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May I ask a silly question? I want to be sure if it is true that the unions of reproducing kernel of all points in Hilbert space spanned it?

i.e. if $K: X \to \mathbb{C}$ is reproducing kernel of H, then $\cup K(.,x) = H \, \, \forall x \in X $?

hala sami
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1 Answers1

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No, this is generally not the case for two reasons:

  • If $K$ is a bounded kernel, i.e., $M = \sup_{x \in X} \lVert K(\,\cdot\,, x) \rVert < \infty$, then clearly $\bigcup_{x \in X} K(\cdot, x) \subseteq \mathrm B[0, M] \neq H.$ This indicates that one rather should consider the set $$ E := \bigcup_{x \in X} \operatorname{span} \{K(\,\cdot\,, x)\}.$$ However, one would like to consider also vectors that are linear combinations of elements in $E$. Thus, $F := \operatorname{span} \{K(\,\cdot\,, x) : x \in X\}$ is really the right set to discuss. Clearly, we have $E \subsetneq F$ in general and $E$ is generally not even a vector space and this is not a consequence of our specific setting rather than basic linear algebra. Indeed, let $H = \mathbb R^2$, $E = \operatorname{span}\{ (1, 0)\} \cup \operatorname{span}\{(0, 1)\}$ and $F = \operatorname{span} \{(1, 0), (0, 1)\}$. Then clearly, $$E = \{(x, y) \in \mathbb R^2 : (x \in \mathbb R \text{ and } y = 0) \text{ or } (x = 0 \text{ and } y \in \mathbb R)\}$$ and $(1, 1) = (1, 0) + (0, 1) \notin E$ whereas $(1,0), (0, 1) \in E$. However, $F = \mathbb R^2$.
  • If $H$ is not finite-dimensional with basis $(e_n)_{n \in \mathbb N}$ (i.e., $H$ is seperable), then $H \neq \operatorname{span} \{e_n : n \in \mathbb N\}$. For example let $H = \ell^2(\mathbb N)$ and $e_n$ the $n$-th unit vector. Then one has $(1/n)_{n \in \mathbb N} \in H$ but $(1/n)_{n \in \mathbb N} \notin \operatorname{span} \{e_n : n \in \mathbb N\}$ as each sequence in $\operatorname{span} \{e_n : n \in \mathbb N\}$ has only finitely many entries that are not equal to zero.

However, one can proof that $H = \overline{F}$, where $F := \operatorname{span} \{K(\, \cdot \,, x) : x \in X\}$:

  • By classical Hilbert space theory, one has $A^{\perp \perp} = \overline A$ for each subspace $A \subseteq H$. Here $A^\perp$ denotes the orthogonal complement of $E$ given by $A^\perp := \{f \in H : \forall g \in A \colon \langle f, g \rangle = 0\}$. This observation yields the following lemma:

Lemma. Let $A \subseteq H$ be a subspace of the Hilbert space $H$. Then $A$ is dense in $H$ if and only if $A^\perp = \{0\}$.

Proof. Suppose first that $A$ is dense in $H$. Then $H = \overline A = A^{\perp \perp}$. Thus, $A^\perp \subseteq (A^{\perp \perp})^{\perp} = H^\perp = \{0\}$. Since $0 \in A^\perp$ by definition, it follows that $A^\perp = \{0\}$.

Conversely, assumme that $A^\perp = \{0\}$. Then $\overline A = A^{\perp \perp} = \{0\}^\perp = H$. Thus $A$ is dense in $H$. $\quad \square$

  • We can now prove finally that $H = \overline F$. By the lemma above we just need to show that $F^\perp = \{0\}$. So let $f \in F^\perp$. Then, by definition, $\langle f, g \rangle = 0$ for all $g \in F$. In particular, the reproducing property of the kernel implies $$f(x) = \langle f, K(\, \cdot \,, x) \rangle = 0$$ since $K(\, \cdot \,, x) \in F$ for all $x \in X$. Thus, $f(x) = 0$ for all $x \in X$ which implies $f = 0$. Thus, $F^\perp \subseteq \{0\}$ and since $0 \in F^\perp$, we obtain $F^\perp = \{0\}$. Thus, $F$ is dense in $H$ or, equivalently, $$\overline F = \overline{\operatorname{span}}\{K(\, \cdot \,, x) : x \in X\} = H.$$

I hope that answers your question in a systematic way :-)

Yaddle
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