(Integral) Operator Norm: Find $||\phi||$ where $\phi : \mathcal{L^1(m)} \to \mathbb{R}$ is defined by $\phi(f) = \int (x - \frac{1}{2}) f(x) dm(x)$ Let $(X, \mathcal{A}) = ([-1, 1], \mathcal{B_{[-1, 1]}})$ be a measurable space equipped with the Lebesgue measure, $m$.
Define $\phi : \mathcal{L^1(m)} \to \mathbb{R}$ by
$\phi(f) = \int (x - \frac{1}{2}) f(x) dm(x)$
To find $||\phi||$, I think the general strategy is to find an upper bound on $\phi$, and then construct a function $f$ such that it is attained.
In our case, we have
$||\phi|| = inf\{c \ge 0: |\phi(f)| \le c ||f||\}$. We have that $|\phi(f)| \le \int |(x - \frac{1}{2})| |f(x)| dm(x)$
I can split this up into two intervals: $[-1, 0]$ and $[0, 1]$.  On $[0, 1]$, we have that $|x - \frac{1}{2}| \le \frac{1}{2}$. On $[-1, 0]$, we have that $|x - \frac{1}{2}| \le \frac{3}{2}$. Hence $|\phi(f)| \le 2 \int |f|$. This is my bound. However, I am unable to finish the proof because I am not sure how to construct a function $f$ which attains this bound.
 A: We have that
\begin{equation}
|\phi(f)| = \int_{[-1,1]}\left|(x-\frac{1}{2})f(x)\right| \le ||x-\frac{1}{2}||_{L^\infty[-1,1]} ||f||_{L^1[-1,1]} \le \frac{3}{2}||f||_{L^1[-1,1]}
\end{equation}
and we deduce that
\begin{equation}
||\phi||\le\frac{3}{2}\qquad\qquad (1)
\end{equation}
Fixing $n\in \mathbb{N}_{>0}$ we consider the functions define as
\begin{equation}
f_n(x) := n\chi_{[-1,-1+1/n]}
\end{equation}
where $\chi$ is the indicator function.
Now we have that
\begin{equation}
f_n\in L^1([-1,1]) \qquad \text{and} \qquad  ||f||_{L^1([-1,1])}=1
\end{equation}
Therefore, Hence, developing the calculations easily obtains that
\begin{equation}
\phi(f_n) = -\frac{3}{2}+\frac{1}{2n}
\end{equation}
Then we have that
\begin{equation}
|\phi(f_n)|\le\left|-\frac{3}{2}+\frac{1}{2n}\right|||f_n||_{L^1([-1,1])}
\end{equation}
So,
\begin{equation}
||\phi|| \ge \left|-\frac{3}{2}+\frac{1}{2n}\right| \qquad \forall n\in \mathbb{N}_{>0} \qquad\qquad (2)
\end{equation}
We can now conclude from the the previous equation (number (1) and (2) ) that
\begin{equation}
||\phi||=\frac{3}{2}
\end{equation}
A: The norm of $\varphi$ is defined as
$$
\|\varphi\|=\sup_{\|f\|_{L^1[-1,1]}}\left|\int_{[-1,1}\Big(x-\frac{1}{2}\Big)
\,f(x)\,dx\,\right|
$$
Clearly,
$$
\left|\int_{[-1,1]}\Big(x-\frac{1}{2}\Big)\,f(x)\,dx\,\right|\le \max_{x\in [-1,1]}] \Big|x-\frac{1}{2}\Big|\,
\left|\int_{[-1,1]} |f(x)|\,dx\,\right|=\frac{3}{2}\|f\|_{L^1[-1,1]}
$$
For $f(x)=\frac{2}{a}\chi_{[-1,-1+a]}$, where $a\in (0,1)$,
clearly, $\|f\|_{L^1[-1,1]}=1$, while
$$
\varphi(f)=\int_{[-1,1]}\Big(x-\frac{1}{2}\Big)\,f(x)\,dx=
\frac{2}{a}\int_{-1}^{-1+a}\Big(x-\frac{1}{2}\Big)\,dx=-\frac{3}{2}+\frac{a}{2},
$$
and hence
$$
|\varphi(f)|=\frac{3}{2}-\frac{a}{2}
$$
and hence
$$
\|\varphi\|\ge \frac{3}{2}-\frac{a}{2},
$$
for all $a\in (0,1)$. Consequently $\|\varphi\|=3/2$.
