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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.