By David R. Morrison, Janos Kolla Summer Research Institute on Algebraic Geometry

**Read or Download Algebraic Geometry Santa Cruz 1995: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz (Proceedings of Symposia in Pure Mathematics) (Pt. 2) PDF**

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**Additional info for Algebraic Geometry Santa Cruz 1995: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz (Proceedings of Symposia in Pure Mathematics) (Pt. 2)**

**Sample text**

Hence, the ρk are bounded. 74). Then F (x, uk )dx → I F (x, u0 ) dx. 2) and satisﬁes u0 2 H − (f (u0 ), u0 ) = 0. 112), uk 2 H = (f (uk ), uk ) + o(1) → (f (u0 ), u0 ) = u0 2 H, showing that uk converges to u0 in H. Hence, the PS condition holds, and the proof is complete. 38. 2). Proof. We have F (x, u) dx ≤ u 2 2 + B, I where W (x) dx. B= I Thus, 2G(u) ≥ u 2 H − u 2 −B = u 2 − B ≥ −B. Consequently, α = inf G > −∞. 111). 37 to reach the desired conclusion. 14 Exercises 1. 2). 2. 10) by |h|. 3.

Proof. 46) for u ∈ C 1 (I). We note that x u(x)2 − u(x )2 = x 2u (y)u(y) dy ≤ x ≤ |2u (y)u(y)| dy x (|u (y)|2 + |u(y)|2 ) dy = u 2 H, x, x ∈ I. I We pick x ∈ I so that 2πu(x )2 = u(y)2 dy. I This can be done by the mean value theorem for integrals. Hence, u(x)2 ≤ 1 u 2π 2 + u 2 H ≤ 1+ 1 2π u 2 H. 46) for u ∈ C 1 (I). Now, I claim that it holds for any u ∈ H. To see this, let u be any function in H. Then there is a sequence {uk (x)} of functions in C 1 (I) such that uk − u H → 0. 46) for functions in C 1 (I), |uj (x) − uk (x)| ≤ K uj − uk H → 0, j, k → ∞.

8), there is a sequence {un } ⊂ C ∞ (I) such that un − u → 0, un − h → 0. Thus, x un (x) − un (0) = un (t)dt. 11, un converges to u uniformly in I. Taking the limit, we have x u(x) − u(0) = h(t)dt. 0 16 Extrema This shows that u is diﬀerentiable at each point and its derivative equals h. 14. 48) then u ∈ H and u = (u ) = u − f. In particular, u is continuous in I and is the derivative of u in the usual sense. Proof. 3), u satisﬁes (u , v ) = −(u − f, v), v ∈ C 1 (I). 49) This means that u has a weak derivative equal to u − f.