Prove that $f$ is a linear combination of the functions $k_{y_i}(x) = min\{x,y_i\}$.

Question

This is exercise 2.4 of "An Introduction to the Theory of Reproducing Kernel Hilbert Spaces" by Paulsen, and it states:

Let $y_0 = 0 < y_1 < \dots < y_n$ be given and let $f: [0,\infty) \to \mathbb R$ be a functions satisfying:

• $f$ is continuous;
• $f(0) =0$;
• $f$ is linear on each of the intervals, $[y_i, y_i+1], 0\leq i \leq n-1$;
• $f$ is constant for $x \geq y_n$.

Prove that $f$ is a linear combination of the functions $k_{y_i}(x) = \min\{x,y_i\}$.

How do you prove this with RKHS theory? Of course, we have that $K(x,y) = \min\{x,y\}$ is a kernel and by Moore's Theorem there exists a unique RKHS $\mathcal H(K)$ of functions on $\mathbb R$. If we now knew that $f$ is in that space, the rest would follow. But how can we conclude that?

Answer 1

Let $V$ be space of all functions $f$ as you described. There are a few easy things to check:

• $V$ is a linear space
• $\dim V = n$ (note that any function $f$ is given by its values in $y_1,\ldots,y_n$)
• $k_{y_1},\ldots,k_{y_n} \in V$
• $k_{y_1},\ldots,k_{y_n}$ are linearly independent (note that any $k_{y_m}$ cannot be a linear combination of $k_{y_1},\ldots,k_{y_{m-1}}$)

By a dimension argument, $k_{y_1},\ldots,k_{y_n}$ forms a basis of $V$, and hence any function $f$ is a linear combination of $k_{y_1},\ldots,k_{y_n}$.