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By Thomas Craig

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Example text

32) with λ = 0. With this choice of λ we have x λ+ = xλ− = x; ˆ thus, the terms with u 0 drop out while the terms involving L can be estimated as before. 32) is zero, while the remaining terms can be estimated as before. Finally, statement (iii) follows from (ii) taking C0 = 0. , L is not twice differentiable and u 0 is not semiconcave, then u may fail to be semiconcave, as shown by the next example. 3 Consider a one-dimensional problem with lagrangian and initial cost given respectively by L(q) = q2 + |q|, 2 u 0 (x) = |x| .

From the deﬁnition it follows that, for any x ∈ A, D − (−u)(x) = −D + u(x) . 2 Let A = R and let u(x) = |x|. Then it is easily seen that D + u(0) = ∅ whereas D − u(0) = [−1, 1]. √ Let A = R and let u(x) = |x|. Then, D + u(0) = ∅ whereas D − u(0) = R. Let A = R2 and u(x, y) = |x| − |y|. Then, D + u(0, 0) = D − u(0, 0) = ∅. 3 Let x ∈ A and θ ∈ Rn . The upper and lower Dini derivatives of u at x in the direction θ are deﬁned as ∂ + u(x, θ ) = lim sup u(x + hθ ) − u(x) h ∂ − u(x, θ ) = u(x + hθ ) − u(x) , h h→0+ ,θ →θ and lim inf h→0+ ,θ →θ respectively.

Proof — Let us set γ (s) = x + s − t1 (y − x), t2 − t 1 s ∈ [t1 , t2 ]. 17), d w(s, γ (s)) = ∂t w + ∇w · γ˙ ds w|γ˙ |2 |∇w|2 ≥ ∂t w − − w 4 nw w|γ˙ |2 ≥− − . 2t 4 It follows that ln w(t2 , y) w(t1 , x) = t2 t1 ≥ t2 t1 d ln w(s, γ (s)) ds ds − n |y − x|2 − 2t 4(t2 − t1 )2 n t2 = − ln 2 t1 − ds |y − x|2 , 4(t2 − t1 ) which proves our statement. 18) holds in a much more general context than the one considered here; our aim was to point out the relationship with semiconcavity induced by the Hopf–Cole transformation.