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For a fixed function $f$ defined on $[-1, 1]$, let's define $$G_m(f)=\sum_{k=1}^m w_k^{(m)}f\left(x_k^{(m)}\right)$$ where $x_1^{(m)}, x_2^{(m)}, \ldots, x_m^{(m)}$ are the $m$ roots of the $m$th Legendre Polynomial and $w_1^{(m)}, w_2^{(m)}, \ldots, w_m^{(m)}$ are the corresponding weights demanded by Gaussian Quadrature.

We know that if $p$ is any polynomial of degree less than or equal to $2n-1$, then $$G_n(p)=\int_{-1}^1p(x)\;dx\,.$$

This has the easy corollary that for any polynomial $p$, there is a natural number $n$ such that $$\int_{-1}^1p(x)\;dx=G_n(p)=G_m(p)$$ for any $m\geq n$. The notation here hides the novelty in this result: $G_n$ is calculated vastly differently from $G_m$. The points and weights used in these approximations are different, but it's true nonetheless.

The question is:

Is the converse true? That is, for a fixed continuous function $f$ on $[-1, 1]$ such that there is a natural number $n$ such that $G_n(f)=G_m(f)$ for all $m\geq n$, is it true that $f$ must be a polynomial?

Of course, if $f$ satisfies this property we have that $$\int_{-1}^1f(x)\;dx=G_m(f)$$ for $m\geq n$. I thought this would be a clever twist of usage with the Weierstrass approximation theorem, but it's been a little more involved than that. I've looked into polynomials of best approximation, but I haven't been able to leverage these effectively. One of the stumbling blocks has been that we don't know $G_m(f^2)=G_n(f^2)$ for all $m\geq n$.

One of the things I've been attempting is to understand the set $$\{p: p(x_k^{(n)})=f(x_k^{(n)})\text{ for all }k\text{ and }\deg(p)\leq 2n-1\}\,.$$ Does this set include a polynomial of best approximation for $f$ with degree less than $2n-1$? I don't know. It does have the interpolating polynomial of degree $n-1$. But what relationships do these polynomials share?

Any help is appreciated.

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    $\begingroup$ I think it should be true not just for continuous functions but also for $L^2[-1,1]$. If we let $p_n$ be the normalized Legendre polynomials, then the weights satisfy $\sum_{k=0}^mw_k^{(m)}p_m(x_k)=0$ except for $\sum_{k=0}^mw_k^{(0)}p_0(x_k)=\sqrt{2}$. So, writing $f=\sum_n\alpha_np_n$ and substituting into your requirements gives $\sum_{n=m+1}^\infty\alpha_n(\sum_{k=0}^mw_k^{(m)}p_m(x_k))=0$. These are equations of the form $<a,b_m>=0$ for an infinite number of such vectors. I don't know enough about the $b_m$ to tell whether they are dense in $\ell^2$, but if so that forces $a=0$ as reqrd. $\endgroup$ – Chrystomath Dec 4 '17 at 16:17
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The question was

Is the converse true? That is, for a fixed continuous function $f$ on $[-1, 1]$ such that there is a natural number $n$ such that $G_n(f)=G_m(f)$ for all $m\geq n$, is it true that $f$ must be a polynomial?

The answer is trivially negative.

Recall that the symmetric quadrature on the interval symmetric wrt. zero has the form $$Q(f)=\sum_{i=1}^n w_i\bigl(f(x_i)+f(-x_i)\bigr)+w_0f(0)$$ (the case $w_0=0$ is possible).

Because each Gauss-Legendre quadrature is symmetric, it vanishes on any odd function. So, $G_m(f)=0$ for any $m$ and any odd function $f$ and, of course, there are many non-polynomial odd functions. Then the condition in question holds but the assertion is false.

Additionally, also the integral vanishes for integrable odd functions. Then the conjecture is false even with stronger assumption $\displaystyle\int_{-1}^1 f(x)\,\text{d}x=G_m(f)$ for any $m\in\Bbb N$.

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  • $\begingroup$ Anyone else kicking themselves? :) $\endgroup$ – Calum Gilhooley Dec 5 '17 at 13:47
  • $\begingroup$ Ah! My intuition was that the approximations would have to get strictly better. Didn't think of the approximations as being exact by a technicality. Yes, I'm kicking myself. Of course, your examples can also be used to provide counterexamples for functions which are strictly positive. $\endgroup$ – Robert Wolfe Dec 5 '17 at 15:32
  • $\begingroup$ @Robert The exactness of the approximation was a kind of by product in my example. Odd functions are not good candidates to be strictly positive. Simply there is no such functions. So, my example could not be extended in a direction you propose. $\endgroup$ – szw1710 Dec 5 '17 at 15:43
  • $\begingroup$ A polynomial with a large enough infimum added to any odd function will result in a strictly positive function with eventually exact Gaussian approximations. $\endgroup$ – Robert Wolfe Dec 5 '17 at 16:03

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