Coercivity type condition $\langle x - y , \mu(x) - \mu(y) \rangle _ {\mathbb R ^d } \leq || x-y || ^2_{\mathbb R ^d } $ implies Lipschitz Let's define the coercivity type condition as :
$$ \forall x,y \in \mathbb R^d : \langle x - y , f(x) - f(y) \rangle _ {\mathbb R ^d } \leq  || x-y || ^2_{\mathbb R ^d } $$
It seems to me that this condition is stronger than locally Lipschitz continuous for $f$. However, I have no clue how to prove it.
If it is stronger than local. Lipschitz, what about Lipschitz continuous?
 A: A conclusion from my comment in the general case $\mathbb{R}^d$. Let $g$ be any decreasing function from $\mathbb{R}$ to $\mathbb{R}$, and define $f : \mathbb{R}^d \mapsto \mathbb{R}^d$ by $f(x) = \big(g(x_i)\big)_{1\le i\le d}$.
Then for any $x,y \in \mathbb{R}^d$, $\langle x - y , f(x) - f(y) \rangle _ {\mathbb R ^d } = \sum \limits_{i=1}^d (x_i-y_i)(g(x_i)-g(y_i)) \le 0 \le ||x-y||_{\mathbb{R}^d}^2$.
Yet if $f$ were locally Lipschitz, then so would its first coordinate $g$ be.

Proof that there exists a decreasing real function $g$ that is nowhere locally Lipschitz (i.e. not Lipschitz on any interval): with $(q_n)_{n \in \mathbb{N}}$ an enumeration of the rational numbers, let $$g(x) = - \sum \limits_{k=1}^{+\infty} \frac{\sqrt[3]{x-q_k}}{2^k (1+\sqrt[3]{|q_k|})}.$$
This function is well defined because for $x \in [-A,A]$, the general term is bounded in absolute value by $\frac{\sqrt[3]{A}+\sqrt[3]{|q_k|}}{2^k (1+\sqrt[3]{|q_k|}} \le \frac{\sqrt[3]{A}+1}{2^k}$, so the sum converges uniformly on $[-A,A]$.
Ad absurdum, assume that $g$ is Lipschitz in some open interval $I$, and let $k \in \mathbb{N}$ such that $q_k \in I$. Since $g$ is defined as a sum of decreasing functions, $$\Big|-\frac{\sqrt[3]{x-q_k}}{2^k(1+\sqrt[3]{|q_k|}} - 0 \Big| \le |g(x)-g(q_k)| \le C|x-q_k|$$ for $x$ in $I$, which is a neighborhood of $q_k$. We conclude that $x \mapsto \sqrt[3]{x-q_k}$ is Lipschitz at $q_k$, which is obviously false.
