Prove that $\|P_h\|\geq\int_{-1}^{1}|h(t)|dt$ using Tietze extension theorem

Question

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.

Answer 1

The components of $[0,1]$ for which $h(t)>\delta$ holds are disconnected from those where $h(t)<-\delta$ because $h$ is continuous. Hence the following is continuous on $A_{\delta}$, so the extension theorem can be used

$f_0(t) = \begin{cases} 1 & t\in A_{\delta} \land h(t)>0 \\ -1 & t\in A_{\delta} \land h(t)<0 \\ \text{Extension} & \text{else} \end{cases}$

Then
$|P_h(f_0) - \int_{-1}^1|h(t)|dt| = |\int_{-1}^{1}f_0(t)h(t)dt - \int_{-1}^1|h(t)|dt| \\\hspace{4.152cm}= \left|\int_{A_\delta}|h(t)|dt + \int_{[-1,1]\setminus A_\delta}f_0(t)h(t)dt - \int_{-1}^1|h(t)|dt\right| \\\hspace{4.152cm}\leq \left|\int_{A_\delta}|h(t)|dt - \int_{-1}^1|h(t)|dt \right|+\left|\int_{[-1,1]\setminus A_\delta}f_0(t)h(t)dt\right| \\\hspace{4.152cm}\leq 4\delta$

because
$\left|\int_{A_\delta}|h(t)|dt - \int_{-1}^1|h(t)|dt\right|=\left|\int_{-1}^1\left(\begin{cases} 0 & t\in A_\delta \\ |h(t)| & t\notin A_\delta \end{cases}\right)dt\right|\leq(1-(-1))\underset{t\notin A_\delta}{\max}|h(t)|=2\delta$

$\left|\int_{[-1,1]\setminus A_\delta}f_0(t)h(t)dt\right| \leq \int_{[-1,1]\setminus A_\delta}|f_0(t)h(t)|dt \leq \int_{-1}^1 \underset{t\notin A_\delta}{\max}|f_0(t)h(t)|dt = 2\delta$

Now by the reverse triangle inequality:
$\left||P_h(f_0)| - \int_{-1}^1|h(t)|dt\right| \leq |P_h(f_0) - \int_{-1}^1|h(t)|dt|\leq 4\delta \qquad\qquad\qquad(*)$

You already proved specifically $|P_h(f_0)|\leq \int_{-1}^{1}|h(t)|dt \|f_0\|_\infty = \int_{-1}^{1}|h(t)|dt$.
Hence we can rewrite the LHS of $(*)$

$4\delta \geq \left||P_h(f_0)| - \int_{-1}^1|h(t)|dt\right| = -|P_h(f_0)|+\int_{-1}^1|h(t)|dt \implies\\ |P_h(f_0)|\geq \int_{-1}^1|h(t)|dt-4\delta $

By definition $|P_h(f_0)|\leq||P_h||\cdot ||f_0||_\infty$ but $||f_0||_\infty=1$ so $||P_h||\geq|P_h(f_0)|\geq\int_{-1}^1|h(t)|dt-4\delta$. For all $\delta$, so $||P_h||\geq\int_{-1}^1|h(t)|dt$

Answer 2

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.