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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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Line 7:
:<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
:<math> f(x) = \langle K_x, f \rangle.
The function
:<math> K(x,y)
is called a reproducing kernel for the Hilbert space. In fact, ''K'' is uniquely determined by the above condition
for every ''f'', where
▲:<math>f(x)=\int_X K(x,y) f(y)\,dy</math>
▲for every ''f'', where ''X'' is often the real numbers or '''R'''<sup>''n''</sup>.
==Bergman kernel==
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