By Otto Schreier, Emanuel Sperner

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**Sample text**

K, X = ∂K is (strictly) convex at qi = xi with respect to the unit normal νi pointing into the interior of Ω. 34) Mi : Πi −→ Πi , i = 1, . . , k, are non-negative semi-definite (resp. 35) i = 2, 3, . . , k + 1. In particular, det dJγ (q0 ) = 0. Proof: Since ∂K is (strictly) convex at qi , it follows from the definitions of ψ˜i (cf. 3) that they are non-negatively semi-definite (resp. positive definite) symmetric linear maps. 34) to define the maps Mi inductively; the definition is correct and Mi ≥ 0 (resp.

K and t, m = 1, . . , n − 1. Having fixed j, there are three possibilities for i. Case 1. i ∈ / Ij ∪ {j}. 11) is 0. Case 2. i ∈ Ij . 10) implies ∂2G (t) (m) ∂uj ∂ui ∂ϕi (0) = −aij (t) ∂uj ∂ϕj +aji (t) ∂uj ∂ϕi (0), (m) (0) ∂ui ∂ϕi (0), vji (m) ∂ui (0), vji . Case 3. i = j. Then ∂2G (t) (m) ∂uj ∂uj (0) = vji , i∈Ij ∂ 2 ϕj (0) (t) (m) ∂uj ∂uj ∂ϕj aji + (t) ∂uj i∈Ij − ∂ϕj aji (t) ∂uj i∈Ij (0), ∂ϕj (m) (0) ∂uj (0), vji ∂ϕj (m) ∂uj (t) (0), vji . Fix an arbitrary vector ξ = (ξj )1≤j≤k,1≤t≤n−1 ∈ (Rn−1 )k .

Uj ) ∈ Rn−1 , vij = aij (qi − qj ). Clearly, aij = aji > 0 and vji = −vij ∈ Sn−1 . For j = 1, . . , k, t = 1, . . , n − 1 and u sufficiently close to 0 we have ∂G (t) ∂uj (u) = i∈Ij ϕj (uj ) − ϕi (ui ) ∂ϕj , (u ) . 3, 0 is a critical point of G. We will prove that the second fundamental form of G at 0 is non-negative defined. 11) ∂uj ∂ui for i, j = 1, . . , k and t, m = 1, . . , n − 1. Having fixed j, there are three possibilities for i. Case 1. i ∈ / Ij ∪ {j}. 11) is 0. Case 2. i ∈ Ij .