ACCEPTED MANUSCRIPT Lemma1Frtu.e】∈.2eae S(u.v)-F(w.o)lls 3(u(1-u)+(1-) PROOF.By direct computation.we obtain: Ib:o-5ll-(boo -2b:n+bzo)+b.-2ba+ban) +(boa-2b1+ba)+(bip-2b1 ba)] H(b03-2b1+b2)+(ban-2b21+) +a-n+bo+-g+h训 l(b:a-2ba2+baz)+(bai-2bz+ba3 )(h+bea) y-5 011d「Bgel ≤路aa 1221 e 1221 0110 B(c) Subetitung te aboveqtynto Eq.(7).webave l(u.v)-F(u.o)ll s 3(u(1-u)+o(1-v))MACCEPTED MANUSCRIPT ACCEPTED MANUSCRIPT Lemma 1 For (u, v) ∈ Ω, we have S(u, v) − F(u, v) ≤ 3(u(1 − u) + v(1 − v))Mb . PROOF. By direct computation, we obtain: b1,0 − b1,0 = 2 3 (b0,0 − 2b1,0 + b2,0) + 1 3 (b1,0 − 2b2,0 + b3,0) ≤ 1 3 (2(b0,0 − 2b1,0 + b2,0) + (b1,0 − 2b2,0 + b3,0)) ≤ Mb , and b1,1 − b1,1 = 2 9 [(b0,0 − 2b1,0 + b2,0)+(b0,0 − 2b0,1 + b0,2)] + 1 4 [(b0,1 − 2b1,1 + b2,1)+(b1,0 − 2b1,1 + b1,2)] + 1 6 [(b0,2 − 2b1,2 + b2,2)+(b2,0 − 2b2,1 + b2,2)] + 7 36[(b1,0 − 2b2,0 + b3,0)+(b0,1 − 2b0,2 + b0,3)] + 1 12[(b1,2 − 2b2,2 + b3,2)+(b2,1 − 2b2,2 + b2,3)] + 1 36[(b3,0 − 2b3,1 + b3,2)+(b0,3 − 2b1,3 + b2,3)] + 1 18[(b3,1 − 2b3,2 + b3,3)+(b1,3 − 2b2,3 + b3,3)] ≤ 2Mb . By symmetry, it follows that: 3 i=0 3 j=0 bi,j − bi,jB3 i (u)B3 j (v) ≤ Mb  B3 0(u) B3 1(u) B3 2(u) B3 3 (u)  ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0110 1221 1221 0110 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ B3 0(v) B3 1(v) B3 2(v) B3 3(v) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = Mb(B3 1(u) + B3 2(u) + B3 1(v) + B3 2 (v)) = 3(u(1 − u) + v(1 − v))Mb . Substituting the above inequality into Eq. (7), we have: S(u, v) − F(u, v) ≤ 3(u(1 − u) + v(1 − v))Mb . 9