Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.8, Problem 3: What is the norm of this functional?

Question

What is the norm of the linear functional $f$ defined on the normed space $C[a, b]$ of all functions defined and continuous on the closed interval $[a,b]$ with the norm defined as $$\Vert x \Vert \colon= \max_{t\in[a,b]} \vert x(t) \vert \; \; \; \forall x \in C[a,b]?$$

Let $$ f(x) \colon= \int_a^{\frac{a+b}{2}} x(t) dt - \int_{\frac{a+b}{2}}^b x(t) dt \; \; \; \forall x \in C[a,b].$$

I know that $f$ is linear and bounded because for all $x \in C[a,b]$, we have $$\vert f(x) \vert = \left\vert \int_a^{\frac{a+b}{2}} x(t) dt - \int_{\frac{a+b}{2}}^b x(t) dt \right\vert \leq \left\vert \int_a^{\frac{a+b}{2}} x(t) dt \right\vert + \left\vert \int_{\frac{a+b}{2}}^b x(t) dt \right\vert \\ \leq \int_a^{\frac{a+b}{2}} \vert x(t) \vert dt + \int_{\frac{a+b}{2}}^b \vert x(t) \vert dt \leq \int_a^{\frac{a+b}{2}} \max_{\tau\in[a,b]} \vert x(\tau) \vert dt + \int_{\frac{a+b}{2}}^b \max_{\tau\in[a,b]} \vert x(\tau) \vert dt \\ = \int_a^b \max_{\tau\in[a,b]} \vert x(\tau \vert dt = (b-a)\Vert x \Vert_{C[a,b]},$$ which shows that $f$ is bounded and that $$\Vert f \Vert \leq b-a.$$

What next? How to arrive at the reverse inequality?

Answer 1

We need to show that $||f||\geq b-a$. To do this take a sequence of continuous functions, which is monotone and converges to the step function \begin{align*} x(t):=\begin{cases}1,& a\leq t\leq\frac{a+b}{2} \\ -1,& \frac{a+b}{2}<t\leq b\end{cases}. \end{align*} As an example we can use piecewise linear functions such that \begin{align*} x_n(t)&=1,\quad t\in\left[a,\frac{a+b}{2}-\frac{1}{n}\right], \\ x_n(t)&=-1,\quad t\in\left[\frac{a+b}{2},b\right]. \end{align*} Then $||x_n||=1$, and using the monotone convergence theorem (or the dominated convergence theorem) we obtain \begin{align*} \lim_{n\to\infty}f(x_n)=f\left(\lim_{n\to\infty}x_n\right)=f(x)=b-a. \end{align*} This shows the desired result.

Answer 2

Here's a rather pedestrian proof. If it seems confusing, just draw the picture. You're making a function that is 1 up until some point just a bit to the left of the midpoint, and -1 from the midpoint to $b$. This isn't continuous, so you force it continuous by connecting the two line segments. This why we needed to give ourselves a bit of room to work with, so we can ensure this connecting line has a well defined slope. This doesn't contribute anything to the integral, and so you end up with something just a bit smaller than $b-a$, but you can make it as close as you like.

Define

$$x(t)=\begin{cases}1,& a\leq t\leq\frac{a+b}{2}-\epsilon\\ \frac{a+b}{\epsilon}-\frac{2}{\epsilon}t-1 ,& \frac{a+b}{2}-\epsilon \leq t \leq \frac{a+b}{2} \\ -1,& \frac{a+b}{2}<t\leq b\end{cases}.$$

Then you get

$$ \begin{align*} f(x) &= \int_a^{\frac{b+a}{2}-\epsilon} dt+ \int_{\frac{b+a}{2}-\epsilon}^\frac{b+a}{2} \left(\frac{a+b}{\epsilon}-\frac{2}{\epsilon}t-1\right) dt +\int_\frac{b+a}{2}^bdt \\ &= b-a-\epsilon \end{align*} $$

So, if you combine that with the work you already did, you have

$$b-a-\epsilon \leq \Vert f \Vert \leq b-a $$

for every $\epsilon >0.$ Since we can make $\epsilon$ arbitrarily small, this implies $\Vert f \Vert = b-a.$