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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 $\mathbb{R}_+$.

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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?

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