How can I prove the following:
Given a polynomial $f(x)$ of third degree, suppose you have another polynomial $S(x)$ which is a B-spline representation of $f(x)$. Prove that it is not possible to make a natural spline polynomial of $S(x)$.
My approach was that since $S(x)$ is not a smooth function because it is in segments, a natural spline representation would not be possible. Also, if you take the same points you used for the B-spline, then it is not even differentiable.
How can I make it a much more formal proof and is my approach correct?
Thanks.
It is unclear to me what you mean with "....make a natural spline polynomial of $S(x)$", so I will assume you meant "$S(x)$ is not a natural cubic spline".
To prove this is pretty simple. Assuming $S(x)$ exactly represents the $[x_0, x_1]$ portion of $f(x)$, if $S(x)$ is a natural cubic spline then its 2nd derivatives should be zero at $x_0$ and $x_1$ (this is by definition). Consequently, we must have $f''(x_0)=f''(x_1)=0$. This condition is not necessarily true and therefore $S(x)$ is not necessarily a natural cubic spline.