Reproducing kernel Hilbert space: Difference between revisions

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Rv. Claim made about finite set X in previous edit is wrong. Nowhere does it say H consists of all functions on X
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fix cause of previous revert
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:<math> f \mapsto f(x) </math>
 
from ''H'' to the complex numbers is continuous for any ''x'' in ''X''. By the [[Riesz representation theorem]], this implies that for given ''x'' there exists an element ''K''<sub>''x''</sub> of ''H'' suchwith thatthe forproperty every function ''f'' in the space,that:
 
:<math> f(x) = \langle K_x, f \rangle. \quad \forall f \in H \quad (*)</math>
 
The function
 
:<math> K(x,y) =\equiv K_x(y) </math>
 
is called a reproducing kernel for the Hilbert space. In fact, ''K'' is uniquely determined by the above condition (*). In some concrete contexts this amounts to saying
 
For example, when ''X'' is finite and ''H'' consists of all complex-valued functions on ''X'', then an element of ''H'' can be represented as an array of complex numbers. If the usual inner product is used, then ''K''<sub>''x''</sub> is the function whose value is 1 at ''x'' and 0 everywhere else.
:<math>f(x)=\int_\Omega K(x,y) f(y)\,dy</math>
 
In other contexts, (*) amounts to saying
for every ''f'', where &Omega; is the appropriate ___domain, often the real numbers or '''R'''<sup>''n''</sup>.
 
:<math>f(x)=\int_\Omegaint_X K(x,y) f(y)\,dy</math>
 
for every ''f'', where &Omega;''X'' is the appropriate ___domain, often the real numbers or '''R'''<sup>''n''</sup>.
 
==Bergman kernel==