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3) be in force. e. in Q p , R 2 . 9) Proof. Define R R + n+1 , n = 0, 1, . . , 2 2 and construct the family of nested and shrinking cylinders Q a20 Rnp , Rn . Consider piecewise smooth cutoff functions 0 ≤ ζn ≤ 1, defined in these cylinders and satisfying the following set of assumptions: Rn = ζn = 1 a0 p 2 Rn+1 , Rn+1 in Q |∇ζn | ≤ ; ζn = 0 on ∂p Q 2n+1 ; R 0 ≤ (ζn )t ≤ a0 p 2 Rn , Rn ; 2p(n+1) a0 p . 6) for the functions (u − kn )+ , with ω ω kn = µ+ − λ+1 − λ+1+n , n = 0, 1, . . , 2 2 in the cylinders Q a20 Rnp , Rn , and with ζ = ζn , reads 0 (u − kn )2+ ζnp dx + sup − a0 2 p Rn

16), ω 4 µ− + > ω 2n+2 ω 4 ω = 2n−1 2n+1 in S. Therefore, 2 ≥ ln 2n−1 2 = (n − 1)2 (ln 2)2 in S. 19), we get x ∈ K R : u(x, t) < µ− + 4 ω 2n+2 ≤C n 2λ(p−2) K R , 4 (n − 1)2 for all t ∈ (−tˆ, 0), and to prove the lemma we choose s∗ = n + 2 with n>1+ 2C λ(p−2) 2 . 4 Reduction of the Oscillation We now state the main result in the context of the first alternative. 4. 3) is in force. e. in Q t, R 8 . Proof. 8) holds, let Rn = R R + n+3 , 8 2 n = 0, 1, . . 15). Take piecewise smooth cutoff functions 0 < ζn (x) ≤ 1, independent of t, defined in KRn and satisfying |∇ζn | ≤ ζn = 1 in KRn+1 ; 2n+4 .

3) is in force. Given ν∗ ∈ (0, 1), there exists s∗ ∈ N, depending only on the data, such that x ∈ K R : u(x, t) < µ− + 4 ω 2s∗ ≤ ν∗ K R ∀t ∈ (−t, 0). , 4 Proof. 8) applied to the function (u − k)− in the cylinder Q(t, R2 ), with the choices k = µ− + ω 4 and c= ω 2n+2 , where n ∈ N will be chosen later. In this cylinder, we have − = ess sup k − u ≤ Hu,k u − µ− − Q(t, R 2 ) − If Hu,k ≤ ω 8, ω 4 ≤ − ω . 16) − the result is trivial for the choice s∗ = 3. 3 that the logarithmic function ψ − (u) is defined in the whole of Q(t, R2 ) and it is given by ψ −H − ,k, { u,k ⎧ ⎪ ⎪ ⎪ ⎨ ln ω 2n+2 } (u) = ⎪ ⎪ ⎪ ⎩ − Hu,k − Hu,k +u−k+ ω if u < k − ω 2n+2 if u ≥ k − ω 2n+2 .