By Charles L. Dodgson

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Dz dz Then, the stress components are calculated by the following formulae: (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) σx(j ) = 2Re[(μ1 )2 Φ1 (z1 ) + (μ2 )2 Φ2 (z2 )], (j ) σy(j ) = 2Re[Φ1 (z1 ) + Φ2 (z2 )], (j ) (j ) (j ) (j ) (j ) (j ) (j ) = −2Re[μ1 Φ1 (z1 ) + μ2 Φ2 (z2 )]. τxy (52) The displacements components become (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) U (j ) = 2Re[p1 ϕ1 (z1 ) + p2 ϕ2 (z2 )], V (j ) = 2Re[q1 ϕ1 (z1 ) + q2 ϕ2 (z2 )], (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (j ) (53) (j ) (j ) (j ) (j ) where pk = c11 (μk )2 + c12 − c16 μk , μk qk = c11 (μk )2 + c22 − c26 μk .

Clearly, (25) (with y = 0) yields g(px) + g(px) − g(x) − g(0) ≤ ε x ∈ V. So, (27) implies that L(px, px) + L(px, px) − L(x, x) (29) ≤ L(px, px) + g(0) − g(px) + L(px, px) + g(0) − g(px) + g(x) − L(x, x) − g(0) + g(px) + g(px) − g(x) − g(0) ≤ 13ε x ∈ V. Since b is biadditive and it is very easy to check that −2 L(px, px) = L(px, px) + L(px, px) − L(x, x) x ∈ V, from (29), we get 2k 2 L(px, px) = L(pkx, pkx) + L(pkx, pkx) − L(kx, kx) ≤ 13ε x ∈ V , k ∈ N, (30) 52 J. Brzd¸ek which means that (22) holds.

Take u ∈ V . Analogously as before, we deduce that there is v ∈ V with pv = −pu. Clearly p(x + u) + p(y + v) = px + py x, y ∈ V . Hence, replacing x by x + u and y by y + v in (16), we have g(x + u + z) − g(x + u) − g(px + py + pz + pw) + g(px + py) + g(y + w + v) − g(y + v) ≤ 2ε (17) x, y ∈ V . It is easily seen that (16) and (17) imply g(x + u + z)− g(x + u) − g(x + z) + g(x) + g(y + w + v) − g(y + w) − g(y + v) + g(y) ≤ g(px + py + pz + pw) − g(px + py) (18) 48 J. Brzd¸ek − g(x + z) − g(y + w) + g(x) + g(y) + g(px + py + pz + pw) − g(px + py) − g(x + u + z) − g(y + w + v) + g(x + u) + g(y + v) ≤ 4ε x, y ∈ V , which with x replaced by x + t yields g(x + t + u + z) − g(x + t + u) − g(x + t + z) + g(x + t) + g(y + w + v) − g(y + w) − g(y + v) + g(y) ≤ 4ε t, x, y ∈ V .