Linear functional

Question

For $a<b \in \mathbb{R}$, let $G=(a,b)$ be a bounded interval. For every $x \in G$, let the generalized function $\delta_x$ be defined by $$ \int_G \delta_x \phi(x)dx = \phi(x) ~~ \text{for every} ~~ \phi \in C^0(\overline{G}) $$ How can I show that for every $x \in G$ and every $p \in [1, \infty]$ the function $\delta_x$ is a linear continuous functional on $W^{1,p}(G)$, that is $\delta_x \in W^{1,p}(G)\text{*}$ for $x \in G$ and every $p \in [1,\infty]$ where $\text{*}$ denotes the dual space.

Answer 1

Here is an answer for the case $p=1$. Let $f\in C^1[a,b]$. First, we choose some $\xi\in (a,b)$ such that $\int_a^b|f(t)|\,dt = (b-a)|f(\xi)|$. Then we have (for all $x\in (a,b)$) \begin{align*} |f(x)|& \le |f(x) - f(\xi)| + |f(\xi)| = \left|\int_\xi^x f'(t)\,dt\right| + \frac 1 {b-a}\int_a^b|f(t)|\,dt\\ &\le \int_{\min\{\xi,x\}}^{\max\{\xi,x\}} |f'(t)|\,dt + \frac 1 {b-a}\int_a^b|f(t)|\,dt\le\max\left\{1,\frac 1 {b-a}\right\}\|f\|_{W^{1,1}}. \end{align*} Due to this inequality, the linear functional $\delta_x : C^1[a,b]\to\mathbb K$ (where $\mathbb K$ is either $\mathbb R$ or $\mathbb C$), defined by $\delta_x(f) := f(x)$, is continuous w.r.t. the $W^{1,1}$-norm. And as $C^1[a,b]$ is dense in $W^{1,1}(a,b)$, it can be uniquely extended to a continuous linear functional on $W^{1,1}(a,b)$.