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1、泛函分析期末综合实验泛函分析期末综合实验实验课程名称实验课程名称 泛函分析综合实验泛函分析综合实验 学院学院 数学与统计学院数学与统计学院 年级年级 2009 专业班专业班 数学与应用数学数学与应用数学 学学 生生 姓姓 名名 王王 磊磊 学学 号号 20092237 开开 课课 时时 间间 2011 至至 2012 学年第学年第 2 学期学期ContentsContents .1 Mathematical Concepts.1 1Q1 it is called the quotient topology induced by .p i.Analysis: A topology associa
2、ted with other structure such as algebra structure always produce unusual properties we are especially interested in. The topology is of course defined by letting it consist of those subsets U of A such that is open in X. It is easy to check that is a topology. The sets and 1( )pU A are open because
3、 and . The other two conditions 1()p 1( )pAX follow from the equations: 11()() JJpUpUUU1111()()iiiipUpUIIii.Examples: Let be the map of the real line R onto the three-point set pdefined by , , Aa b c 0,( )0,0.aif xp xbif xcif x b)Let be a quotient map; let A be subspace of X that is saturated with :
4、pXArespect; let be the map obtained by restricting.p:( )qAp Ap 1.If A is either open or closed in X, then q is a quotient map. 2.If is either an open map or a closed map, then is a quotient mappq i.Analysis: 1.Step1. We verify first the following two equations:if ;11( )( )qVpV( )Vp Aif ;()( )( )p UA
5、p Up AUXTo check the first equation, we note that since and A is ( )Vp Asaturated, is contained in A. It follows that both and 1( )pV1( )pVequal all points of A that are mapped by into V. To check the 1( )qVp second equation, we note that for any two subsets U and A of X, we have the inclusion ()( )
6、( )p UVp Up ATo prove the reverse inclusion, suppose , for ( )( )yp up auUand . Since A is saturated, A contains the set , so that in aA1( ( )pp aparticular A contains u. Then , where .( )yp uuUAStep2. Now suppose A is open or is open. Given the subset V of , p( )p Awe assume that is open in A and s
7、how that V is open in .1( )qV( )p ASuppose first that A is open. Since is open in A and A is open in X, 1( )qVthe set is open in X. Since , the latter set is open 1( )qV11( )( )qVpVin X, so that V is open in X. Since and is open 11( )( )qVpV1( )qVin A, we have for some set U open in X. Now 1( )pVUAb
8、ecause is surjective; then 1( )p pVVp1( )()( )( )Vp pVp UAp Up AThe set is open in Y because is an open map;hece V is open in ( )p Up ( )p AMathematics in papersThe function of a language is to employ logical structures and real-world references to convey, process, assign meaning. Whenever we want d
9、escribe an mathematical idea, we use mathematical language of which aim to resolve or eliminate ambiguous and fallacious reasoning. This is great crucial important in papers filled with symbols, equations and examples.1Example 1.1. Problem. Let the sequence an be defined by the recurrence formula a
10、= 0, and a = a + 2, for n 2. Prove that a converges.a)Proof: First, we want give a bound of a:and the equation holds if 2na 0na Suppose we have 2na 12222nnaaWe have by induction.2na Hence, 122nnnnnaaaaa The conclusion is trival. b)Applying logical structure into our proof One of the most famous logi
11、cal structures is perhaps Syllogism. For example: If A THEN B SINCE C IS A HENCE B There exist even more complicated and useful logical structures. But, what do we mean by usefull ? In my opinion, this term refer to accurate. 2Example 1.2.Problem. For , prove that the alternating group is the only 5
12、n nAnormal subgroup of the symmetric group .nS Last semester, I have done a perfect proof on it. Q:2nnSAnnAS, is a simple group by the theorem on finite simple and their classification.5n nASuppose there is a nHSHence eHA a)We want to show .HAA Applying ,provided that,nnnHSAS :AAHH ( )aHa HaHhaAHH ,
13、 we have KerAHHAA By Fundamental Homorphism Theorems. Hcnce, ,HAA Hence HAeb)Similarly, by the function defined above, employing and .nnASnnHS /ImA Ker/ /AeHAA /AAHHBy Theorem of Lanrange, we have 1 2nnnSA HAHSHHence, , .21H HeCombine (a) and (b), we conclude that the alternating group is the only n
14、ormal nAsubgroup of the symmetric group nS3Prove the inequalities: 11112122,1123nnnnKHint:Use integrals a)Proof: i.(Integrals)1212(1)nndxnnnxQ1212(1)11nndxnnnnxThe conclusion hence is trivial. ii.(Fundamental transformation)Note 1222(1)21nnnnnnAnd 1222(1)21nnnnnn The conclusion is trivial.Mathematical in ApplicationWe want to use special mathematical models to illustrate how to apply mathematics into application. Optimizing parking lots is a good example. In recent years, along with Chinese urban economic development, urbanization an
