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If the inner product between two functions is the $\int f(x)g(x)dx$ and it's equal to $\int f(x)g(x)dx = f(x)$, what conditions must $g(x)$ satisfy in order for this equality to hold?

jtb
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Filippo Portera
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The space in which $$<f,g> = \int f(x)g(x) \mathrm dx$$ is called the (real) $L^2$ space. In $L^2$ the representer of the evaluation functional is the dirac delta function, as $$ \int f(x) \delta(x-y) \mathrm d x = f(y)$$ however $\delta \notin L^2$, so $L^2$ is not a RKHS!

Riccardo Sven Risuleo
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  • I havent' understood if Wikipedia is authoritative in this matter, but in https://en.wikipedia.org/wiki/Reproducing_kernel_Hilbert_space somebody says that: <f(.), k(x,.)> = f(x) – Filippo Portera Feb 14 '19 at 14:35
  • yes indeed, but when you write $<f,g>$ you are taking an inner product, which means that both $f$ and $g$ belong to the same vector space. if there is a representer of the evaluation functional $k(x,\cdot)$ inside of the vector space, then this is a RKHS. In the case of $L^2$, the representer is $\delta(x-\cdot)$ and this is not in $L^2$. So $L^2$ is not a RKHS. – Riccardo Sven Risuleo Feb 14 '19 at 14:49
  • sorry, what is "a representer of the evaluation functional k(x,⋅) inside of the vector space"? Do you have any reference (article, book, etc... ) that exaplains this matter? – Filippo Portera Feb 14 '19 at 18:47
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    Wikipedia itself is enough: if $f\in \mathcal H$ and $f(x) = L_xf = <f,K_x>$ for some $K_x \in \mathcal H$, then $\mathcal H$ is a RKHS. In the case of $L^2$, $\delta \notin L^2$, so it is not a RKHS. – Riccardo Sven Risuleo Feb 15 '19 at 14:59