“…24)ˆT 0ˆMˆT 0ˆM |u(x, t) − v(y, s)|(∂ t φ + ∂ s φ) + q(u(x, t), v(y, s), x, t), grad x g φ g + q(v(y, s), u(x, t), y, s), grad y g φ g dv g (y) ds dv g (x) dt ≥ 0 with q(u, k, x, t) := sgn(u − k) (f (u, x, t) − f (k, x, t)) .We setφ(x, t, y, s) := ψ(t)ψ(x)ωǭ(t − s)κ ǫ (x, y), where ψ ∈ C ∞ 0 ((0, T )) with ψ ≥ 0 andψ ∈ C ∞ 0 (M ) withψ ≥ 0, ωǭ(s) := ω( s ǫ ) with ω ∈ C ∞ (R), supp(ω) ⊂ (−1, 1), ω ≥ 0 and´R ω(s)ds = 1 and κ ǫ (x, y) := 1 ǫ n κ dg(x,y) ǫ with κ ∈ C ∞ (R),supp(κ) ⊂ (−1, 1), κ ≥ 0 and R n κ(|z|)dz = 1. Thus, (4.24) yieldŝ T 0ˆM ψ ′ (t)ψ(x)ˆT 0ˆM ωǭ(t − s)κ ǫ (x, y)|u(x, t) − v(y, s)| dv g (y) ds dv g (x) dt +ˆT 0ˆM ψ(t)ˆT 0ˆM ωǭ(t − s) q(u(x, t), v(y, s), x, t), grad x g κ ǫ (x, y) gψ (x) + q(u(x, t), v(y, s), x, t), grad x gψ (x) g κ ǫ (x, y) + q(v(y, s), u(x, t), y, s), grad y g κ ǫ (x, y) gψ (x) dv g (y) ds dv g (x) dt ≥ 0.With the same argumentation as in[20,[1714][1715][1716][1717][1718] we obtain by subsequently lettingǭ and ǫ tend to zerôT 0ˆM ψ ′ψ |u − v| + ψ sgn(u − v) grad gψ , f (u) − f (v) g dv g dt ≥ 0. Settingψ := 1 − R δ we get withLemma 4.8 for δ ց 0 (4.25) MˆT 0 ψ ′ |u − v| dt dv g ≥ˆ∂ MˆT 0 ψ sgn(T u − T v) f (T u) − f (T v), N g dt dv g .…”