Let $\textbf{x} = (x_1, \dots, x_n)^T \in \mathbb{R}^n$ and $k \in \mathbb{N}$. We define $$ X := \begin{pmatrix} 1 & x_1 & \cdots & x_1^k \\ \vdots & \vdots & & \vdots \\ 1 & x_n & \cdots & x_n^k \end{pmatrix} \in \mathbb{R}^{n \times (k+1)} $$ and a function $b:\mathbb{R} \rightarrow \mathbb{R}^{k+1}$ with $b: x_0 \mapsto (1, x_0, \dots, x_0^k)^T$. Then, we can compute the equivalent kernel $$l(x_0) = b(x_0)^T(X^T X)^{-1} X^T$$ for any $x_0 \in \mathbb{R}$. The question is: how can we prove that the sum of the components of $l(x_0)$ is $1$ for all $x_0 \in \mathbb{R}$ and $k \in \mathbb{N}$?
Further notes:
This question arises in the context of local polynomial regression and it is a simplified version of Exercise 6.2 in "Elements of Statistical Learning".
I tried induction over $k$ and used a formula for blockwise inversion with no luck.
The sum of the components of $l(x_0)$ can be expressed as $l(x_0) \mathbf{1}_{n}$. After defining $S_k := \sum_{i=1}^n x_i^k$ we can rewrite the sum as follows \begin{align*} l(x_0) \mathbf{1}_{n} &= b(x_0)^T (X^TX)^{-1} X^T \mathbf{1}_{n} \\ &= b(x_0)^T (X^TX)^{-1} (S_0, \dots, S_k)^T \\ &= b(x_0)^T \begin{pmatrix} S_0 & S_1 & \cdots & S_k \\ \vdots & \vdots & \ddots & \vdots \\ S_k & S_{k+1} & \cdots & S_{2k} \end{pmatrix}^{-1} \begin{pmatrix} S_0 \\ \vdots \\ S_k \end{pmatrix} \\ &\overset{(*)}{=} b(x_0)^T e_1 \\ &= 1 \;. \end{align*} At step $(*)$ we used the definition of the inverse of a matrix, i.e. $A^{-1} A = I = (e_1, \dots, e_n)$ for all invertible matrices $A$.