The following is taken from Efficient Large-Scale Filter/Filterbank Design via LMI Characterization of Trigonometric Curves by Hoang Duong Tuan, Tran Thai Son, Ba-Ngu Vo, and Truong Q. Nguyen
Consider the curve $$ C_{a,b}: =\{(1, \cos(t), \cos(2 t), \ldots, \cos(n t))^{T}: \cos t\in[\cos a, \cos b]\subset[-1,1]\}\subset \mathbb R^{n+1} $$ (I will only consider $a = \pi$, $b = 0$, so we take all $0 \le t < 2 \pi$) and define \begin{align*} T_{k}(t) & = \begin{pmatrix} 1 \\ \cos(t) \\ \ldots \\ \cos(k t) \end{pmatrix} \begin{pmatrix} 1 & \cos(t) & \ldots & \cos(k t) \end{pmatrix}^T \\ & = \begin{pmatrix} 1 & \cos(t) & \ldots & \cos(k t) \\ \cos(t) & \frac{1}{2}(\cos(2 t) + 1) & \ldots & \frac{1}{2}[\cos((k + 1)t) - \cos((k - 1)t)] \\ \cos(kt) & {1\over 2}\left[\cos((k+1)t)+\cos((k-1)t)\right] & \ldots& {1 \over 2} (\cos(2kt)+1) \end{pmatrix}. \end{align*} The matrix $T_k(y)$ is obtained from $T_k(t)$ via $\cos(ht) \longleftrightarrow y_{h}$ for $0 \le h \le k$, that is, we have $$ T_k(y) := \begin{pmatrix} y_0 & y_1 & \ldots & y_k \\ y_1 & \frac{1}{2}(y_{2 k} + 1) & \ldots & \frac{1}{2}(y_{k + 1} - y_{k - 1}) \\ y_k & {1\over 2}\left(y_{k + 1}+y_{k - 1}\right) & \ldots& {1 \over 2} (y_{2k}+1) \end{pmatrix}. $$ Furthermore, let $T_{k, \ell}(t) := \cos(\ell t) T_k(t)$ and $T_{k, \ell}(y)$ be obtained via the same transformation. Lastly, let $$ F_{k}^{a,b}(y) = (\cos(b) +\cos(a)){T}_{1,k-1}(y)-{1\over 2}T_{2,k-1}(y)-\left({1\over 2}+\cos(a) \cos(b)\right)T_{k-1}(y) $$
The authors now claim that (Thm. 3, bottom right on p. 4396) $$ \text{conv}(C_{a, b}) = \begin{cases} \left\{(y_{0}, y_{1}, \ldots, y_{n})^T:T_{k}(y)\ge 0, \ F_{k}^{a,b}(y)\ge 0, y_{0}=1\right\}, & n=2k \\ \left\{(y_{0}, y_{1}, \ldots, y_{n})^T:\cos(b) T_{k}(y)\ge T_{1,k}(y)\ge\cos(a) T_{k}(y), y_{0}=1\right\}, & n=2k+1. \end{cases} $$
My question: If $n = 3$, we have $k = 1$ and (in the special case that $a = \pi$, $b = 0$ and thus $\cos(a) = - 1$, $\cos(b) = 1$) we have \begin{align} \text{conv}(C_{\pi, 0}) & = \{ (1, y_1, y_2, y_3): T_1(y) \ge y_1 T_1(y) \ge - T_1(y) \} \\ & = \{ (1, y_1, y_2, y_3): (1 \pm y_1) T_1(y) \ge 0 \} \\ & = \{ (1, y_1, y_2, y_3): T_1(y) \ge 0 \} \\ & = \{ (1, y_1, y_2, y_3): y_2 + 1 \ge y_1^2 \}, \end{align} but as $y_3$ is not present in any entry of $T_1$, this would mean that $y_3$ can be arbitrary, which is clearly false. What am I missing?