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An operator $G$ is $\beta$-cocoercive when $⟨Gz-Gw,z-w⟩\ge \beta \Vert Gz-Gw{\Vert }^2$ with $\beta >0$. This strengthens monotonicity and implies a Lipschitz constant $1/\beta$ by the Cauchy-Schwarz inequality. Gradient maps of smooth convex functions provide an important converse through the Baillon–Haddad theorem.

Here $L$-smooth means that $f$ is differentiable on the entire Euclidean space and has an $L$-Lipschitz gradient, with $L>0$. Together with convexity, this implies $1/L$-cocoercivity:First, integration of the Lipschitz gradient along a line segment proves the descent lemma:Indeed, subtract the linear term and integrate $⟨\nabla f(v+t(w-v))-\nabla f(v),w-v⟩$ for $0\le t\le 1$; its absolute value is at most $Lt\Vert w-v{\Vert }_2^2$.

Fix $u$ and define $q(w)=f(w)-⟨\nabla f(u),w⟩$. This is a convex function, has the same Lipschitz gradient constant, and $\nabla q(u)=0$, so $u$ is a global minimizer. Put $d=\nabla f(v)-\nabla f(u)=\nabla q(v)$. The descent lemma at $v$ with $w=v-d/L$ givesThereforeInterchange $u,v$ and add the inequalities. The function values cancel, leaving the asserted cocoercivity. This is a proof of the Baillon–Haddad theorem without assuming a second derivative. Convexity is essential: $f(u)=-\Vert u{\Vert }_2^2/2$ has a $1$-Lipschitz gradient but its gradient is not cocoercive.