Let's consider $(n + 1) -$evenly distributed nodes on interval $[-1, 1]$ and we want to interpolation odd function $f$. I want to prove, that interpolation polynomial:
$$L_nf(x) = \sum_{j = 0}^n f(a_j)\prod_{k = 0, k \ne j}^n \frac{x - a_k}{a_j - a_k}$$
has to be an odd function.
My work so far
Let's firstly notice, that nodes $a_i$ can be written in a form of $a_i = \frac{2i}{n} - 1$. We want to prove that $L_nf(-x) = -L_nf(x)$. Let's rewrite left equality:
$$L_nf(-x) = \sum_{j = 0}^n f(a_j) \prod_{k = 0, k \ne j}^n \frac{-x-a_k}{a_j - a_k}=$$
$$= \sum_{j = 0}^n f(a_j)\prod_{k = 0, k \ne j}^n\frac{-x-\frac{2k}{n} + 1}{\frac{2j}{n} - 1 - \frac{2k}{n} + 1}= \sum_{j = 0}^n f(a_j) \prod_{k = 0, k \ne j}^n \frac{-xn - 2k + n}{2(j - k)} = $$ $$(-1)^n \sum_{j = 0}^n f(a_j) \prod_{k = 0, k \ne j}^n \frac{xn + 2k - n}{2(j - k)}$$
whereas,
$$-L_nf(x) = (-1)\sum_{j = 0}^n f(a_j) \prod_{k = 0, k \ne j}^n \frac{x - 2k + n}{2(j - k)}$$
And here - I'm not sure how to prove the equality of those two expressions. Also - somehow we have to use fact, that function $f$ is odd, but I don't see room for it. Could you please help me with this problem?
This claim relies on the fact that the minimal degree interpolation polynomial $p$, as obtained using the Lagrange interpolation formula, is unique for the given function table.
As $f(x)=-f(-x)$, the same $p$ also interpolates $-f(-x)$. Reversing the operations, here the symmetry of the sampling nodes sequence is used, this means that also $-p(-x)$ interpolates $f$. Due to the uniqueness this implies $p(x)=-p(-x)$.