Compute the norm of functional $f\in (W^{2,1}[0,1])^*$ given by $$f(u)=\int_0^{1/2} u(t)\,dt.$$
My attempt The norm of $f$ is computed by \begin{align} \|f\|=\sup_{u\ne 0} \frac{f(u)}{\|u\|_{W^{1,2}[0,1]}}. \end{align} Since \begin{align} f(u)\le \int_0^1|u(t)|\,dt\le \|u\|_{W^{1,2}[0,1]} \end{align} we know $\|f\|\le 1$. Then I guess $\|f\|=1$ and want to find a $u$ such that \begin{align} \|u\|_{W^{1,2}[0,1]} = \int_0^{1/2}u(t)dt. \end{align} But I met some problem. Naively I want to set $$ u(t)=\begin{cases} 1 & 0\le t\le 1/2;\\ 0 & 1/2< t\le 1. \end{cases} $$ But it doesn't satisfy $u\in W^{2,1}[0,1]$. Then I want to construct a smooth $u$ such that $u\equiv 1$ on $[0,1/2]$ and $u\equiv 0$ on $[1/2+\epsilon,1]$. But I found that $$ \int_0^1|u'(t)|dt\ge u(1)-u(0)=1. $$ So this also fails.
Now I guess $\|f\|\ne 1$, and I got stuck on the precise computation. Any helps? Thanks in advance!
I thought I found it out. Set $u\equiv 1$ we know $\|f\|\le 1/2$. Then we show $\|f\|=1/2$. By density, it suffices to show that $$ \int_0^{1/2}u(t)\,dt\le\frac12\|u\|_{W^{1,2}[0,1]}\quad\forall u\in C^\infty(0,1).\quad (*) $$ Set $m=\inf_{(0,1)}|u|$ and $M=\sup_{(0,1)}|u|$. Since $$ u(x_1)-u(x_2)=\int_{x_1}^{x_2}u'(t)\,dt\le \int_0^1|u'(t)|\,dt $$ we easily know $$ M-m\le \|u'\|_{L^1[0,1]}. $$ Then $$ \|u\|_{L^1[0.5,1]}+\|u'\|_{L^1[0,1]}\ge \frac12m+(M-m)\ge \frac12M\ge \|u\|_{L^1[0,0.5]} $$ and hence $$ \|u\|_{W^{1,2}[0,1]}\ge \|u\|_{L^1[0,0.5]}+\|u\|_{L^1[0.5,1]}+\|u'\|_{L^1[0,1]} \ge 2\|u\|_{L^1[0,0.5]} \ge 2\int_0^{1/2}u(t)\,dt. $$ Hence we get $(*)$, which complete the proof of $\|f\|=1/2$.