Finding the norm of the linear functional $f(x)=\int_{-1}^0 x(t) \, dt - \int_0^1 x(t) \, dt$ on $C[-1,1]$

Question

This is a basic question of functional analysis, but I want to know how to...

Find the norm of the linear functional $f$ defined on $C[-1,1]$ by $$f(x)=\int_{-1}^0 x(t) \, dt - \int_0^1 x(t) \, dt.$$

If anything, I know of the theorem from my textbook (FA by Kreszig, page 105):

Let us choose the space $C[a,b]$. Then $f$ is defined by $$f(x) = \int_a^b x(t) \, dt. \tag{$x \in C[a,b]$}$$ We see that $f$ is linear. We prove that $f$ is bounded and has the norm $\|f\|=b-a$. We obtain $$|f(x)|=\left|\int_a^b x(t) \, dt \right| \le (b-a) \max_{t \in[a,b]} |x(t)| = (b-a) \|x\|.$$ Taking the supremum over all $x$ of norm $1$, we obtain $\|f\| \le b-a$. To get $\|f\| \ge b-a$, we choose the particular $x=x_0=1$, note that $\|x_0\|=1$ and use the previous formula $|f(x)| \le \|f\| \|x\|$: $$\|f\| \ge \frac{|f(x_0)|}{\|x_0\|}=|f(x_0)|=\int_a^b dt = b-a.$$

Can I use what was said above (copied from textbook) to apply it to the $f(x)$ I'm given? i.e. $$\|f(x)\|=\left\|\int_{-1}^0 x(t) \, dt - \int_0^1 x(t) \, dt \right\|$$ I don't really have the slightest idea on this though.

Answer 1

Every continuous linear functional $\Phi$ on $C[-1,1]$ can be written as $$ \Phi(f) = \int_{-1}^{1}f(t)\,d\mu(t) $$ where $\mu$ is a function of bounded variation on $[-1,1]$. When normalized so that $\mu(x+0)=\mu(x)$ for $-1 \le x < 1$, then $\|\Phi\|=\mbox{Var}(\mu)$ is the total variation of $\mu$. For your functional, $\mu$ is the continuous function $$ \mu(t) = \int_{-1}^{t}\{\chi_{[-1,0]}(s)-\chi_{[0,1]}(s)\}\,ds, $$ a function of bounded variation whose total variation $\mbox{Var}(\mu)$ is the $L^{1}$ norm of the density function: $$ \|\mu\| = \int_{-1}^{1}|\chi_{[-1,0]}(s)-\chi_{[0,1]}(s)|\,ds = 2. $$ That is, $|\Phi(f)| \le 2\|f\|$ for all $f \in C[-1,1]$, and $2$ is the smallest bound.

Answer 2

Here you can estimate more: \begin{align}|f(x)| &= \left|\int_{-1}^0x(t)\,dt-\int_0^1x(t)\,dt\right| \\ &\le\left|\int_{-1}^0x(t)\,dt\right|+\left|\int_0^1x(t)\,dt\right| \\ &\le\sup_{t\in[-1,0]}|x(t)|+\sup_{t\in[0,1]}|x(t)| \\ &\le 2\cdot \sup_{t\in[-1,1]}|x(t)| \\ &=2\|x(t)\|_\infty\end{align} thus $\|f\|\le 2$

Now you must show $\|f\|\ge 2$, and you will obtain $\|f\|=2$.

I thought about this some more, and to my knowledge there is no $x^*\in C[-1,1]$ s.t $\|x^*\|_\infty=1$ and $|f(x)|=2$, such an example would be $y(t)=\left\{ \begin{array}{ll} 1 & \mbox{if $-1\le t\le 0$};\\ -1 & \mbox{if $0\lt t\le 1$}.\end{array} \right. $ but of course here $y\not\in C[-1,1]$, but if we consider the sequence $x^*_n(t)=\left\{ \begin{array}{lll} 1 & \mbox{if $-1\le t\le \frac{-1}{n}$};\\ -nt & \mbox{if $\frac{-1}{n}\lt t\lt\frac{1}{n}$};\\ -1 & \mbox{if $\frac{1}{n}\le t\le 1$}.\end{array} \right.$ Here $x^*_n\in C[-1,1]$ for each $n\in\Bbb N$.

Then we see that (by definition of sup) $\|f\|=\sup_{x\in C[-1,1]:\|x\|=1}|f(x)|\ge|f(x^*_n)|\,\forall n\in \Bbb N$.

Since (if my calculations are correct) $|f(x^*_n)|=|2+\frac{1}{n^2}-\frac{2}{n}|$, we obtain $\|f\|\ge|2+\frac{1}{n^2}-\frac{2}{n}|\,\forall n\in\Bbb N\Rightarrow \|f\|\ge 2$, now of course combining the fact that $\|f\|\le 2$ and $\|f\|\ge 2\Rightarrow \|f\|=2$.