Example of a monotone Lipschitz operator that is not cocoercive?

Question

Let $\mathcal{X}$ be a real Hilbert space, let $x,y\in\mathcal{X}$, and let $L\in\left]0,+\infty\right[$. I am looking for an operator $T\colon\mathcal{X}\to\mathcal{X}$ which is \begin{align} \text{monotone:}&\quad \quad \langle x-y\,|\,Tx-Ty\rangle\geq 0 \tag{1}\\ \text{and $L$-Lipschitz:}&\quad \quad L \|x-y\|\geq \|Tx-Ty\|,\tag{2}\\ \end{align} but not cocoercive. As a reminder, $T$ is cocoercive if there exists $\beta\in\left]0,+\infty\right[$, such that \begin{equation} \langle x-y\,|\,Tx-Ty\rangle\geq \beta\|Tx-Ty\|^2. \tag{3} \end{equation} It is clear that (3) implies (1) and (2). However, I have heard that the converse is not true. I'm having trouble cooking up a counterexample.

Answer 1

There are some natural finite-dimensional examples. Take for $X$ the usual Euclidean $\mathbb{R}^n$, and for $T$ a nonzero skew-symmetric matrix, so that $T$ is linear and for each vector $x$, $\langle x|\,Tx\rangle =0$.

Therefore, as $T$ is nonzero, (1) holds but (3) doesn’t – and $T$ is bounded.

Note that if you require some sort of strict monotonicity for $T$ (ir even coerciveness) then $T=\epsilon I+S$ with $S$ nonzero skew-symmetric and $\epsilon$ small enough will be a counterexample for a similar reason.