Characterization of positive elements of the subspace spanned by $1,~z$ and $\bar z$ in $C(\mathbb T)$

Question

Let $C(\mathbb T)$ denotes the space of continuous function on the unit circle $\mathbb T$ on the complex plane and $X \subset C(\mathbb T)$ is the subspace spanned by $1,~z$ and $\bar z$, where $z$ is the coordinate function. Now show that an element $a1+bz+c \bar z$ of $X$ is positive if and only if $c=\bar b$ and $a\ge 2|b|.$

An element $x \in X$ is positive if $x =\bar x$ and spectrum of $x$ is contained in $\mathbb R_{\ge 0}$. Now notice that, $$ \begin{aligned} a1+bz+c \bar z = \bar{a1+bz+c \bar z} \implies & a=\bar a \text{ and } b=\bar c\\ \implies & a \in \mathbb R ~\text{ and } ~ c=\bar b. \end{aligned} $$ Here I proved that $c=\bar b$ but I am unable to show the second condition that $a \ge 2|b|$. Please help me to solve this. Thanks for you time and cooperation.

Answer 1

Put $z=-\frac {\overline b} {|b|}$. This is a point on $\mathbb T$ and you get $a \geq -2|b|$ from the hypothesis and the fact that $c=\overline b$. [The case $b=0$ is left to you].

Answer 2

Note that $$ a+bz+\bar b\bar z=a+2\mathrm{Re}(bz)\geq a-2|bz|=a-2|b| $$ for all $z\in\mathbb T$ with equality if $z=\frac{\bar b}{|b|}$ for $b\neq 0$ and $z\in\mathbb T$ arbitrary if $b=0$. Thus $$ \min_{z\in\mathbb T}(a+bz+\bar b\bar z)=a-2|b|. $$