TITLE: Prove that $\|P_h\|\geq\int_{-1}^{1}|h(t)|dt$ using Tietze extension theorem
QUESTION [5 upvotes]: For each $h\in C([-1,1])$, define the linear functional $P_h:C([-1,1])\to \mathbb{R}$ by
$$
P_h(f)=\int_{-1}^{1}f(t)h(t)dt,\qquad f\in C([-1,1]).
$$
Here is the Banach space $C([-1,1]):=\{f:[-1,1]\to \mathbb{R}:f\textrm{ continuous}\}$, endowed with $\| f\|_\infty=\sup_{t\in [-1,1]}|f(t)|$ for $f\in C([-1,1])$.
Problem: Prove that $\| P_h\|=\int_{-1}^{1}|h(t)|\,dt$.
First, we have
$$
|P_h(f)|\leq \int_{-1}^{1}|f(t)h(t)|dt\leq \int_{-1}^{1}|h(t)|dt \|f\|_\infty\implies \|P_h\|\leq \int_{-1}^{1}|h(t)|dt
$$
To prove the other inequality, I got the hint as follows: For each $\delta>0$, let $A_\delta=\{t\in [-1,1]:|h(t)|\geq \delta\}$. Define $f_0$ suitably on $A_\delta$ and use Tietze extension theorem. 

Tietze extension theorem: 
   If $K$ is a compact Hausdorff space and $K_0$ is a closed subset of
   $K$ then for each $f_0\in C(K_0)$ there exists $f\in C(K)$ such that
  $f|_{K_0}=f$ and $\|f\|_\infty = \|f_0\|_\infty$.

Do you have an idea which map $f_0$ should look like defined on $A_\delta$?
Edit: I asked a comrade about this, and he got the hit from our lecturer. He said about putting $f_0(t)=\frac{h(t)}{|h(t)|}$. This is well-defined on $A_\delta$. Then, he claimed something like
$$
\| P_h(f_0)\|=\left | \int_{-1}^{-1}f_0(t)h(t) dt \right |\geq \int_{A_\delta}\frac{h(t)}{|h(t)|}h(t)dt=\int_{A_\delta}\frac{|h(t)|^2}{|h(t)|}dt=\int_{-1}^{-1}|h(t)|dt.
$$
I really doubt about this claim because it doesn't look right. Is there a way fix it? I guess I want something like 
$$
\|P_h\|\|f\|_\infty=\|P_h\|\|f_0\|_\infty\geq \| P_h(f_0)\|\geq \|f\|_\infty \int_{-1}^{-1}|h(t)|dt
$$
for some $f\in C([-1,1])$ by Tietze extention theorem.

REPLY [2 votes]: I'm going to approach the problem differently, just in case you might be interested. 
Define $\text {sgn }x$ to be $1$ on $(0,\infty),$ $0$ at $0,$ and $-1$ on $(-\infty,0).$ Let $Z$ denote the zero set of $h.$ Then the map $x\to d(x,Z)$ is continuous. Furthermore, the functions
$$f_n(x) = (d(x,Z)/2)^{1/n}\text { sgn }(h(x)),\, n=1,2,\dots$$
all belong to $C[-1,1].$ Note that $|f_n(x)|\le 1$ for all $x\in [-1,1].$ Moreover, for any fixed $x\in [-1,1],$ we have $f_n(x) \to \text {sgn }(h(x)).$
We are set up to use the dominated convergence theorem. From the above,
$$\lim_{n\to \infty}\int_{-1}^1 f_n(x)h(x)\, dx = \int_{-1}^1 \lim_{n\to \infty}f_n(x)\cdot h(x)\, dx $$ $$= \int_{-1}^1\text {sgn }(h(x))\cdot h(x)\, dx = \int_{-1}^1|h(x)|\, dx.$$
Thus $\|P_h\|$ is at least $\int_{-1}^1|h(x)|\, dx.$ Since we have the other inequality, $\|P_h\|=\int_{-1}^1|h(x)|\, dx$ as desired.