Show that a such set of Hahn-Banach extensions is infinite Let $M$ be the subspace of $C([-1,1])$ consisting of all even functions. Let $\psi:M\to \mathbb{R}$ be the linear functional given by $\psi(f)=\int_{-1}^{1}f(t)dt$. 
Show that there are infinitely many bounded linear extensions $F$ of $\psi$ to $C([-1,1])$ such that $\|F\|=\|\psi\|$.
Hint: First show that $P_h$ is an extension of $\psi$ if $h(t)+h(-t)=2$ for all $t\in [-1,1]$.
It is related to the posts (1) and (2). I tried to use $h(t)=|t+1|$, because then it satisfies $h(t)+h(-t)=2$ on $[-1,1]$. But I do not know how to answer this problem in general. If the post needs some more information, let me know.

Choose $\psi = P_h$. With $h(t)+h(-t)=2$, we get for even $f$
$\psi(f) = P_h(f)=\int_{-1}^1f(t)h(t)dt=\int_{-1}^0f(t)h(t)dt+\int_{0}^1f(t)h(t)dt
\\\hspace{5.78cm}=\int_{-1}^0f(-t)(2-h(-t))dt+\int_{0}^1f(t)h(t)dt
\\\hspace{5.78cm}=\int_{0}^1f(t)(2-h(t))dt+\int_{0}^1f(t)h(t)dt
\\\hspace{5.78cm}=\int_{0}^1(f(t)(2-h(t))+f(t)h(t))dt
\\\hspace{5.78cm}=2\int_{0}^1f(t)dt=\int_{-1}^1f(t)dt$
Which proves the hint. Consequently $h(t)=1+t^n$ work for any positive odd $n$, and not two choices of $n$ makes $P_h(f)$ the same, since
$P_h(t\mapsto t)=\int_{-1}^1 t\cdot h(t)dt= \frac{2}{n+2}$
