [a,b]\times ]-\varepsilon,\varepsilon[ \rightarrow \mathbb{R}^n$ is called variation of the map $ \gamma_0: [a,b]\rightarrow \mathbb{R}^n $ such that $ \gamma(x,0)=\gamma_0(x) $ for all $ x\in [a,b] $.