Reproducing kernel Hilbert space and Dirac evaluation functional

An integral transform between functions f∈Xf\in X and g∈Xg \in X is:

g(x)=∫baA(x,y)f(y)dyg(x) = \int_{a}^{b}A(x,y)f(y)dy

The identity transform is:

f(x)=∫baδ(x−y)f(y)dyf(x) = \int_{a}^{b}\delta(x-y)f(y)dy

First question: In what spaces the expression f(x0)=∫baδ(x0−y)f(y)dyf(x_0) = \int_{a}^{b}\delta(x_0-y)f(y)dy is a correct evaluation functional?

In a reproducing kernel Hilbert space HH there is the following evaluation functional for f,k(x0,⋅)∈Hf, k(x_0,\cdot) \in H:

f(x0)=∫bak(x0,y)f(y)dyf(x_0) = \int_{a}^{b}k(x_0,y)f(y)dy

δ(x0−y)\delta(x_0-y) isn’t a reproducing kernel in L2L^2 space where f(x0)=∫baδ(x0−y)f(y)dyf(x_0) = \int_{a}^{b}\delta(x_0-y)f(y)dy is valid, because δ(x0−y)∉L2\delta(x_0-y) \notin L^2. So L2L^2 isn’t a reproducing kernel Hilbert space because the evaluation functional in this space isn’t written in terms of functions ∈\in L2L^2.

Second question: Is an evaluation functional unique for a given space, particularly for a given Hilbert space? Does it mean that RKHS don’t have evaluation functionals in terms of a Dirac function?




In a RKHS the evaluation functional must be bounded. L2([a,b])L^2([a,b]) can be modified to become a RKHS. Let X={f,f(a)=0,f′∈L2([a,b])}X = \{ f, f(a) = 0,f’ \in L^2([a,b])\}. It is clearly isomorphic to L2([a,b])L^2([a,b]), and the evaluation functional is Lx(f)=f(x)=∫x0f′(x)dxL_x(f) = f(x) = \int_0^x f'(x)dx, which is bounded with respect to ‖f′‖L2([a,b])\|f’\|_{L^2([a,b])} (but not with respect to ‖f‖L2([a,b])\|f\|_{L^2([a,b])})
– user1952009
2 days ago