From Kreyszig's Introductory Functional Analysis with Applications:
Find the norm of the linear functional $f$ defined on $\mathscr{C}[-1,1]$ by
$$f(x) = \int_{-1}^0x(t)\, dt – \int_0^1 x(t)\, dt$$
Alright, so I've determined the norm is bounded by the following
\begin{align}
|f(x)|
&= \left| \int_{-1}^0x(t)\, dt – \int_0^1 x(t)\, dt \right| \\ \\
&\leq \left| \int_{-1}^0x(t)\, dt \right| + \left| \int_0^1 x(t)\, dt \right| \\ \\
&\leq \int_{-1}^0 \left| x(t) \right| \, dt + \int_0^1 \left| x(t) \right|\, dt \\ \\
&= \int_{-1}^1 | x(t) | \, dt \\ \\
&\leq 2 \max_{t\in [-1, 1]}| x(t) | \\ \\
&= 2 \|x\|
\end{align}
Taking the supremum over all $\|x\| = 1$ we see that $\|f\| \leq 2$.
Now, the book states that $\|f\| = 2$, and from examples I've seen, the easiest (and a standard) approach is to cleverly choose $x_0 \in \mathscr{C}[-1,1]$ such that $\|x_0\| = 1$ and $|f(x_0)| = 2$. But looking at the geometry of the situation, it doesn't seem obvious to me that such an $x_0$ could exist. Then again, I think the definition of supremum doesn't require that such an $x_0$ exists.
So from here I'm thinking, that we need to find a sequence of functions $(x_n)$ in $\mathscr{C}[-1, 1]$ such that
\begin{align}
x_n(t) \longrightarrow \left\{
\begin{array}
. 1, \quad -1 < t< 0 \\
-1, \quad 0 <1 < 1
\end{array} \right.
\end{align}
or something with equivalent properties—where the norm is $1$ and the functional evaluates to $2$. Examples of such sequences of functions is are
$$x_n(t) = \left[\sin(\pi t)\right]^{\frac{1}{2n+1}} \quad \text{or} \quad x_n(t) = t^{\frac{1}{2n+1}}$$
Where you can see that at $x = 1/2$ we have $\sin(\pi x) = 1$, implying that $\|x_n\| = 1$ for all $n = 1, 2, …$
Next it will take some more work but I think we show that for any $\epsilon > 0$ we can find an $N(\epsilon)$ such that for all $n > N$ we have
$$2 – |f(x_n)| < \epsilon$$
which would imply the supremum of all such $f(x_n)$ is $2$. Searching for this $N(\epsilon)$ the way I've set it up has been very tedious. This leads me to believe I'm either going about that part of the problem wrong or inefficiently. Any hints would be nice, thank you.
Best Answer
Your idea about finding a sequence is right, as $f$ doesn't reach it's norm. But I think it's easier to take $x_n$ such that we can find integral explicitly. Something like $$x_n(t) = \begin{cases} 1& t < -\frac{1}{n}\\ -n \cdot t& -\frac{1}{n} \leqslant t \leqslant \frac{1}{n}\\ -1& t > \frac{1}{n}\\ \end{cases}$$
Then $f(x_n)$ is not hard to calculate exactly, and even easier to be seen greater then $2 - \frac{2}{n}$.