How to show non existence of an operator $A: V \to V^*$

Question

Let $V$ be a Banach space , $p \in (1,2)$ and $\mu >0.$ How can I prove that there does not exist an operator $A:V \to V^*$ such that $$\langle Au-Av, u-v\rangle \geq \mu \Vert u-v \Vert ^p,\qquad u, v \in V. $$

Answer 1

Pick an arbitrary point $v\in V$ such that $\|v\|=1$.

Let $n$ be an arbitrary integer. (Our goal is a proof by contradiction as $n\to\infty$).

We consider the points $v_k:= \tfrac{k}{n} v$ for $k=0,\ldots, n$. Then, by assumption we have $$ \langle A v_{k+1} - Av_k,v_{k+1}-v_k \rangle \geq \mu \| v_{k+1}-v_k\|^p $$ for every $k=0,\ldots,n-1$.

Summing up those terms will lead to the inequality $$ \tfrac 1n \langle Av - A0 , v \rangle \geq \mu n^{1-p}. $$

For sufficiently large $n$, this cannot be true.