What is the norm of this bounded linear functional? Let $a$, $b$ be two arbitrary but fixed real numbers such that $a < b$, let $C[a,b]$ denote the normed space of all continuous real (or complex) valued functions defined on $[a,b]$ with the maximum norm, and let $f$ be the functional defined on $C[a,b]$ as $$f(x):= \int_{a}^{b} x(t) x_0(t) dt $$ for all $x$ in $C[a,b]$, where $x_0$ is a fixed element of $C[a,b]$. 
Now $f$ is evidently linear and bounded. How to compute the norm of $f$? What is this norm? 
And what is the norm of the bounded linear functional $g$ defined as follows: $$g(x):= \int_{a}^{(a+b)/2} x(t) \cdot dt - \int_{(a+b)/2}^{b} x(t) \cdot dt$$ for all $x$ in $C[a,b]$? 
 A: Since every $x \in C[a, b]$ is continuous on a compact set, it's bounded.
For any $x \in C[a, b]$, we have:
$$
|f(x)| = \left|\int_a^b x(t) x_0(t) \, dt \right| \le \int_a^b |x(t)| \cdot |x_0(t)| \, dt \le \|x\| \int_a^b |x_0(t)| \, dt
$$
Thus, $f$ is bounded and its norm satisfies:
$$
\|f\| \le \int_a^b |x_0(t)| \, dt \newcommand{sgn}{\operatorname{sgn}}
$$
In fact, equality holds. To see this, consider the sign function $\hat x(t) = \sgn(x_0(t))$. By Lusin's theorem, there is exists a sequence of functions $x_n \in C[a, b]$ such that $\|x_n\| \le 1$ and $x_n(t) \to \hat x(t)$ as $n \to \infty$ for every $t \in [a, b]$. By the dominated convergence theorem, we have:
\begin{align}
\lim_{n \to \infty} f(x_n) &= \lim_{n \to \infty} \int_a^b x_n(t) x_0(t) \, dt \\
&= \int_a^b \lim_{n \to \infty}  x_n(t) x_0(t) \, dt \\
&= \int_a^b \sgn(x_0(t)) x_0(t) \, dt \\
&= \int_a^b |x_0(t)| \, dt
\end{align}

For the second linear functional you added, approximate the following function via Lusin's theorem in a similar manner:
$$
\hat x(t) = \begin{cases} 1 & \text{if } t \le \frac{a+b}{2} \\
-1 & \text{if } t > \frac{a+b}{2}
\end{cases}
$$
