Studying functional analysis, I was asked if I could determine the norm of the following functional $l:C[0,\pi]\rightarrow \mathbb R$ $$l(x)=x(0)-x(\pi/4)+\int\limits_{0}^\pi x(s)\sin sds$$
I'm familiar with several examples of finding norm of integral operators, but I don't understand: how should I manage the $x(0)-x(\pi/4)$ part at the same time? Any hint appreciated!
The naive bound is to do $$ \Big|x(0)-x(\pi/4)+\int\limits_{0}^\pi x(s)\sin s\,ds\Big| \leq 2\|x\|+\|x\|\int_0^\pi\sin s\,ds=4\|x\|. $$ Can we achieve the $4$? That would require to take the value $-1$ at $\pi/4$ without substantially departing from $x=1$. Given $n$, let $$ x(s)=\begin{cases} 1,&\ s\in[0,\pi]\setminus[\frac\pi4-\frac1n,\frac\pi4+\frac1n]\\[0.3cm] -1+2n|\frac\pi4-s|,&\ s\in [\frac\pi4-\frac1n,\frac\pi4+\frac1n] \end{cases} $$ Note that $\sin s\geq0$ on the whole interval we are considering. Then $x\in C[0,\pi]$, $\|x\|=1$, so $x(s)\geq-1$ for all $s$, and \begin{align} l(x) &=2+\int_0^{\frac\pi4-\frac1n}\sin s\,ds+\int_{{\frac\pi4+\frac1n}}^\pi\sin s\,ds+\int_{\frac\pi4-\frac1n}^{\frac\pi4+\frac1n}x(s)\,\sin s\,ds\\[0.3cm] &=4-\int_{\frac\pi4-\frac1n}^{{\frac\pi4+\frac1n}}\sin s\,ds+\int_{\frac\pi4-\frac1n}^{\frac\pi4+\frac1n}x(s)\,\sin s\,ds\\[0.3cm] &\geq4-2\Big(\frac\pi4+\frac1n-\Big(\frac\pi4-\frac1n\Big)\Big)\\[0.3cm] &=4-\frac4n. \end{align} As this can be done for any $n$, we get that $\|l\|=4$.