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.
Best Answer
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.