Property of norm in a dual space

Question

Let $$E:=\{x\in C^1([0,1]):x(0)=x(1)=0\}.$$ Then $(E,\|\cdot\|)$ is a normed space if one sets $$\|x\|=\sup_{t\in[0,1]}\{|x'(t)|\}.$$ Choose $s\in[0,1]$ and define $g_s:E\rightarrow\mathbb R$ as $g_s(x)=x'(s)$. Clearly, $g_s$ is a continuous linear functional. I'd like to prove that $$t\neq s\Rightarrow \|g_t-g_s\|_*=2.$$ Here, $\|\cdot\|_*$ stands for the usual norm in $E'$: $$\|g\|_*=\sup_{\{\|x\|=1\}}|g(x)|.$$ A simple calculation using triangle inequality give us $t\neq s\Rightarrow \|g_t-g_s\|_*\leq2$. But I couldn't find $x\in E$ with $\|x\|=1$ and $|x'(t)-x'(s)|=2$. If someone could help me, I'd be grateful. Anyway, thanks for your attention!

Answer 1

Say $0<t<s<1$ then define the piecewise linear function $$\hat{x}(r)=\begin{cases}0&,0\leq r\leq t\\r-t&,t\leq r\leq \frac{s+t}{2}\\-r+s&\frac{s+t}{2}\leq r\leq s\\0&s\leq r\leq 1\end{cases}$$ $\hat{x}$ is an integrable function and thus one can use the mollifier $$\eta(r)=\begin{cases}Ce^{\frac{1}{r^2-1}}&-1<r<1\\0& else\end{cases}$$ where $C$ is chosen so that $\int_{-\infty}^{\infty}\eta(r)dr=1$. Set $\eta_{\varepsilon}(r)=\frac{1}{\varepsilon}\eta(\frac{r}{\varepsilon})$ and define $$\hat{x}^{\varepsilon}(r)=\int_0^1\eta_{\varepsilon}(r-u)\hat{x}(u)du.$$ One can show that $\hat{x}^{\epsilon}$ are smooth functions on $(\varepsilon,1-\varepsilon)$ and that $\hat{x}^{\varepsilon}\rightarrow \hat{x} a.e.$. Furthermore $$|\frac{d\hat{x}^{\varepsilon}(r)}{dr}|=|\int_0^1\eta_{\varepsilon}'(r-u)\hat{x}(u)du|=|\int_0^1 \eta_{\varepsilon}(r-u)\hat{x}'(u)du|\leq \sup_{u\in(0,1)}|\hat{x}'(u)|\leq 1$$ where we used partial integration, the fact that the functions disappear outside $[t,s]$, $\eta_{\varepsilon}$ is symmetric in $r$ and $u$ and that $\int_0^1\eta_{\varepsilon}(r)dr=1$. Now see that $$|g_t(\hat{x}^{\varepsilon})-g_s(\hat{x}^{\varepsilon})|\rightarrow |\hat{x}'(t)-\hat{x}'(s)|=1-(-1)=2$$ as $\varepsilon\rightarrow 0$ to conclude $||g_t-g_s||_*=2.$