Does every positive continuous function have a non-negative interpolating polynomial of every degree?

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 $$f(x_i)=p_n(x_i),\forall i=0,1,...,n$$.

• 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. – Jochen Wengenroth Oct 25 '18 at 12:33
• 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 $n$"? – Wlodek Kuperberg Oct 25 '18 at 18:52
• @WlodekKuperberg: sorry, I edited ... you can also see my answer – user521337 Oct 26 '18 at 1:16
• For each $n$ there is the "worst" example for which the conjecture works. Thus, I believe that the conjecture always works. – Wlod AA Oct 27 '18 at 7:45

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 $$p_n(x)\ge f(x)-r/2\ge r-r/2=r/2>0,\forall x \in [a,b], \forall n \ge n_0$$.

• This also follows from my comment above. – Jochen Wengenroth Oct 26 '18 at 5:43
• This proves the claim for large n under the additional assumption that $f>0$, but not for all $n$. – Christian Remling Oct 26 '18 at 17:26
• @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 $f>0$ ... – user521337 Oct 27 '18 at 0:51

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.