Let $N\_1:=\chi\_{[0,1]}$ be defined as this characteristic function and $N\_n:=N\_{n-1}\*N\_1$ then this leads to polynomials with support $[0,n]$. These splines are well-studied [click for wikipedia](http://en.wikipedia.org/wiki/Spline_wavelet) My question is now: If you take the polynomial $N\_n|\_{[i,i+1]}$ and consider its unique smooth extension on $[0,n]$ which we denote $f\_n,$ then I guess that $f\_1,..,f\_n$ are linearly independent. 


Example: 


For $n=2$ we get $N\_2(x):=x$ on $[0,1]$ and $N\_2(x):=-x+2$ on $[1,2]$ and $N\_2$ is zero anywhere else. Now, it is true that $f\_1(x):=x$ and $f\_2(x):=-x+2$ are linearly independent as functions on $[0,2]$. 


Is there a way to prove that this is true for all $n$? Due to their recursive definition, there is at least for me, no obvious way to do this.



Splines linearly independent
Your conjecture is true. Assume that $n$ is the smallest integer such that these polynomials are linearly dependent. Then there exist $a\_0,\dots,a\_{n-1}$ with
$$f(x):=\sum\_{i=0}^{n-1} a\_i N\_n(x+i)=0$$
for all $x\in [0,1]$. Let
$$g(x):=\sum\_{i=0}^{n-1} a\_i N\_{n-1}(x+i).$$
By the definition of the splines we have
$$f(x)=(N\_1\*g)(x)=0$$
for all $x\in [0,1]$. It follows that 
$$\int\_0^1 g(y)dy=0 \quad \text{and} \int\_0^x g(y)dy=\int\_1^{1+x} g(y)dy$$ 
for all $x\in [0,1]$. It follows that
$$g(1+x)=g(x) \text{ for all } x\in [0,1]$$
and
$$g^{(i)}(0)=g^{(i)}(1) \text{ for all } i\in \{0,\dots,n-3\}$$
Let $p$ be the smooth extension of $g|\_{[0,1]}$. Assume that $p\neq 0$ and let $a\_kx^k$ be the leading term of $p$. Then $0=g^{(k-1)}(1)-g^{(k-1)}(0)=k!a\_k$ which is a contradiction. Hence $g(x)=0$ for all $x\in [0,1]$. As $N\_n|\_{[n-1,n]}\neq 0$ at least one of the coefficient $a\_0,\dots,a\_{n-2}$ is nonzero. It follows that
$$\{ N\_{n-1}|\_{[i,i+1]}|i\in \{0,\dots,n-2\}\}$$
are linear dependent which is a contradiction to the choice of $n$ being minimal.


