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For a jointly smooth convex function $f(x,y)$, update $y$ implicitly by its proximal minimization at fixed $x$, then update $x$ explicitly with the gradient at the mixed point $(x,y^{+})$. The combined map is $(\tau L/2)$-averaged for $0<\tau L<2$, as follows by applying cocoercivity at two mixed points. Its fixed points are precisely the minimizers. If a minimizer exists, averaged iteration proves convergence.

For $f(x,y)=h(x,y)+\varphi (y)$, where $h$ is convex with an $L_h$-Lipschitz gradient and $\varphi$ is convex, use a proximal $y$-step for the full section and an explicit $x$-step for $h$. Monotonicity of the two selected subgradients of $\varphi$ combines with cocoercivity of $\nabla h$ to make the full update firmly nonexpansive for $0<\tau L_h\le 1$. Existence of a minimizer then gives convergence, even though the full objective is not smooth.

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.

Assume $\mathrm{argmin}f\ne \varnothing$, as is necessary for convergence to a minimizer. The printed hypotheses alone omit this condition: $f(x,y)=x$ in scalar $x$ is convex with an $L$-Lipschitz gradient for any prescribed $L>0$, but has no minimizer, and the update sends $x^k=x^0-k\tau$ to $-\infty$.

Under existence, define the full update $T(x,y)=(x^{+},y^{+})$. The proximal $y$-step is uniquely defined by strong convexity of its quadratic term and satisfiesThe partial subdifferential in this equation belongs to the scalar function $y\mapsto f(x,y)$; differentiating a subdifferential itself would not define the printed proximal minimization.

We prove directly that the alternating proximal-gradient operator $T$ is averaged. For two inputs $z=(x,y)$ and $\tilde{z}=(\tilde{x},\tilde{y})$, setThe input difference is $(p,q+\tau b)$, the output difference is $(p-\tau a,q)$, and the difference of the two update residuals is $\tau (a,b)$. Exact expansion givesThe first inequality uses cocoercivity at the mixed points $w,\tilde{w}$. Therefore $T$ is $(\tau L/2)$-averaged for the requested $0<\tau L\le 1$. In particular, the endpoint $\tau L=1$ is covered: $T$ is firmly nonexpansive, not merely nonexpansive.

A fixed point has $y^{+}=y$ and $x^{+}=x$, so the two optimality equations give ${\nabla }_{y}f=0$ and ${\nabla }_{x}f=0$. Conversely, at a minimizer both gradients vanish and neither step moves. By convexity,The assumed existence of a minimizer supplies a fixed point. The Browder convergence theorem for averaged operators now givesNo uniqueness of the minimizer is needed. The same estimate even proves averagedness for $0<\tau L<2$ in this smooth setting, but the requested interval already suffices. The existence condition and the use of the gradient at $(x^k,y^{k+1})$ are both essential to the proof as presented.