Show that $\left\{\varphi\in{L^p(\mu)}':f(x)\ge\langle\varphi,f\rangle\text{ for all }f\in L^p(\mu)\right\}=\{\delta_x\}$

Question

Let $(E,\mathcal E,\mu)$ be a measure space and $$\delta_xf:=f(x)\;\;\;\text{for }f:E\to\mathbb R$$ denote the evaluation functional at a fixed $x\in E$ with $\mu(\{x\})>0$. Now let $p\in[1,\infty]$. How can we show that $$\left\{\varphi\in{L^p(\mu)}':f(x)\ge\langle\varphi,f\rangle\text{ for all }f\in L^p(\mu)\right\}\tag1$$ is precisely $\{\delta_x\}$?

Obviously, $\delta_x$ is contained in $(1)$. Moreover, the condition in $(1)$ implies some kind of operator norm bound on the $\varphi$ contained in $(1)$. How can we conclude?

Answer 1

Let $X$ be a real vector space and $u,v$ two linear maps $X \to \Bbb{R}$.

If $\forall f \in X, \langle u,f\rangle \ge 0$ then $u = 0$.

Proof : if $u \ne 0$ then for some $f \in X$,$\langle u,f\rangle\ne 0$ so that $\langle u,f\rangle$ or $\langle u,-f\rangle$ is not $\ge 0$.

If $\forall f \in X, |\langle u,f\rangle| \le |\langle v,f\rangle|$ then $u = C v$ where $C\in [-1,1]$

Proof : take $g\in X,\langle v,g\rangle=1$ then $|\langle u,f-\langle v,f\rangle g\rangle| \le |\langle v,f-\langle v,f\rangle g\rangle|=0$ so that $u= \langle u, g\rangle v$