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.
 A: First calculate $\|\psi\|$. We have:
$$|\psi(f)| = \left|\int_{-1}^1f(t)\,dt\right| \le \int_{-1}^1 |f(t)|\,dt \le \int_{-1}^1 \|f\|_\infty\,dt  =2\|f\|_\infty$$
so $\|\psi\| \le 2$.
For the even function $f \equiv 1$ we have $$\|\psi\| \ge \frac{|\psi(f)|}{\|f\|_\infty} = 2 
$$
so we conclude $\|\psi\| = 2$.
Consider the bounded linear functionals $F_1, F_2 : C[-1,1] \to \mathbb{R}$ given by:
$$F_1(f) = \int_{-1}^1 f(t)\,dt$$
$$F_2(f) = 2\int_{0}^1 f(t)\,dt$$
$F_1$ and $F_2$ are both extensions of $\psi$. Indeed, for an even function $f$ we have:
$$F_2(f) = 2\int_{0}^1 f(t)\,dt = \int_{-1}^1 f(t)\,dt = \psi(f) = F_1(f)$$
And also we have $\|F_1\| = \|F_2\| = 2 = \|\psi\|$ which is shown similarly as for $\psi$:
$$|F_1(f)| = \left|\int_{-1}^1f(t)\,dt\right| \le \int_{-1}^1 |f(t)|\,dt \le \int_{-1}^1 \|f\|_\infty\,dt = 2\|f\|_\infty$$
$$|F_2(f)| = \left|2\int_{0}^1f(t)\,dt\right| \le 2\int_{0}^1 |f(t)|\,dt \le 2\int_{0}^1 \|f\|_\infty\,dt =2\|f\|_\infty$$
Therefore, we get $\|F_1\|, \|F_2\| \le 2$. Since they extend $\psi$, we also have the reverse inequality.
So, $F_1$ and $F_2$ are both Hahn-Banach extensions for $\psi$.
Now, it is a known result that if there are two different Hahn-Banach extensions of a functional, then there are infinitely many Hahn-Banach extensions.
Namely, for every $\alpha \in [0,1]$ the functional $\alpha F_1 + (1 - \alpha) F_2$ is also a Hahn-Banach extension of $\psi$.
A: 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
$P_h(t\mapsto t)=\int_{-1}^1 t\cdot h(t)dt= \frac{2}{n+2}$
