lajos/Numerator of the stability function

In this note we are partially solving the \(MonRad\)(numerator of stability function)\(=2s\) conjecture for \(p=2\), in the interval \(3\le s\le 8\).

The first three coefficients of the numerator of the stability function with approximation order \(p=2\), and a non-linear inequality between them. By repeatedly using the formulae for the derivative of the determinant and the trace

\[ \left(\det(\Phi(\cdot))\right)\prime = \det(\Phi(\cdot))\cdot \textrm{tr}(\Phi^{-1}(\cdot)\Phi\prime(\cdot)) \]

and

\[ \left(\textrm{tr}(\Phi(\cdot))\right)\prime = \textrm{tr}(\Phi\prime(\cdot)) \]

(where \(\Phi:\mathbb{R}\to \mathbb{R}^{s\times s}\) is a smooth (and for the first formula, invertible) matrix function), the cyclic invariance of the trace together with the order conditions (6.35a\(-\)b.), we see that the numerator of the stability function, \(\det(I-z(A-eb^T))\), can be written in the following form (for the notation \(\tau\) and \(\tau_2\), see https://mathwiki.kaust.edu.sa/lajos/General%20trace%20inequalities):

\[ P(z)=1+(1-\tau)z+\frac{1}{2}\left((1-\tau)^2-\tau_2\right)z^2+\sum_{k=3}^s a_k z^k. \] Of course there are further restrictions on the real coefficients \(a_k\), but for the present proof the above class of polynomials will be sufficient. We also introduce \(a_0:=1\), \(a_1:=1-\tau\), \(a_2:=\frac{1}{2}(a_1^2-\tau_2)\), and tacitly assume that the numerator and the denominator of the stability function have no common roots. (I can justify this last assumption in the \(s=3\) case, based on the Griend-Kraaijevanger’s \(R_{m/n,2}\) estimates.) Since we suppose \(A\ge 0\), this trivially implies \(\tau\ge 0\) and \(\tau_2\ge 0\). Then the trace inequality \(\tau^2\le s \tau_2\) implies \(-\tau_2\le -\frac{\tau^2}{s}\), so \(a_2=\frac{1}{2}(a_1^2-\tau_2)\le \frac{1}{2}(a_1^2-\frac{\tau^2}{s})\), but \(\tau=1-a_1\), hence

\[\frac{1}{2}\left(a_1^2-\frac{(1-a_1)^2}{s}\right)\ge a_2.\]

The above consideration shows that in the present situation we can apply the following general result.

Proposition. Pick an integer \(s\) with \(3\le s\le 8\) and let \(P(z):=1+\sum_{k=1}^s a_k z^k\) with real coefficients \(a_k\). Suppose \(P^{(k)}(-2s)\ge 0\) for all \(k=0,1, \ldots, s\), further, that \(\frac{1}{2}\left(a_1^2-\frac{(1-a_1)^2}{s}\right)\ge a_2.\) Then \(P(z)=\left(1+\frac{z}{2s}\right)^s\).

Proof. By https://mathwiki.kaust.edu.sa/lajos/Bounds%20on%20the%20coefficients%20of%20a%20polynomial%20having%20all%20derivatives%20non-negative with \(r=2s\), we have \(a_n\ge 0\) for all \(n=1, 2, \ldots, s\).

Now we apply the following general identity.

Lemma. Fix any integer \(s\ge 2\) and choose an arbitrary \(P(z):=1+\sum_{k=1}^s a_k z^k\) with real coefficients. Then \[ \sum_{k=0}^s \frac{(2s)^k (s-k)(s-k-1)}{k! s(s-1)}P^{(k)}(-2s)=1-4a_1+\frac{8s}{s-1}a_2. \]

Proof of the Lemma. Direct verification without any tricks. LATEX IT, or more precisely, find a generalization. I’ve proved this identity for all \(s\ge 2\), but I will be able to make use of it below only if \(s\le 8\). But as I see, my current proof has a kind of flexibility that would allow other non-negative linear combinations of \(P^{(k)}(-2s)\) to be considered that could be used to extend the validity of the Proposition to larger \(s\)-intervals.

Since the \(P\)-derivatives on the left-hand side are non-negative by assumption, the whole left-hand side is non-negative. Hence we’ve derived

\[ 1-4a_1+\frac{8s}{s-1}a_2\ge 0, \] together with the earlier \(a_1\ge 0\), \(a_2\ge 0\) and the assumption \(\frac{1}{2}\left(a_1^2-\frac{(1-a_1)^2}{s}\right)\ge a_2\). \(\quad\quad\quad\)(*)

From the first inequality we estimate \(a_2\) as \(a_2\ge \frac{s-1}{8s}(4a_1-1)\), so \(\frac{1}{2}\left(a_1^2-\frac{(1-a_1)^2}{s}\right)\ge \frac{s-1}{8s}(4a_1-1)\). But this inequality can be factorized as

\[ \frac{\left(2 a_1-1\right) \left(2 a_1(s-1)+5-s\right)}{8 s}\ge 0, \]

showing (by using \(s>1\) only) that either

\[ a_1\le \frac{1}{2}\quad\textrm{and}\quad a_1\le\frac{s-5}{2(s-1)}\quad\quad\quad\quad\quad (1) \]

or

\[ a_1\ge \frac{1}{2}\quad\textrm{and}\quad a_1\ge\frac{s-5}{2(s-1)}\quad\quad\quad\quad\quad (2). \]

We now show that (1) can not occur for \(3\le s\le 8\): for any \(1 < s < 9\), we have \(\frac{s-5}{2(s-1)}=\frac{1}{2}-\frac{2}{s-1} < \frac{1}{4}\), therefore \(a_1 < \frac{1}{4}\). But then (*) says that

\[ 0\le a_2\le \frac{1}{2}\left(a_1^2-\frac{(1-a_1)^2}{s}\right)=\frac{a_1^2 (s-1)+2 a_1-1}{2 s}, \] so \(a_1^2 (s-1)+2 a_1-1\ge 0\), implying either \(a_1\le\frac{1}{1-\sqrt{s}} < 0\) or \(a_1\ge\frac{1}{1+\sqrt{s}}>\frac{1}{1+\sqrt{9}}=\frac{1}{4}\). The first case is ruled out by \(a_1\ge 0\), and the second one is by \(a_1 < \frac{1}{4}\).

Hence the only possibility is (2) above. But then \(\frac{1}{2}\le a_1 \le \frac{\binom{s}{1}}{r^1}=\frac{s}{2s}=\frac{1}{2}\) by https://mathwiki.kaust.edu.sa/lajos/Bounds%20on%20the%20coefficients%20of%20a%20polynomial%20having%20all%20derivatives%20non-negative, so we have \(P(z)=\left(1+\frac{z}{2s}\right)^s.\)