By Masaaki Yoshida

This quantity concentrates on hypergeometric features, ranging from the ordinary point of equivalence kinfolk to the exponential functionality. subject matters coated contain: the configuration house of 4 issues at the projective line; elliptic curves; elliptic modular services and the theta services; areas of six issues within the projective undeniable; K3 surfaces; and theta services in 4 variables.

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**Extra resources for Hyper Geometric Functions, My Love: Modular Interpretations of Configuration Spaces**

**Example text**

Since x0 ∈ Ω is arbitrary and xε → x0 as ε → 0, it follows that the set of x’s for which J 2,+ u(x) = ∅ is dense in Ω. 2,+ u(x), where X (h) Fix p ∈ Rn and suppose {( p, X m )}∞ m → X as 1 ⊆ J m → ∞. Fix ε > 0 and m(ε) ∈ N such that |X m(ε) − X | ≤ ε. 19) 28 2 Second Definitions and Basic Analytic Properties of the Notions as δ → 0. 19) and then letting ε → 0, we obtain lim sup z→0 u(z + x) − u(x) − p · z − 21 X : z ⊗ z |z|2 ≤ 0, which says that ( p, X ) ∈ J 2,+ u(x). Consequently, (f) follows.

17), since the left hand side vanishes, for z = εw we have o(ε2 ) X − D 2 u(x) : w ⊗ w ≥ . ε2 By letting ε → 0, we see that all the eigenvalues of X − D 2 u(x) are non-negative. Hence, X − D 2 u(x) ≥ 0 in S(n). Consequently, (c) follows by setting A := X − D 2 u(x). (f) Follows from (e) be choosing A := t I , t > 0, where I the identity of Rn×n . Indeed, we have that the ray Du(x), D 2 u(x) + t I : t ≥0 is contained in J 2,+ u(x). Rn . (g) We simplify things by assuming in addition that u ∈ C 0 (Ω) and Ω The general case follows by covering Ω with an increasing sequence of bound open sets.

By definition, u ε is semiconvex with semiconvexity constant 1/ε. The conclusion now follows by invoking Alexandroff’s theorem. (h) Let ( p, X ) ∈ J 2,+ u ε (x). By Theorem 8 of Chap. 2, there is a ψ ∈ C 2 (Rn ) such that u ε − ψ ≤ (u ε − ψ)(x) with Dψ(x) = p and D 2 ψ(x) = X . Hence, for all z, y ∈ Ω and x ε ∈ X (ε), we have u(y) − |y − z|2 |y − z|2 − ψ(z) ≤ sup u(y) − − ψ(z) 2ε 2ε y∈Ω = (u ε − ψ)(z) ≤ (u ε − ψ)(x) = u(x ε ) − |x ε − x|2 − ψ(x). 2ε Hence, we have the inequality u(y) − |x ε − x|2 |y − z|2 − ψ(z) ≤ u(x ε ) − − ψ(x).