By Michael Bildhauer

The writer emphasizes a non-uniform ellipticity situation because the major method of regularity concept for options of convex variational issues of forms of non-standard progress conditions.

This quantity first makes a speciality of elliptic variational issues of linear development stipulations. the following the proposal of a "solution" isn't visible and the viewpoint should be replaced a number of instances in an effort to get a few deeper perception. Then the smoothness houses of ideas to convex anisotropic variational issues of superlinear development are studied. inspite of the elemental variations, a non-uniform ellipticity situation serves because the major software in the direction of a unified view of the regularity conception for either sorts of problems.

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Additional resources for Convex Variational Problems: Linear, Nearly Linear and Anisotropic Growth Conditions

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15. 1 on f the dual problem (P ∗ ) admits a unique solution σ. Proof. We fix a weak L2 -limit σ of the δ-approximation. 6 this limit is known to be a solution of (P ∗ ). Suppose by contradiction that the dual problem admits a second maximizer σ ˜ = σ. Now, Ω is divided into four parts, Ω = Σ 0 ∪ Σ 1 ∪ Σ2 ∪ Σ3 , where we have by definition σ(x) = σ ˜ (x) for x ∈ Σ0 , σ(x) = σ ˜ (x) , σ(x) ∈ U , σ ˜ (x) ∈ U for x ∈ Σ1 , σ(x) = σ ˜ (x) , σ(x) ∈ U , σ ˜ (x) ∈ ∂U for x ∈ Σ2 , |Σ3 | = 0 . 4, ii). 1 it is also seen to be strictly convex on U .

25 Since a lower semicontinuous convex function is continuous on straight lines to the boundary and since ε is not depending on m, we may pass in (21) to the limit m → ∞ and obtain for all x ∈ Σ2 f ∗ κ(x) = f ∗ σ(x) + σ ˜ (x) 2 < 1 ∗ 1 σ (x)) . f (σ(x)) + f ∗ (˜ 2 2 The same computations as outlined in (18) give (19), we have shown Ω = Σ0 ∪ (Ω − Σ0 ) with |Ω − Σ0 | = 0 , and the theorem is proved by the definition of Σ0 . 3 Partial C 1,α - and C 0,α -regularity, respectively, for generalized minimizers and for the dual solution In the general setting of vector-valued variational problems with linear growth full regularity cannot be expected – even if we additionally assume that D2 f (Z) > 0 holds for any matrix Z ∈ RnN .

We 3 Variational integrands with (s, μ, q)-growth 50 like to remark that the exponent −μ/2 occurring on the left-hand side of iii) is the best possible one. This is evident if we consider Y parallel to Z. The second inequality of iii) again is immediate. Next we are going to prove iv): observing 1 |D2 Φ(Z)| = sup D2 Φ(Z)(Y, Y ) ≤ 2 |Y |=1 1 + t2 |Z|2 −μ 2 dt 0 we get |Z| |D2 Φ(Z)| |Z|2 ≤ 2 |Z| 1 + s2 −μ 2 ∞ ds ≤ 2 |Z| 0 1 + s2 −μ 2 ds , 0 the last integral being finite on account of μ > 1. 10. 9 provides a regular class of integrands with linear growth (with some appropriate choice of μ > 1).

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