2
$\begingroup$

Recall that a (reproducing kernel hilbert space) RKHS has two equivalent definitions:

1) Its a Hilbert space of functions $\mathcal{H}$ (i.e. vector space with an inner product $\langle \cdot, \cdot \rangle$ and a norm derived from it $|| \cdot ||_{\langle \cdot, \cdot \rangle}$) and an evaluational functional that is continuous.

2) Equivalently, its a reproducing kernel $K: X \times X \rightarrow \mathbb{R}$ (still a Hilbert space $\mathcal{H}$) with the the properties that $K(x,\cdot) = K_x(\cdot) \in \mathcal{H}$ and the reproducing property $\langle f, K_x \rangle = f(x) $.

Thus I was wondering, does a RKHS always has a distance function defined on its space? Does it always have a distance space because it always has a norm defined on its space? How do you go proving that it always has a distance function because it always has a norm? It feels it should because norm and distance are nearly identical in the mathematical definition but was unsure on how to prove it.

$\endgroup$
2
$\begingroup$

Every Hilbert space has a natural notion of distance (metric), namely $$d(a,b)=\|a-b\|$$ (This is also true for every normed space; indeed, one of the axioms of a norm is essentiall the triangle inequality for $d$.)

Starting with the definition of Hilbert space norm as $\|x\|^2=\langle x,x\rangle$, one can prove the triangle inequality by using Cauchy-Schwarz, as follows: $$ \|x+y\|^2 = \langle x+y,x+y\rangle \le \|x\|^2+\|y\|^2+2\|x\|\|y\| = (\|x\|+\|y\|)^2 $$ In terms of $d$, this gives $$ d(a,c) = \|a-c\| = \|a-b+b-c\|\le \|a-b\|+\|b-c\| = d(a,b)+d(b,c) $$

$\endgroup$
  • $\begingroup$ Seems a perfectly good answer to me, @Pinocchio are you satisfied? Can you tag it as a good answer and close this question? $\endgroup$ – Næreen Feb 10 '16 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.