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 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.


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.

  • $\begingroup$ What do you mean by "its unique smooth extension"? $\endgroup$ – Dirk May 19 '15 at 6:43

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.

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