$g$ ($h$) interpolates $f$ at $x_0,\ldots x_{n-1}$ ($x_{1},\ldots,x_n$), then $g(x)+\frac{x_0-x}{x_n-x_0}[g(x)-h(x)]$ interpolates $f$

Question

If $g$ interpolates the function $f$ in $x_0, \ldots x_{n-1}$ and if $h$ interpolates the function $f$ in $x_{1}, x_{2}, \ldots, x_n$, then $$ g(x) + \frac{x_0 - x}{x_n - x_0}\big[ g(x) - h(x)\big] $$ interpolates $f$ in $x_{0}, \ldots, x_{n}$.

I do not know how to show this though. The exercise notes that $g$ and $h$ need not be polynomials.

So if $g$ interpolates $f$ in $x_0, \ldots, x_{n-1}$, then $g(x_i) = y_i = f(x_i)$. Similar for $h$. The difference $g(x) - h(x)$ gives $g(x_n) - h(x_0)$ since they share every other point $x_i$ for $0 < i < n$. From here, I am not sure where to go.

Answer 1

Let $$ \zeta(x) := g(x) + \frac{x_0 - x}{x_n - x_0}\big[ g(x) - h(x)\big] $$ We want to show $\zeta(x_i) = f(x_i)$ for $i \in \{0, \ldots, n\}$.

For $i = 0$ we have $$ \zeta(x_0) = g(x_0) + \underbrace{\frac{x_0 - x_0}{x_n - x_0}}_{= 0}\big[ g(x_0) - h(x_0)\big] = g(x_0) = f(x_0). $$ For $i \in \{1, \ldots, n - 1\}$ we have $f(x_i) = g(x_i) = h(x_i)$ and, therefore, $$ \zeta(x_i) = g(x_i) + \frac{x_0 - x_i}{x_n - x_0}\underbrace{\big[ g(x_i) - h(x_i)\big]}_{= 0} = g(x_i) = f(x_i). $$ Finally, for $i = n$ we have $$ \zeta(x_n) = g(x_n) + \underbrace{\frac{x_0 - x_n}{x_n - x_0}}_{= -1}\big[ g(x_n) - h(x_n)\big] = g(x_n) - (g(x_n) - h(x_n)) = h(x_n) = f(x_n), $$ which concludes the proof.