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$L_2$ space is the space of square integrable functions. It is a Hilbert space, but it is not RKHS. How to show it is not RKHS? The reason is that the kernel would need to be the Dirac delta function. But it is not in $L_2$.

$$f(x)=\int_z \delta(x-z) f(z)dz.$$

$\int_z \delta(z)^2 dz$ is not finite.

I don't understand the above. Why "the kernel would need to be the Dirac delta function"? The below is the slide that I am watching from the YouTube:

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Jonathen
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  • actually, $f(x)$ makes no sense for $L^2$; the elements of $L^2$ are classes of functions and, for a given $x\in\mathbb R$, $f(x)$ may have different values for different $f$'s from the same class – user8268 Dec 20 '23 at 22:09
  • @whpowell96 No, I need a way that answer the question from the view of Dirac delta function. – Jonathen Dec 21 '23 at 01:00
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    I feel like the use of the Dirac delta function here is confusing you. If a space is an RHKS, then pointwise evaluation is a continuous linear functional. $L^2$ (equivalence classes of) functions are not bounded in general, so pointwise evaluation is not continuous. – whpowell96 Dec 21 '23 at 01:33

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The idea is that for $f \in C_c^{\infty}(\mathbb{R})$, $f(x) = \int \delta_x(y)f(y)\,dy$, so asking whether the evaluation map $f \mapsto f(x)$ is continuous (i.e, has a unique continuous extension from $C_c^\infty$ to $L^2$) is the same as asking whether $\delta_x \in L^2$. This is equivalent to asking whether $\hat{\delta_x} \in L^2$. Since $\hat{\delta}_x(w) = e^{iwx}$, $\hat{\delta_x} \notin L^2$.

Mason
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