(Integral) Operator Norm: Find $||\phi||$ where $\phi : \mathcal{L^1(m)} \to \mathbb{R}$ is defined by $\phi(f) = \int (x - \frac{1}{2}) f(x) dm(x)$

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}