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Jan 29, 2021 at 17:03 vote accept lrnv
Jan 29, 2021 at 17:03 comment added lrnv Yes you are right, I see it now.
Jan 29, 2021 at 16:55 comment added Iosif Pinelis @lrnv : You have $\sum_i \widehat{a_{n,i}}^2=\infty$. So, the very definition $\hat f(x): = \sum_i\widehat{a_i} g_i(x)$ makes no sense. So, you don't even have a $\hat f$.
Jan 29, 2021 at 16:53 history edited Iosif Pinelis CC BY-SA 4.0
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Jan 29, 2021 at 16:48 comment added lrnv Again, thanks for the details. So to clarify, $\widehat{a_i} -> a_i$ but there is no way of making $\lVert f - \hat{f} \rVert_2^2$ go to 0 ? (according to point 2.). It it still strenge to me, as $\mathbb E( \lVert f - \hat f \rVert_2^2) \to 0$ and $\lVert f - \hat f \rVert_2$ is always positive..
Jan 29, 2021 at 16:41 comment added Iosif Pinelis @lrnv : (i) It does not make sense to say that $\widehat{a_i}$ converges weakly to $a_i$. However, it is true that $\widehat{a_{n,i}}\to a_i$ (as $n\to\infty$) if the function $g_i$ is bounded and continuous; this follows because the empirical probability measure $\mu_n$ converges weakly to the probability measure $\mu$ with density $f$. (ii) If $b_{n,i}\to b_i$ for each $i$, this does not in general imply that $\sum_i b_{n,i}\to\sum_i b_i$; e.g. consider $b_{n,i}:=1(n=i)$ and $b_i:=0$.
Jan 29, 2021 at 16:26 comment added lrnv Thanks for the extensive explanation. So, by your point 3., am i right to say that $\widehat{a_i}$ converges weakly to $a_i$ ? Which means that for all $i$, $(a_i - \widehat{a_i})^2 \to 0$. Then, why the fuck does the sum for all $i$ not converge... Dont get me wrong, I got your argument, it convinced me, but I still do not feel it.
Jan 29, 2021 at 16:12 history edited Iosif Pinelis CC BY-SA 4.0
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Jan 29, 2021 at 16:07 history edited Iosif Pinelis CC BY-SA 4.0
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Jan 29, 2021 at 15:58 history answered Iosif Pinelis CC BY-SA 4.0