$ \newcommand{\real}{\mathbb{R}} $ A $\color{red}{\text{(strictly)}}$ positive definite kernel $k: \real^d\times \real^d \to \real$ satisfies for all $x_i \in \real^d$, $a=(a_1,\dots, a_n)\in \real^d$ $$ \sum_{i=1}^n a_i a_j k(x_i, x_j) \ge 0 \quad \color{red}{>0 \iff a\neq 0} $$ The reproducing kernel hilbertspace (RKHS) $H=H(k)$ of kernel $k$ is then defined as the closure of $$ S = \Bigl\{ u: \real^d\to\real \;:\; u = \sum_{i=1}^n a_ik(x_i, \cdot), a_i \in \real, x_i\in \real^d, n\ge 1\Bigr\} $$ where the closure is with respect to the norm $\|\cdot\|_H$ induced by the dot product defined by the bilinear extension of $$ \langle k(x_i, \cdot), k(x_j, \cdot\rangle_H = k(x_i, x_j). $$ I have two questions:
- Is this norm $\|\cdot\|_H$ really a norm (i.e. does positive definiteness hold, in the sense that $\|u\|_H = 0$ implies $u=0$?
- If yes in general or assuming the kernel is strictly positive definite, how can you show that? I would think that this should be wrong if the kernel is not strictly positive definite, but maybe I don't see something important and literature (e.g. Adler (2007)) calls $\|\cdot\|_H$ a norm without proof). And I could not find an actual proof.
I would assume that $\| \cdot \|{H}$ is defined as $\| x \|{H} := \sqrt{\langle x, x \rangle_{H}}$ with $\langle \cdot, \cdot \rangle_{H}$ being an inner product. Since for an inner product, we have $\langle x, x\rangle = 0$ if and only if $x = 0$, we conclude that: $$ \| x \|{H} = 0 \Longleftrightarrow \langle x, x\rangle = 0 \Longleftrightarrow x = 0 $$ which proves that $\| x \|{H}$ is positive definite. The property is thus independent of the kernel.
However, in the context of function spaces, it is crucial to note that "=" here means "almost everywhere".