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Let $f:[a,b] \to (0,\infty)$ be a continuous function. Then is it necessarily true that for every $n\ge 1$, we can find $n+1$ distinct points $\{x_0,x_1,...,x_n\}$ in $[a,b]$ such that the interpolating polynomial $p_n(x)$ of $f$ at those points is non-negative on $[a,b]$ i.e. $p_n(x) \ge 0,\forall x \in [a,b]$ ?

NOTE: $p_n(x)$ is the unique, degree at most $n$, polynomial such that $f(x_i)=p_n(x_i),\forall i=0,1,...,n$.

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  • $\begingroup$ A non-conclusive idea: If $p$ is the best approximation to $f$ among polynomials of degree $n$ (in the uniform norm), the difference $f-p$ changes signs at least $n+2$ times (by a theorem of Chebychev). The intermediate value theorem thus gives you $n+1$ points where $f=p$. However, $p$ need not be positive. $\endgroup$ – Jochen Wengenroth Oct 25 '18 at 12:33
  • $\begingroup$ Suppose $f(x)=0$ for all $x\in [a,b]$, then what would be your "unique degree $n$ polynomials $p_n(x)$"? Did you mean to say "degree at most $n$"? $\endgroup$ – Wlodek Kuperberg Oct 25 '18 at 18:52
  • $\begingroup$ @WlodekKuperberg: sorry, I edited ... you can also see my answer $\endgroup$ – user521337 Oct 26 '18 at 1:16
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    $\begingroup$ For each $n$ there is the "worst" example for which the conjecture works. Thus, I believe that the conjecture always works. $\endgroup$ – Wlod AA Oct 27 '18 at 7:45
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I think I can prove the following : For every continuous function $f:[a,b]\to (0,\infty)$, $\exists n_0>1$ such that for every $n\ge n_0$, there are $n+1$ distinct points in $[a,b]$ such that the interpolating polynomial $p_n(x)$ of $f$, at those points, satisfy $p_n(x) >0,\forall x \in [a,b]$.

Firstly, let $c\in[a,b]$ be such that $f(c)=\inf_{x\in[a,b]} f(x)$. Let $r:=f(c)$. Then $f(x)\ge r >0,\forall x \in [a,b]$.

Now to prove my claim, let us recall the following standard theorem (due to Marcinkiewicz I believe ?) from the theory of polynomial interpolation:

For any function $f(x)$ continuous on an interval $[a,b]$, there exists a table of nodes for which the sequence of interpolating polynomials $p_n(x)$ converges to $f(x)$ uniformly on $[a,b]$.

And then , for these sequence of polynomials, we have that $\exists n_0 \in \mathbb N$ such that $ f(x)-p_n(x)\le |f(x)-p_n(x)|\le r/2,\forall x \in [a,b],\forall n\ge n_0$, then $p_n(x)\ge f(x)-r/2\ge r-r/2=r/2>0,\forall x \in [a,b], \forall n \ge n_0$.

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  • $\begingroup$ This also follows from my comment above. $\endgroup$ – Jochen Wengenroth Oct 26 '18 at 5:43
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    $\begingroup$ This proves the claim for large n under the additional assumption that $f>0$, but not for all $n$. $\endgroup$ – Christian Remling Oct 26 '18 at 17:26
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    $\begingroup$ @ChristianRemling: yes I know it isn't for all $n$ ... I didn't say it was a complete answer ... and b.t.w. I really am interested in $f>0$ ... $\endgroup$ – user521337 Oct 27 '18 at 0:51
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This question was recently studied in that paper:

F. Charles, M. Campos-Pinto, B. Després, Algorithms for positive polynomial approximation, hal-01527763,

assuming that the function $f$ is Lipschitz on the interval. Let $[a,b]=[0,1]$. The interpolating polynomials $p_{n}$ are seeked in the form \begin{align*} p_{n}(x) & =a_{p}^{2}(x)+x(1-x)b_{p-1}^{2}(x),\quad n=2p+1,\\ p_{n}(x) & =xa_{p}^{2}(x)+(1-x)b_{p}^{2}(x),\quad n=2p, \end{align*} with $a_{p}$ and $b_{p}$ polynomials of degree $p$, and $b_{p-1}$ a polynomial of degree $p-1$, which are classical representations for non-negative polynomials in $[0,1]$. The interpolation points are chosen to be $0$, $1$, and $n-1$ points in $(0,1)$ which are obtained through a converging fixed point algorithm for a map defined on $(0,1)^{n-1}$.

Some details when $n=2$ or $n=3$ : first, let $f(y)=g^{2}(y)$ with $g(y)>0$.

  • when $n=2$, $p_{2}(x)$ is seeked in the form $a_{1}^{2}(x)+x(1-x)b_{0}^{2}$. One sets $a_{1}(0)=g(0)>0$ and $a_{1}(1)=-g(1)<0$ so that there exists $\alpha\in(0,1)$ with $a_{1}(\alpha)=0$. It then suffices to choose $b_{0}$ such that $f(\alpha)=\alpha(1-\alpha)b_{0}^{2}$.

  • when $n=3$, $p_{3}(x)$ is seeked in the form $xa_{1}^{2}(x)+(1-x)b_{1}^{2}(x)$. One shows that there exists $0<\alpha<\beta<1$ such that \begin{align} a_{1}(1) & =g(1), \quad a_{1}(\alpha) =-g(\alpha)/\sqrt{\alpha}, \quad a_{1}(\beta) =0,\\ b_{1}(0) & =-g(0), \quad b_{1}(\alpha) =0, \quad b_{1}(\beta) =g(\beta)/\sqrt{1-\beta}, \end{align} which is checked through direct calculations.

The general case $n\geq4$ consists in generalizing the above conditions and showing that there exists $n-1$ interpolation points in $(0,1)$ that satisfy these conditions.

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