# How to prove Fekete / Markov-Lukasz theorem: nonnegative univariate polynomial on [-1,1] can be decomposed accoording to even/odd-ness of degree

I've been a bit stuck on these few problems. I'm trying to prove the following statements but I'm just not sure what to or how to start:

For a univariate polynomial $$f\in \mathbb{R}[x]$$ of degree $$d$$. Assume that $$f\geq 0$$ on the domain [-1,1]. Show that

• if $$d$$ is even, then $$f$$ has decompositions:

(1) $$f=s_0 + (1-x^2)s_1$$ where $$s_0$$ and $$s_1$$ sums of squares with deg($$s_0)\leq d$$ and deg($$s_1)\leq d-2$$

(2) $$f=(1+x)s'_0 + (1-x)s'_1$$ where $$s'_0$$ and $$s'_1$$ sums of squares with deg($$s'_0),deg(s'_1)\leq d$$

• if $$d$$ is odd, then $$f$$ has decompositions:

(1) $$f=(1+x)z_0 + (1-x)z_1$$ where $$z_0$$ and $$z_1$$ sums of squares with deg($$z_0),deg(z_1)\leq d-1$$

(2) $$f=z'_0 + (1-x^2)z'_1$$ where $$z'_0$$ and $$z'_1$$ sums of squares with deg($$z'_0)\leq d+1$$ and deg($$z'_1)\leq d-1$$

I have a feeling i need to use the Goursat transform. Which is defined as the univariate transform $$G(f)(x) = (1+x)^df(\frac{1-x}{1+x})$$ for some univariate polynomial $$f\in\mathbb{R}[x]$$ of degree $$d$$. I can (if nessesary) use that if $$f\geq 0$$ on [-1,1] $$\iff G(f)\geq 0$$ on $$\mathbb{R}_+$$.

I can only answer the (1) versions:

So, assuming $$f\geq 0$$ on [-1,1] $$\Rightarrow^{Q2.1}$$ $$G(f)(x) = (1+x)^df\big(\frac{1-x}{1+x}\big)\geq0$$ on $$\mathbb{R}_+$$. Then from Polya's theorem, we know we can rewrite $$G(f)$$ to be of the form $$G(f) = s_0 + xs_1$$ for some SOS polynomials $$s_0,s_1$$ with deg($$s'_0)\leq d$$ and, deg$$(s'_1)\leq d-1$$. Now we substitute $$y:= \frac{1-x}{1+x}$$ (so $$x=\frac{1-y}{1+y}$$), as \begin{align*} \lim\limits_{x\to 0} \frac{1-x}{1+x} &= \frac{1}{1} = 1\\ \lim\limits_{x\to \infty} \frac{1-x}{1+x} &= \lim\limits_{x\to 0} \frac{-x}{x} = -1 \end{align*} So $$y$$ is within the domain [-1,1]. Substituting this into the Goursat transform, we attain: $$\big(\frac{2}{1+y}\big)^d f(y) = s_0 + \frac{1-y}{1+y}s_1$$

Now we distinguish the two cases of $$d$$:

ODD In this case we can rewrite $$d=2m+1$$ for some $$m\in\mathbb{N}$$. Then \begin{align*} 2^df(y) &= (1+y)^{2m+1}s_0 + (1-y)(1+y)^{2m} s_1 \\ \Rightarrow\quad f(y) &= (1+y)\Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0\Big) +(1-y)\Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_1\Big) \end{align*} And because $$s_0$$ is a SOS, they can be rewritten as $$\sum\limits_{i=1}^{m_0} q_{0,i}^2$$ with deg$$(q_{0,i})\leq d$$. And so \begin{align*} \frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0 &= \frac{((1+y)^m)^2}{\sqrt{2^m}^2}\sum\limits_{i=1}^{m_0} q_{0,i}^2\\ &= \sum\limits_{i=1}^{m_0} (\frac{(1+y)^m}{\sqrt{2^m}} q_{0,i})^2 =: z_0 \end{align*} Which is again a SOS. Same holds for $$\Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_1\Big)$$. So for $$d$$ is odd, $$f$$ is of the form $$f(y) = (1+x)z_0 + (1-x)z_1$$ with $$z_0$$ and $$z_1$$ SOS

EVEN In this case we can write $$d=2m$$. Thus \begin{align*} 2^df(y) &= (1+y)^{2m}s_0 + (1-y)(1+y)^{2m-1} s_1 \\ \Rightarrow\quad f(y) &= \Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0\Big) +(1-y)(1+y)\Big(\frac{((1+y)^{m-1})^2}{\sqrt{2^m}^2} s_1\Big)\\ &= \Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0\Big) +(1-y)(1+y)\Big(\frac{((1+y)^{m-1})^2}{\sqrt{2^m}^2} s_1\Big)\\ &= \Big(\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0\Big) +(1-y^2)\Big(\frac{((1+y)^{m-1})^2}{\sqrt{2^m}^2} s_1\Big)\\ \end{align*}

As for the $$d$$ odd case, these can be rewritten $$\frac{((1+y)^m)^2}{\sqrt{2^m}^2} s_0$$ and $$\frac{((1+y)^{m-1})^2}{\sqrt{2^m}^2} s_1$$ as sums of square functions. Therefore, for $$d$$ even, we have that $$f$$ is of the form

$$f=s_0 + (1-x^2)s_1$$ with $$s_0$$ and $$s_1$$ SOS

Only thing i have not been able to show is that these $$s_0,s_1,z_0$$ and $$z_1$$ have proper degrees. Ideas, anyone?