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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.

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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 $$ d(a,c) = \|a-c\| = \|a-b+b-c\|\le \|a-b\|+\|b-c\| = d(a,b)+d(b,c) $$

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  • $\begingroup$ 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
    Commented Feb 10, 2016 at 10:25

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