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I'm confused about Orthonormal basis for $L^2(\mathbb{R})$ by Hermite Polynomials of the answer. ( The satement $ A_{n,m(n)} \rightarrow -\int f(x)^2 e^{-x^2} dx$ )

The following hold?

Let $f \in L^2(\mathbb{R})$ satisfying $ \int_{-\infty}^{\infty} x^nf(x) e^{-x^2}dx=0 \ \forall n \in \mathbb{Z}_{\ge 0}$, and

$(p_n(x))$ be polynomial sequence. Then

$$\lim_{n \to \infty}\int_{-n}^{n}f(x) p_n (x) e^{-x^2}dx =0$$

Any advise would be appreciated.

lyn
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  • Might be harder than I think, but this looks like we're just going to use linearity of the integral and then make the estimate $| \int_{-n}^n - \int_{-\infty}^\infty | < \epsilon$. – Charles Hudgins Jul 27 '23 at 11:57
  • @Charles Hudgins Thank you for your comment. The difficult part is to use $\int_{-\infty}^{\infty} x^nf(x) e^{-x^2}dx=0 \ \forall n \in \mathbb{Z}_{\ge 0}$ – lyn Jul 28 '23 at 10:22
  • An important bit of advice here is not to overstate the claims made in the proof you are asking about. They never claim anything equivalent to $\lim_{n \to \infty}\int_{-n}^{n}f(x) p_n (x) e^{-x^2}dx =0$. Instead they use separate indices $n, m$ for the degree of the polynomial and the limits of integration, and when they do make $m$ a function of $n$, it is a carefully chosen function, not a baldface assumption that $m =n$. – Paul Sinclair Jul 28 '23 at 13:52
  • @Paul Sinclair I think they are essentially the same thing, so I wrote in a brief way. – lyn Jul 29 '23 at 03:58
  • And I am telling you that you are very much mistaken about that. Your "brief" way is not at all equivalent, turning something that is straightforward to prove into something that I strongly suspect is false. – Paul Sinclair Jul 29 '23 at 05:38
  • @Paul Sinclair What I wrote is simpler and easier. – lyn Jul 29 '23 at 06:10
  • Simpler, yes. Easier? How so? You cannot figure it out! If you would give proper consideration to the actual statements, you would realize they work, while your "easier" version doesn't. – Paul Sinclair Jul 29 '23 at 12:34
  • @Paul Sinclair I can't understand why you say that. Please tell me how to use it. – lyn Jul 29 '23 at 13:08
  • In the other thread, $n$ is the index of the functions $g_n$ which have compact support but converge to the integrand. Effectively, $n$ controls the limits of integration. $m$ is the index for the polynomial approximations of the $g_n$. Effectively, $m$ controls the degree of the polynomials. By having them separate, they have the freedom to pick for each $n$, whatever value of $m$ was sufficient. But your version demands that degree and width must march in lock-step together. That lack of freedom is what makes your version much harder, not easier. – Paul Sinclair Jul 29 '23 at 19:22
  • @Paul Sinclair My version is easier because it is a special case; $ supp(g_n) \sim { x | -n < x < n }$. – lyn Jul 30 '23 at 04:59
  • No, that is part of why it is simpler. And if that had been all you did, I would agree that it is at least not harder. However the other simplification you made (and conveniently ignored in that comment) makes the matter much harder. – Paul Sinclair Jul 30 '23 at 13:09
  • @Paul Sinclair I answered all. what's left? – lyn Jul 30 '23 at 14:48
  • Your answer is only relevant to the first two sentences of my comment. You completely ignored the rest. – Paul Sinclair Jul 30 '23 at 16:57
  • @Paul Sinclair No. I answered about it. – lyn Jul 30 '23 at 17:02
  • With that manifestly false claim, I fail to see any point in continuing this conversation. – Paul Sinclair Jul 31 '23 at 03:26
  • I never said $m=n$. – lyn Jul 31 '23 at 14:38

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