Download Analysis and Geometry: MIMS-GGTM, Tunis, Tunisia, March by Ali Baklouti, Aziz El Kacimi, Sadok Kallel, Nordine Mir PDF

By Ali Baklouti, Aziz El Kacimi, Sadok Kallel, Nordine Mir

This e-book contains chosen papers offered on the MIMS (Mediterranean Institute for the Mathematical Sciences) - GGTM (Geometry and Topology Grouping for the Maghreb) convention, held in reminiscence of Mohammed Salah Baouendi, a most famed determine within the box of a number of advanced variables, who kicked the bucket in 2011. All examine articles have been written through best specialists, a few of whom are prize winners within the fields of complicated geometry, algebraic geometry and research. The ebook bargains a important source for all researchers attracted to fresh advancements in research and geometry.

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Additional info for Analysis and Geometry: MIMS-GGTM, Tunis, Tunisia, March 2014. In Honour of Mohammed Salah Baouendi

Example text

6) 50 H. 6), we infer that 2N L 1 (R ) gk,λ 1, which ends the proof of the right hand side of the assertion. Once observed that the function u can be recast under the form u = (log |D|)−k gλ (log |D|)k u , we end the proof of the result. 12. Taking advantage of the fact that the spectrum of the function u is included in eλ C with λ ≥ 1, we find that u(ξ) = φ(λ−1 log |ξ|) u(ξ), where φ is a function of D(R) chosen as above. Therefore (log |D|)k u = (log |D|)k gλ u , where gλ (ξ) = φ(λ−1 log |ξ|).

Suffridge, Boundary behavior of rational proper maps. Duke Math. J. 60(1), 135–138 (1990) 5. P. D’Angelo, Several Complex Variables and the Geometry of Real Hypersurfaces (CRC Press, Boca Raton, 1992) 6. P. D’Angelo, Proper holomorphic maps between balls of different dimensions. Mich. Math. J. 35(1), 83–90 (1988) 7. P. D’Angelo, Proper holomorphic mappings, positivity conditions, and isometric imbedding. J. Korean Math. Soc. 40, 341–371 (2003). May 8. P. D’Angelo, Hermitian analogues of Hilbert’s 17th problem.

Hence (10e) maps the sphere to Q(2, 1). The map (10d) can be found in a similar manner as for (10f), except that we replace (16) with |z 3 f 1 (w)|2 + | f 2 (w)|2 = |z f 3 (w)|2 + | f 4 (w)|2 . (21) Now we must replace |z|6 with (1 − |w|2 )3 . Doing so leads to the following analogue of (17). √ | f 1 (w)|2 + | 3w 2 f 1 (w)|2 + | f 2 (w)|2 + |w f 3 (w)|2 = √ | 3w f 1 (w)|2 + |w 3 f 1 (w)|2 + | f 3 (w)|2 + | f 4 (w)|2 . (22) We must find a 4-by-4 unitary matrix, but solving the equations is possible.

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