I am trying to show that two interpolants are identical.
Suppose that $x_k$, $k=1,\dots,2^d$ are points on the vertices of a rectangle $D$ in $d$-dimensions. Assume that $f(x_k)$ is available for all values of $x_k$.
I have 2 interpolants, namely:
$\tilde{f_1}(x) = \sum_{k=1}^{2^d} w_k(x)f(x_k)$
and
$\tilde{f_2}(x) = \sum_{k=1}^{2^d} v_k(x)f(x_k)$.
These 2 interpolants satisfy the following properties:
1) $\tilde{f_1}(x_k) = \tilde{f_2}(x_k) = f(x_k), \forall \, k.$
2) $w_k(x), v_k(x) \ge 0$ and $\sum_k w_k(x) = \sum_k v_k(x) = 1$, i.e. each interpolant is a convex combination of the known values of $f$.
3) if $f$ is linear, i.e. $f(x) = a^Tx + b$ where $a \in \mathbb{R}^d$ then $f = \tilde f_1 = \tilde f_2$ on the rectangular domain $D$.
In 1-d and 2-d, it is easy to show via geometric arguments that these 2 interpolants are identical. However, it is hard to visualize the problem in higher dimensions.
Has anyone encountered such a problem?
It is not straightforward to show that $w_k = v_k$ for all $k$ and so I am wondering if there's just a theorem that I did not know which implies the statement I want to show.
It does not seem to be true.
Consider $d = 2$, $D$ being unit square and $w_k(\xi, \eta)$ being bilinear interpolation basis $$\begin{aligned} w_1(\xi, \eta) &= (1 - \xi)(1 - \eta)\\ w_2(\xi, \eta) &= \xi(1 - \eta)\\ w_3(\xi, \eta) &= \xi\eta\\ w_4(\xi, \eta) &= (1 - \xi)\eta. \end{aligned} $$
To define $v_k$ let's split $D$ into two triangles (to be certain, along $\xi = \eta$). $$\begin{aligned} v_1(\xi, \eta) &= \begin{cases} 1 - \xi,&\xi > \eta\\ 1 - \eta,&\xi \leq \eta \end{cases}\\ v_2(\xi, \eta) &= \begin{cases} \xi-\eta,&\xi > \eta\\ 0,&\xi \leq \eta \end{cases}\\ v_3(\xi, \eta) &= \begin{cases} \eta,&\xi > \eta\\ \xi,&\xi \leq \eta \end{cases}\\ v_4(\xi, \eta) &= \begin{cases} 0,&\xi > \eta\\ \eta-\xi,&\xi \leq \eta \end{cases} \end{aligned} $$
Every basis satisfies delta property, that is $$ v_i(x_k) = w_i(x_k) = \delta_{ik}. $$ Thus, 1) and 2) are satisfied.
Direct checking for $f(\xi, \eta) = \alpha \xi + \beta \eta + \gamma$ shows that 3) holds also.