Theorem 5.5, pp. 18–19 — one relaxed ADMM step on (1.2) is one DRS step on (Af) + (Bg)(b − ·) with γ = 1/β, with (i)–(v)

Initializing...

Theorem 5.5, pp. 18–19 — one relaxed ADMM step on (1.2) is one DRS step on (Af) + (Bg)(b − ·) with γ = 1/β, with (i)–(v) · Prove2Me