2015南京大学考博真题泛函分析
2015年南京大学考博英语真题及详解【圣才出品】
2015年南京大学考博英语真题及详解Part I Vocabulary and Structure (20%)Directions: There are 20 incomplete sentences in this part. For each sentence there are four choices marked A, B, C and D respectively. Choose the ONE thatbest complete the sentences. Then blacken your answer in thecorresponding letter on your Answer Sheet with a single line through thecenter.1. The ambassador was accused of having _____ on domestic affairs.A. trespassedB. encroachedC. entrenchedD. invaded【答案】A【解析】句意：大使被指控干涉国内事务。
该题为近义词辨析，选项中的四个词均有侵犯的意思。
trespass为不及物动词，一般与介词on搭配使用，意思为“擅自进入；侵犯，侵害；打扰”，强调非法侵入，符合题意。
encroach意思为“蚕食；侵占”，强调侵入并占领。
entrench意思为“用壕沟围绕或保护…；牢固地确立…”，强调在某处站稳脚跟。
invade一般用作及物动词，指“侵入，攻占；侵袭”。
2. The goal is to use crops, weeds, and even animal waste _____ the petroleum that fuels much of American manufacturing.A. in terms ofB. in favor ofC. in spite ofD. in place of【答案】D【解析】句意：目标是使用农作物、杂草甚至动物粪便来代替石油为美国制造业提供能源。
泛函分析答案
泛函分析答案泛函分析题1_3列紧集p191.3.1 在完备的度量空间中，求证：为了子集A是列紧的，其充分必要条件是对?ε > 0，存在A的列紧的ε网．证明：(1) 若子集A是列紧的，由Hausdorff定理，ε > 0，存在A的有限ε网N．而有限集是列紧的，故存在A的列紧的ε网N．(2) 若?ε > 0，存在A的列紧的ε/2网B．因B列紧，由Hausdorff定理，存在B的有限ε/2网C．因C ?B ?A，故C为A的有限ε网．因空间是完备的，再用Hausdorff定理，知A是列紧的．1.3.2 在度量空间中，求证：紧集上的连续函数必是有界的，并且能达到它的上、下确界．证明：设(X, ρ)是度量空间，D是紧子集，f : D→ 是连续函数．(1) 若f无上界，则?n∈ +，存在x n∈D，使得f (x n) > 1/n．因D是紧集，故D是自列紧的．所以{x n}存在收敛子列x n(k) →x0∈D (k→∞)．由f的连续性，f (x n(k))→f (x0) (k→∞)．但由f (x n) > 1/n知f (x n)→ +∞(n→∞)，所以f (x n(k))→ +∞ (k→∞)，矛盾．故f有上界．同理，故f有下界．(2) 设M = sup x∈D f(x)，则?n∈ +，存在y n∈D，使得f (y n) > M- 1/n．{y n}存在子列y n(k) →y0∈D (k→∞)．因此f ( y0 ) ≥M．而根据M的定义，又有f ( y0 ) ≤M．所以f ( y0 ) = M．因此f能达到它的上确界．同理，f能达到它的下确界．1.3.3 在度量空间中，求证：完全有界的集合是有界的，并通过考虑l 2的子集E = {e k }k≥ 1，其中e k = { 0, 0, ..., 1, 0, ... } (只是第k 个坐标为1，其余都是0 )，来说明一个集合可以是有界的但不完全有界的．证明：(1) 若A是度量空间(X, ρ)中的完全有界集．则存在A的有限1-网N = { x0, x1, x2, ..., x n }．令R = ∑1 ≤j≤nρ(x0, x j) + 1．则?x∈A，存在某个j使得0 ≤j≤n，且ρ(x, x j) < 1．因此，ρ(x, x0) ≤ρ(x, x j) + ρ(x j, x0) ≤ 1 + ∑1 ≤j≤nρ(x0, x j) = R．所以A是度量空间(X, ρ)中的有界集．(2) 注意到ρ(e k , e j) = 21/2 ( ?k ≠ j )，故E中任意点列都不是Cauchy列．所以，E中任意点列都没有收敛子列(否则，该收敛子列就是Cauchy列，矛盾)．因此，E不是列紧集．由l 2是完备的，以及Hausdorff定理，知E不是全有界集．但E显然是有界集．1.3.4 设(X, ρ)是度量空间，F1, F2是它的两个紧子集，求证：?x i ∈F i( i = 1, 2)，使得ρ(F1, F2) = ρ(x1, x2)．其中ρ(F1, F2) = inf {ρ(x, y) | x∈F1, y∈F2 }证明：由ρ(F1, F2)的定义，?n∈ +，?x i(n)∈F i( i = 1, 2)，使得ρ(x1(n), x2(n)) < ρ(F1, F2) + 1/n．因F1, F2紧，故不妨假设{x1(n)}, {x2(n)}都是收敛列．设它们的极限分别为x1, x2，则ρ(x1, x2) ≤ρ(F1, F2)．因此ρ(F1, F2) = ρ(x1, x2)．1.3.5 设M是C[a, b]中的有界集，求证集合{F(x) =?[a, x]f(t) dt | f∈M }是列紧集．证明：设A = {F(x) =?[a, x]f(t) dt | f∈M }．由M有界，故存在K > 0，使得?f∈M，ρ( f, 0) ≤K．先证明A是一致有界的和等度连续的．F∈A，存在f∈M，使得F(x) =?[a, x]f(t) dt．由于ρ(F, 0) = max x∈[a,b] | F(x) | = max x∈[a, b] | ?[a, x]f(t) dt |≤ max x∈[a, b] | f(t) | · (b -a ) = ρ( f, 0) · (b -a ) ≤K (b -a )．故A是一致有界的．ε > 0，?s, t∈[a, b]，当| s-t| < ε/K时，F∈A，存在f∈M，使得F(x) =?[a, x]f(u) du．| F(s) -F(t) | = | ?[s, t]f(u) du | ≤ max u∈[a, b] | f(u) | · | s -t |= ρ( f, 0) · | s -t | ≤K · (ε/K) = ε．故A是等度连续的．由Arzela-Ascoli定理，A是列紧集．1.3.6 设E = {sin nt}n≥ 1，求证：E在C[0, π]中不是列紧的．证明：显然E是一致有界的．根据Arzela-Ascoli定理，我们只要证明E不是等度连续的即可．我们的想法是找一个E中的点列f n，以及[0, π]中的两个点列s n 和t n，使得| s n -t n | → 0，但| f n(s n)-f n(t n)|不收敛于0．事实上，这是可以做到的，只要令f n (u) = sin (2n u)，s n = (π/2)(1 + 1/(2n))，t n = (π/2)(1 - 1/(2n))．则s n + t n = π；s n -t n = π/(2n)→ 0(n→∞)．因此，| f n(s n)-f n(t n)| = 2 | sin (2n s n) - sin (2n t n) |= 2 | sin (n (s n -t n)) cos (n (s n + t n)) |= 2 | sin (π/2) cos (n π) | = 2．所以，E不是等度连续的．进而，E在C[0, π]中不是列紧的．1.3.7 求证S空间的子集A是列紧的充要条件是：?n∈ +，?C n> 0，使得x = (ξ1, ξ2, ..., ξn, ...)∈A，都有| ξn | ≤C n( n = 1, 2, ...)．证明：(?) 设x k = (ξ1(k), ξ2(k), ..., ξn(k), ...) ( k = 1, 2, ... )是A中的点列．存在{x k}的子列{x1, k}使得其第1个坐标ξ1(1, k)收敛；存在{x1, k}的子列{x2, k}使得其第2个坐标ξ2(2, k)收敛；如此下去，得到一个{x k}的子列的序列，第( j +1)个子列是第j个子列的子列，且第j个子列的第j个坐标是收敛的．选取对角线构成的点列{x j, j}，则{x j, j}是{x k}的子列，且每个坐标都收敛．根据习题1.2.1的证明可知，S空间的点列收敛的充要条件是坐标收敛．故{x j, j}是收敛点列．所以，A是列紧的．(?) 我们只要证明，?n∈ +，A中的点的第n个坐标所构成的集合是有界集．若不然，设A中的点的第N个坐标所构成的集合是无界的．则存在A中的点列x k = (ξ1(k), ξ2(k), ..., ξn(k), ...) ( k = 1, 2, ... )，使得| ξN(k) | > k．显然，{ ξN(k) }无收敛子列，故{ x k }也无收敛子列，这与A列紧相矛盾．这样就完成了必要性的证明．1.3.8 设(X, ρ)是度量空间，M是X中的列紧集，映射f : X →M满足ρ( f (x1), f (x2)) < ρ( x1, x2 )(?x1, x2∈M, x1≠x2)．求证：f在X中存在唯一的不动点．证明：(1) 首先证明cl(M)是紧集．为此只要证明cl(M)列紧即可．设{ x n }是cl(M)中的点列，则存在M中的点列{ y n }使得ρ( x n, y n) < 1/n．因M列紧，故{ y n }有收敛子列{ y n(k)}，设y n(k) →u∈cl(M)．显然{ x n(k)}也是收敛的，并且也收敛于u∈cl(M)．所以cl(M)是自列紧的，因而是紧集．(2) 令g(x) = ρ( x, f (x))，则g是X上的连续函数．事实上，由ρ( f (x1),f (x2)) < ρ( x1, x2 )可知f : X →M是连续的，因而g也连续．由习题1.3.2知存在x0∈cl(M)，使得g(x0) = inf {ρ( x, f (x)) | x∈cl(M) }．(3) 若g(x0) > 0，则ρ( x0, f (x0)) > 0，即x0≠f (x0)．故ρ( x0, f (x0)) = g(x0) ≤g( f (x0)) = ρ( f (x0), f ( f (x0))) < ρ( x0, f (x0))，矛盾．所以，必有g(x0) = 0，即ρ( x0, f (x0)) = 0，因此x0就是f的不动点．1.3.9 设(M, ρ)是一个紧距离空间，又E?C(M)，E中的函数一致有界并且满足下列的H?lder条件：| x(t1) -x(t2) | ≤Cρ(t1, t2)α(?x∈E，?t1, t2∈M )，其中0 < α≤ 1，C > 0．求证：E在C(M)中是列紧集．证明：由H?lder条件易知E是等度连续的．又E中的函数一致有界，由Arzela-Ascoli定理知E是C(M)中的列紧集．[第3节完] 泛函分析题1_4线性赋范空间p391.4.1 在2维空间 2中，对每一点z = (x, y)，令|| z ||1 = | x | + | y |；|| z ||2 = ( x 2 + y 2 )1/2；|| z ||3 = max(| x |, | y |)；|| z ||4 = ( x 4 + y 4 )1/4；(1) 求证|| · ||i( i = 1, 2, 3, 4 )都是 2的范数．(2) 画出( 2, || · ||i )( i = 1, 2, 3, 4 )各空间中单位球面图形．(3) 在 2中取定三点O = (0, 0)，A = (1, 0)，B= (0, 1)．试在上述四种不同的范数下求出?OAB三边的长度．证明：(1) 正定性和齐次性都是明显的，我们只证明三角不等式．设z = (x, y), w = (u, v)∈ 2，s = z + w= (x + u, y + v )，|| z||1 + || w||1 = (| x | + | y |) + (| u | + | v |) = (| x | + | u |) + (| y | + | v |)≥ | x + u | + | y + v | = || z+ w||1．( || z||2 + || w||2 )2 = ( ( x 2 + y 2 )1/2 + ( u 2 + v 2 )1/2 )2= ( x 2 + y 2 ) + ( u 2 + v 2 ) + 2(( x 2 + y 2 )( u 2 + v 2 ))1/2 ≥ ( x 2 + u 2 ) + ( y 2 + v 2 ) + 2( x u+ y v )= ( x + u )2 + ( y + v)2 = ( || z+ w||2 )2．故|| z||2 + || w||2 ≥ || z+ w||2．|| z||3 + || w||3 = max(| x |, | y |) + max(| u |, | v |)≥ max(| x | + | u |, | y | + | v |) ≥ max(| x + u |, | y + v |) = || z+ w||3．|| ·||4我没辙了，没找到简单的办法验证，权且用我们以前学的Minkowski不等式(离散的情况，用H?lder不等式的离散情况来证明)，可直接得到．(2) 不画图了，大家自己画吧．(3) OA = (1, 0)，OB = (0, 1)，AB = (- 1, 1)，直接计算它们的范数：|| OA||1 = 1，|| OB||1 = 1，|| AB||1 = 2；|| OA||2 = 1，|| OB||2 = 1，|| AB||2 = 21/2；|| OA||3 = 1，|| OB||3 = 1，|| AB||3 = 1；|| OA||4 = 1，|| OB||4 = 1，|| AB||4 = 21/4．1.4.2 设c[0, 1]表示(0, 1]上连续且有界的函数x(t)全体．?x∈c[0, 1]，令|| x || = sup{| x(t) | | 0 < t≤ 1}．求证：(1) || ·||是c[0, 1]空间上的范数．(2) l∞与c[0, 1]的一个子空间是等距同构的．证明：(1) 正定性和齐次性都是明显的，我们只证明三角不等式．|| x || = sup{| x(t) | | 0 < t≤ 1}．|| x || + || y || = sup{| x(t) | | 0 < t≤ 1} + sup{| y(t) | | 0 < t≤ 1}≥ sup{| x(t) + y(t) | 0 < t≤ 1} = || x + y ||．所以|| ·||是c[0, 1]空间上的范数．(2) 任意取定(0, 1]中的一个单调递减列{a k }，满足(i) a1 = 1；(ii) lim k→∞a k = 0．显然，在每个[a k + 1, a k]上为线性函数的f∈c[0, 1]是存在的．设X = { f∈c[0, 1] | f在每个[a k + 1, a k]上为线性函数}．容易验证X是c[0, 1]的子空间．定义? : X →l∞，f #? ( f ) = ( f (a1), f (a2), ...)．则? : X →l∞是线性双射，且|| ? ( f ) ||∞= sup k ≥ 1 | f (a k) | = sup0 < t≤ 1 { | f (t ) | } = || f ||．所以，? : X →l∞是等距同构．因此，l∞与c[0, 1]的一个子空间是等距同构的．1.4.3 在C1[a, b]中，令|| f ||1 = (?[a, b] ( | f(x) |2 + | f’(x) |2) dx )1/2 (?f∈C1[a, b])．(1) 求证：|| · ||1是C1[a, b]上的范数．(2) 问(C1[a, b], || · ||1)是否完备？证明：(1) 正定性和齐次性都是明显的，和前面的习题一样，只验证三角不等式．我们先来证明一个比较一般的结果：若线性空间X上的非负实值函数p, q都满足三角不等式：p(x) + p(y) ≥p(x +y)，q(x) + q(y) ≥q(x +y)，?x, y∈X；则函数h = ( p2 + q2 )1/2也满足三角不等式．事实上，?x, y∈X，由Minkowski不等式，我们有h(x) + h(y) = ( p(x)2 + q(x)2 )1/2 + ( p(y)2 + q(y)2 )1/2≥ (( p(x)+ p(y))2 + ( q(x) + q(y))2 )1/2 ≥ ( p(x + y)2 + q(x + y)2 )1/2 = h(x + y)．回到本题：若令p( f ) = (?[a, b] | f(x) |2dx )1/2，q( f ) = (?[a, b] | f’(x) |2dx )1/2，则( p( f ) + p( g ))2 = ((?[a, b] | f(x) |2dx )1/2 + (?[a, b] | g(x) |2dx )1/2)2= ?[a, b] | f(x) |2dx + 2(?[a, b] | f(x) |2dx )1/2 · (?[a, b] | g(x)|2dx )1/2 + ?[a, b] | g(x) |2dx≥?[a, b] | f(x)|2dx + 2 ?[a, b] | f(x) | · | g(x)| dx + ?[a, b] | g(x)|2dx = ?[a, b] ( | f(x) | + | g(x)| )2dx ≥?[a, b] ( | f(x) + g(x)| )2dx = ( p( f + g ))2．所以有p( f ) + p( g ) ≥p( f + g )．特别地，p( f’) + p( g’) ≥p( f’+ g’)，即q( f ) + q( g ) ≥q( f + g )．因此，线性空间C1[a, b]上的非负实值函数p, q都满足三角不等式．根据开始证明的结论，|| · ||1也满足三角不等式．所以，|| · ||1是C1[a, b]上的范数．(2) 在C1[- 1, 1]中，令f n(x) = (x2 + 1/n2 )1/2 ( ?x∈[- 1, 1] )．则f’n(x) = 2x (x2 + 1/n2 )-1/2 ( ?x∈[- 1, 1] )．显然，f n(x)几乎处处收敛于| x |，f’n(x)几乎处处收敛于2sign( x )．因此，f n(x)依测度收敛于| x |，f’n(x)依测度收敛于2sign( x )．则f’n(x) = 2x (x2 + 1/n2 )-1/2 ( ?x∈[- 1, 1] )．显然，f n(x)几乎处处收敛于| x |，f’n(x)几乎处处收敛于2sign( x )．因此，f n(x)依测度收敛于| x |，f’n(x)依测度收敛于2sign( x )．故在L2[- 1, 1]中，f n(x) → | x |，f’n(x) → 2sign( x )．因此，它们都是L2[- 1, 1]中的基本列，故[- 1, 1] | f n(x) -f m(x) |2 dx → 0(m, n→∞)；[- 1, 1] | f’n(x) -f m’(x) |2 dx → 0(m, n→∞)．故|| f n-f m ||1 = (?[- 1, 1] ( | f n(x) -f m(x) |2 + | f’n(x) -f m’(x) |2 ) dx )1/2→ 0 (m, n→∞)．即{ f n }是C1[- 1, 1]中的基本列．下面我们证明{ f n }不是C1[- 1, 1]中的收敛列．若不然，设{ f n }在C1[- 1, 1]中的收敛于f∈C1[- 1, 1]．因|| f n-f ||1 = (?[- 1, 1] ( | f n(x) -f(x) |2 + | f’n(x) -f’(x) |2 ) dx )1/2≥ (?[- 1, 1] | f n(x) -f(x) |2dx )1/2，故在L2[- 1, 1]中，f n(x) →f．而在前面已说明L2[- 1, 1]中，f n(x) → | x |；由L2[- 1, 1]中极限的唯一性以及f的连续性，知f(x) = | x |．这样就得到f?C1[- 1, 1]，矛盾．所以，{ f n }不是C1[- 1, 1]中的收敛列．这说明C1[- 1, 1]不是完备的．对一般的C1[a, b]，只要令f n(x) = (x - (a + b )/2)2 + 1/n2 )1/2( ?x∈[a, b] )就可以做同样的讨论，就可以证明C1[a, b]不是完备空间．1.4.4 在C[0, 1]中，对每个f∈C[0, 1]，令|| f ||1 = (?[0, 1] | f(x) |2dx )1/2，|| f ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2．求证：|| · ||1和|| · ||2是C[0, 1]中的两个等价范数．证明：(1) 在习题1.4.3的证明中已经包含了|| · ||1是C[0, 1]中的范数的证明．下面我们证明|| · ||2是C[0, 1]中的范数，我们仍然只要验证三角不等式．|| f ||2 + || g ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 + (?[0, 1] ( 1 + x) | g(x) |2dx )1/2= || (1 + x)1/2f(x) ||1 + || (1 + x)1/2g(x) ||1≥ || (1 + x)1/2f(x) + (1 + x)1/2g(x) ||1= || (1 + x)1/2 ( f(x) + g(x) ) ||1≥ (?[0, 1] (1 + x) | f(x) + g(x) |2dx )1/2= || f + g ||2．所以，|| · ||2也是C[0, 1]中的范数．(2) 我们来证明两个范数的等价性．?f∈C[0, 1]|| f ||1 = (?[0, 1] | f(x) |2dx )1/2 ≤ (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 = || f ||2，|| f ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 ≤ 2 (?[0, 1] | f(x) |2dx )1/2 = 2 || f ||1．因此两个范数等价．1.4.5 设BC[0, ∞)表示[0, ∞)上连续且有界的函数f(x)全体，对每个f ∈BC[0, ∞)及a > 0，定义|| f ||a = (?[0, ∞) e-ax | f(x) |2dx )1/2．(1) 求证|| ·||a是BC[0, ∞)上的范数．(2) 若a, b > 0，a≠b，求证|| ·||a与|| ·||b作为BC[0, ∞)上的范数是不等价的．证明：(1) 依然只验证三角不等式．|| f ||a + || g ||a = (?[0, ∞) e-ax | f(x) |2dx )1/2 + (?[0, ∞) e-ax | g(x) |2dx )1/2= || e-ax/2f(x)||L2 + || e-ax/2g(x)||L2≤ || e-ax/2f(x)+ e-ax/2g(x)||L2= || e-ax/2 ( f(x)+ g(x))||L2= (?[0, ∞) e-ax | f(x)+ g(x) |2dx )1/2= || f + g ||a，所以|| ·||a是BC[0, ∞)上的范数．(2) 设f n(x)为[n, +∞)上的特征函数．则f n∈BC[0, ∞)，且|| f n||a = (?[0, ∞) e-ax | f n(x) |2dx )1/2 = (?[n, ∞) e-ax dx )1/2 = ((1/a)e-an)1/2．同理，|| f n||b = ((1/b)e-bn)1/2．故若a < b，则|| f n||a/|| f n||b = (b/a)1/2e-(b -a)n/2→ +∞ (n→+∞)．因此|| ·||a与|| ·||b作为BC[0, ∞)上的范数是不等价的．1.4.6 设X1, X2是两个B*空间，x1∈X1和x2∈X2的序对(x1, x2)全体构成空间X = X1?X2，并赋予范数|| x || = max{ || x1 ||1, || x2 ||2 }，其中x = (x1, x2)，x1∈X1，x2∈X2，|| · ||1和|| ·||2分别是X1和X2的范数．求证：如果X1, X2是B空间，那么X也是B空间．证明：(1) 先验证|| · ||的三角不等式．设x = (x1, x2), y = (y1, y2)∈X1?X2，则|| x + y || = || (x1 + y1, x2 + y2) || = max{ || x1 + y1 ||1, || x2 + y2 ||2 }≤ max{ || x1 ||1 + || y1 ||1, || x2 ||2 + || y2 ||2 }≤ max{ || x1 ||1, || x2 ||2 } + max{ || y1 ||1, || y2 ||2 }= || (x1, x2) || + || (y1, y2) ||= || x || + || y ||，而|| · ||的正定性和齐次性是显然的，所以，|| · ||是X1?X2的范数．(2) 设X1, X2是B空间，我们来证明X也是B空间．设x(n) = (x1(n), x2(n))是X = X1?X2中的基本列，则|| x(n) -x(m) || = max{ || x1(n) -x1(m) ||1, || x2(n) -x2(m)||2 } ≥ || x1(n) -x1(m) ||1，故{x1(n)}是X1中的基本列，同理，{x2(n)}是X2中的基本列．因X1, X2是B空间，故{x1(n)}和{x2(n)}分别是X1, X2中的收敛列．设x1(n) →x1∈X1，x2(n) →x2∈X2，令x = (x1, x2)．则|| x(n) -x || = max{ || x1(n) -x1 ||1, || x2(n) -x2 ||2 }≤ || x1(n) -x1 ||1 + || x2(n) -x2 ||2→ 0 (n→∞)．所以，|| x(n) -x ||→ 0 (n→∞)．即{ x(n) }为X = X1?X2中的收敛列．所以X = X1?X2也是B空间．1.4.7 设X是B*空间．求证：X是B空间，必须且只须对?{x n}?X，∑n≥ 1 || x n || < +∞?∑n≥ 1x n 收敛．证明：(?) ?{x n}?X，记S n = ∑1 ≤j≤n x j，B n = ∑1 ≤j≤n || x n ||，则|| S n + p-S n || = || ∑1 ≤j≤n + p x j -∑1 ≤j≤n x j ||= || ∑n +1 ≤j≤n + p x j ||≤∑n +1 ≤j≤n + p || x j ||= B n + p-B n → 0，(n→∞)．故{ S n }为X中的Cauchy列．由X完备，故{ S n }为X中的收敛列，即∑n≥ 1x n 收敛．(?) 反证法．若(X, ρ)不完备，设(Y, d )为(X, ρ)的一个完备化．不妨设(X, ρ)是(Y, d )的子空间，则存在y∈Y \ X．因cl( X ) = Y，故?n∈ +，存在x n∈X，使得d(x n, y) < 1/2n．则ρ(x n, x m) = d(x n, x m) ≤d(x n, y) + d(x m, y) ≤ 1/2n+ 1/2m → 0，因此{x n}是X中的Cauchy列，但不是收敛列．令z n = x n+1-x n，S n = ∑1 ≤j≤n z j；则z n, S n∈X．因|| z n || = || x n+1-x n || = ρ(x n+1, x n) ≤d(x n+1, y) + d(x n+1, y) ≤ 1/2n+1+ 1/2n < 1/2n - 1，故∑n≥ 1 || z n || < +∞．而S n = ∑1 ≤j≤n z j = ∑1 ≤j≤n ( x j+1-x j ) = x n+1-x1；故∑n≥ 1z n 在中不收敛．矛盾．1.4.8 记[a, b]上次数不超过n的多项式全体为n．求证：?f(x)∈C[a, b]，存在P0(x)∈ n，使得max a ≤x≤b| f(x) –P0(x) | = min{ max a ≤x≤b| f(x) –P(x) | | P∈ n }．证明：注意到 n是B*空间C[a, b]中的n+1维子空间．{1, x, x2, ..., x n}是 n中的一个向量组，把它看成C[a, b]中的一个有限向量组．根据定理p35, 1.4.23，对任意?f(x)∈C[a, b]，存在最佳逼近系数{λ0, λ1, ..., λn}，使得|| f(x) –∑0 ≤j≤n λj x j || = min{ || f(x) –∑0 ≤j≤n a j x j || | (a0, a1, ..., a n)∈ n+1}．令P0(x) = ∑0 ≤j≤n λj x j 就得到要证明的结论．1.4.9 在 2中，对?x = (x1, x2)∈ 2，定义范数|| x || = max(| x1 |, | x2 |)，并设|| x0–λ e1 ||．e1 = (1, 0)，x0 = (0, 1)．求a∈ 适合|| x0–a e1 || = minλ∈并问这样的a是否唯一？请对结果作出几何解释．解：g(λ) = || x0–λ e1 || = || (0, 1) –λ(1, 0)|| = || (–λ, 1)|| = max(| λ |, 1) ≥ 1，故g(λ) 当| λ| ≤ 1时取得最小值1．所以a = 0满足要求．显然满足要求的a不是唯一的．从几何上看就是某线段上的点到某定点的距离都是1．1.4.10 求证范数的严格凸性等价于下列条件：|| x + y || = || x || + || y || ( ?x≠θ, y≠θ) ?x = c y ( c > 0)．证明：(?) 设范数是严格凸的，若x, y ≠θ满足|| x + y || = || x || + || y ||，事实上，我们总有|| (x/|| x ||) || = || (y/|| y ||) || = 1．因x, y ≠θ，故|| x || + || y || > 0，所以|| x + y || ≠ 0．于是|| x ||/|| x + y || + || y ||/|| x + y || = 1．假若x/|| x || ≠y/|| y ||，由严格凸性，得到|| (|| x ||/|| x + y ||)(x/|| x ||) + (|| y ||/|| x + y ||)(y/|| y ||) || < 1，即|| (( x + y )/|| x + y ||) || < 1，矛盾．因此必然有x/|| x || = y/|| y ||，即x = (|| x ||/|| y ||) y．(?) 设?x, y ≠θ，|| x + y || = || x || + || y ||蕴涵x = c y ( c > 0)．下面证明范数是严格凸的．设x≠y，且|| x || = || y || = 1，又设α, β∈(0, 1)，且α + β= 1．我们知道|| α x + β y || ≤ || α x || + || β y || = α || x || + β|| y || = α + β= 1．假若|| α x + β y || = 1，根据我们的条件，就得到α x = c (β y)，其中c > 0．那么，就有|| α x || = || c (β y) ||，而|| x || = || y || = 1，所以α= c β；故x = y，这就与x≠y相矛盾．所以必然有|| α x + β y || < 1，即范数是严格凸的．1.4.11 设X是线性赋范空间，函数? : X → 1称为凸的，如果不等式( λ x + (1 -λ) y ) ≤λ?( x ) + (1 -λ)?( y ) ( ? 0 ≤λ≤ 1)成立．求证凸函数的局部极小值必然是全空间的最小值．证明：设x0是凸函数?的一个局部极小点．如果存在x∈X，使得?( x ) < ?( x0)，则? t ∈(0, 1)，( t x + (1 -t ) x0) ≤t ?( x ) + (1 -t )?( x0) < t ?( x0) + (1 -t )?( x0) = ?( x0)．而对x0的任意邻域U，都存在t ∈(0, 1)，使得t x + (1 -t ) x0∈U．这就与x0是局部极小点相矛盾．因此?x∈X，都有?( x0) ≤?( x )，即x0是?的最小点．1.4.12 设(X, || · ||)是一线性赋范空间，M是X的有限维子空间，{e1, e2, ..., e n}是M的一组基，给定g∈X，引进函数F : n → 1．对?c = (c1, c2, ..., c n)∈ n，规定F(c) = F(c1, c2, ..., c n) = || ∑1 ≤i≤n c i e i-g ||．(1) 求证F是一个凸函数；(2) 若F的最小值点是c = (c1, c2, ..., c n)，求证f = ∑1 ≤i≤n c ie i给出g在M中的最佳逼近元．证明：(1) 设c = (c1, c2, ..., c n), d = (d1, d2, ..., d n)∈ n, λ∈[0, 1]，则F(λ c + ( 1 -λ) d ) = || ∑1 ≤i≤n ( λ c i + ( 1 -λ) d i ) e i-g || = || λ∑1 ≤i≤n c i e i + ( 1 -λ) ∑1 ≤i≤n d i e i- (λ g+ ( 1 -λ)g )|| = || λ(∑1 ≤i≤n c i e i -g) + ( 1 -λ) ( ∑1 ≤i≤n d i e i-g )||≤λ|| ∑1 ≤i≤n c i e i -g || + ( 1 -λ) || ∑1 ≤i≤n d i e i-g ||= λ F(c)+ ( 1 -λ)F(d)，故F是一个凸函数．(2) 因为{e1, e2, ..., e n}是M的一组基，故M中的每个元h都可表示为h = ∑1 ≤i≤n d i e i，其中d = (d1, d2, ..., d n)∈ n．因为F(c) ≤F(d)，故|| f-g || = F(c) ≤F(d) = || h-g ||．那么f就是g在M中的最佳逼近元．1.4.13 设X是B*空间，X0是X的线性子空间，假定?c∈(0, 1)使得?y∈X，有inf { || y–x || | x ∈X0 } ≤c || y ||．求证：X0在X中稠密．证明：设y∈X，?ε > 0，x1∈X0，s.t. || y–x1 || < c || y || + ε /4．x2∈X0，s.t. || (y–x1) –x2 || < c || y–x1 || + ε /8．x3∈X0，s.t. || (y–x1 –x2 ) –x3 || < c || y–x1 –x2 || + ε /16．如此下去，可得到一个X0中的点列{ x n }，满足|| y–∑1 ≤j≤n +1x j|| < c || y–∑1 ≤j≤n x j|| + ε /2n + 2(?n∈ +)．那么，我们可以用数学归纳法证明|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)．当n = 1时，|| y–x1 || < c || y || + ε /4．结论成立．当n = 2时，|| (y–x1) –x2 || < c || y–x1 || + ε /8< c (c || y || + ε /4) + ε /8 < c 2 || y || + ε (1/4 + 1/8)，结论成立．当n≥ 3时，若|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)成立，则|| y–∑1 ≤j≤n +1x j|| < c || y–∑1 ≤j≤n x j|| + ε /2n + 2< c (c n || y || + ε (∑1 ≤j≤n 1/2j + 1)) + ε /2n + 2< c n+1 || y || + ε (∑1 ≤j≤n 1/2j + 1)) + ε /2n + 2< c n+1 || y || + ε (∑1 ≤j≤n+ 11/2j + 1))，因此结论也成立．由数学归纳法原理，?n∈ +，|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)．因为c∈(0, 1)，故存在N∈ +，使得c N || y || < ε /2．令x = ∑1 ≤j≤N x j，则x∈X0．且|| y–x || < ε /2 + ε (∑1 ≤j≤N 1/2j + 1) < ε．所以，X0在X中稠密．[张峰同学的证明] 反证法．若不然，则cl(X0)是X的真闭线性子空间．用Riesz引理，存在y∈X，使得|| y || = 1，且inf { || y–x || | x ∈ cl(X0)} > c．故对此y∈X，有inf { || y–x || | x ∈X0 } > c || y ||，矛盾．1.4.14 设C0表示以0为极限的实数全体，并在C0中赋以范数|| x || = max n≥1| ξn |，( ?x = (ξ1, ξ2, ..., ξn, ...)∈C0 )．又设M = {x = (ξ1, ξ2, ..., ξn, ...)∈C0 | ∑n ≥1 ξn/2n = 0}．(1) 求证：M是C0的闭线性子空间．(2) 设x0= (2, 0, 0, ...)，求证：inf z ∈M || x0–z || = 1，但?y∈M，有|| x0–y || > 1．证明：(1) 显然M ≠?，容易直接验证M是C0的线性子空间．若x k = (ξ1(k), ξ2(k), ..., ξn(k), ...)为M中的点列，且x k→x = (ξ1, ξ2, ..., ξn, ...)∈C0．则?ε > 0，存在N∈ +，使得?k > N，|| x k -x || < ε．此时，?n∈ +，有|ξn -ξn(k)| ≤ max n≥1| ξn -ξn(k) | = || x k -x || < ε．| ∑n ≥1 ξn/2n | = | ∑n ≥1 ξn/2n-∑n ≥1 ξn(k)/2n | = | ∑n ≥1 (ξn -ξn(k))/2n |≤∑n ≥1 |ξn -ξn(k)|/2n≤∑n ≥1 ε/2n = ε．所以，∑n ≥1 ξn/2n = 0，即x = (ξ1, ξ2, ..., ξn, ...)∈M．所以M是C0的闭线性子空间．(2) x0= (2, 0, 0, ...)，?z = (ξ1, ξ2, ..., ξn, ...)∈M，|| x0–z || = max{| 2 -ξ1 |, | ξ2 |, | ξ3 |, ... }．如果| 2 -ξ1 | > 1，则|| x0–z || > 1．如果| 2 -ξ1 | ≤ 1，则| ξ1 | ≥ 1，我们断言{| ξ2 |, | ξ3 |, ... }中至少有一个大于1者．否则，假若它们都不超1，因为ξn → 0 (n→∞)，故它们不能全为1．由∑n ≥1 ξn/2n = 0知| ξ1 |/2 = | ∑n ≥2 ξn/2n | ≤∑n ≥2 | ξn | /2n < ∑n ≥2 1/2n = 1/2，这样得到| ξ1 | < 1，矛盾．故{| ξ2 |, | ξ3 |, ... }中至少有一个大于1者．因此也有|| x0–z || > 1．综上所述，但?y∈M，有|| x0–y || > 1．由此，立即知道inf z ∈M || x0–z || ≥ 1．下面证明inf z ∈M || x0–z || ≤ 1．n∈ +，令z n= (1 - 1/2n, -1, -1, ..., -1, 0, 0, ...)．( z n从第2个坐标开始有连续的n个-1，后面全部是0 )，则(1 - 1/2n)/2 - 1/4 - 1/8 - ... - 1/2n + 1 = 0，因此z n∈M．此时，|| x0–z n || = max{| 1 + 1/2n|, | 1/4|, | 1/8|, ... } = 1 + 1/2n．故inf z ∈M || x0–z || ≥ inf n || x0–z n || = inf n (1 + 1/2n ) = 1．所以，inf z ∈M || x0–z || = 1．1.4.15 设X是B*空间，M是X的有限维真子空间，求证：?y∈X，|| y|| = 1，使得|| y–x || ≥ 1 ( ?x ∈M )．证明：取定z∈X \ M，令Y = span{z} + M．记S = { y∈Y | || y || = 1 }．则M是Y的真闭子空间，而S是Y中的单位球面．由Riesz引理，?n∈ +，存在y n∈S，使得d( y n, M ) ≥ 1 - 1/n．因为Y也是有限维的，故其中的单位球面为自列紧集．存在{y n}的收敛子列．不妨设y n(k) →y∈S．则d( y n(k), M ) ≥ 1 - 1/n(k)，故有d( y, M ) ≥ 1．即|| y–x || ≥ 1 ( ?x ∈M )．1.4.16 若f是定义在区间[0, 1]上的复值函数，定义ωδ( f ) = sup{| f (x) –f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}．如果0< α≤ 1对应的Lipschitz空间Lipα，由满足|| f || = | f(0) | + supδ > 0{δ–αωδ( f )} < +∞的一切f组成，并且以|| f ||为模．又设lipα = { f∈Lipα| lim δ→ 0 δ–αωδ( f ) = 0}．求证Lipα是B空间，而且lipα是Lipα的闭子空间．证明：(1) 显然，C1[0, 1]?Lipα，因此Lipα不空．对区间[0, 1]上的复值函数f, g，?λ∈ ，我们有ωδ( f + g ) = sup{| f (x) + g (x) – f (y) –g (y) | | ?x, y∈[0, 1], | x–y | ≤δ}≤ sup{| f (x) – f (y) | + | g (x) –g (y) | | ?x, y∈[0, 1], | x–y | ≤δ}≤ωδ( f ) + ωδ( g )．ωδ( λ f ) = sup{|λ f (x) –λ f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}= | λ| sup{| f (x) –f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}= | λ| ·ωδ( f )．若f, g∈Lipα，λ∈ ，则|| f + g || = | f(0) + g(0) | + supδ > 0{δ–αωδ( f + g ) }≤ | f(0) | + | g(0) | + supδ > 0{δ–α(ωδ( f ) + ωδ( g )) }= | f(0) | + | g(0) | + supδ > 0{δ–αωδ( f ) + δ–αωδ( g ) }≤ | f(0) | + | g(0) | + supδ > 0{δ–αωδ( f ) }+ supδ > 0{ δ–αωδ( g ) }= || f || + || g || < +∞．|| λ f || = | λ f(0) | + supδ > 0{δ–αωδ( λ f )}= | λ| · | f(0) | + | λ| · supδ > 0{δ–αωδ( f )}= | λ| · || f || < +∞．因此，f + g, λ f∈Lipα，且上述两个不等式表明|| · ||有齐次性和三角不等式．显然，|| f || ≥ 0．当|| f || = 0时，| f(0) | + supδ > 0{δ–αωδ( f )} = 0，意味着f(0) = 0，且ωδ( f ) = 0(?δ> 0)．而ωδ( f ) = 0(?δ> 0)则意味着f为常值．所以，f = 0．即|| · ||有正定性．综上所述，Lipα是B*空间．(2) 我们首先证明集合Lipα?C[0, 1]．f∈Lipα，?x, y∈[0, 1]，x ≠y，记δ = | x -y |．则| f (x) –f (y) | ≤ωδ( f )．而δ–αωδ( f ) ≤ supδ > 0{δ–αωδ( f n-f m) } ≤ || f ||，所以，| f (x) – f (y) | ≤ || f || δα= || f || · | x -y |α，故f∈C[0, 1]．我们再证明，?f∈Lipα，|| f ||C≤ || f ||，其中|| ·||C是C[0, 1]范数．事实上，?x∈[0, 1]，| f (x) | ≤ | f (0) | + | f (x) – f (0) |，故|| f ||C = max x∈[0, 1] | f (x) | ≤ | f (0) | + max x∈[0, 1] | f (x) –f (0) |≤ | f (0) | + sup x∈(0, 1] | f (x) –f (0) |/| x |α≤ | f (0) | + sup x∈(0, 1] { δ–αωδ( f ) } ≤ || f ||．这说明，如果{ f n }是Lipα中的基本列，则它也必是C[0, 1]中的基本列．而C[0, 1]是完备的，故存在f∈C[0, 1]，使得{ f n }一致收敛于f．而{ f n }作为Lipα中的基本列，有|| f n-f m || = | f n(0) -f m(0) | + supδ > 0{δ–αωδ( f n-f m) } → 0 (n, m→∞)，因此?ε > 0，?N∈ +，使得?n, m > N，有| f n(0) -f m(0) | + supδ > 0{δ–αωδ( f n-f m) } < ε．因此supδ > 0{δ–αωδ( f n-f m) } < ε．故?δ > 0，ωδ( f n-f m) < εδα．即?x, y∈[0, 1]，| x -y | ≤δ，都有| ( f n(x) -f m(x)) - ( f n(y) -f m(y)) | < εδα．令m→∞，得到| ( f n(x) -f(x)) - ( f n(y) -f(y)) | ≤εδα．因此，sup {| ( f n(x) -f(x)) - ( f n(y) -f(y)) | | x, y∈[0, 1]，| x -y | ≤δ}≤εδα．即?δ > 0，ωδ( f n-f ) ≤εδα．故supδ > 0{δ–αωδ( f n-f ) } ≤ε．同样地，对不等式| f n(0) -f m(0) | < ε令m→∞，就得到| f n(0) -f(0) | ≤ε．所以，| f n(0) -f(0) | + supδ > 0{δ–αωδ( f n-f ) } ≤ 2ε．这说明f n-f∈Lipα．而f n∈Lipα，故f = ( f -f n ) + f n∈Lipα．而前面的式子也表明|| f -f n || ≤ 2ε．因此|| f n-f || → 0 (n→∞)，即{ f n }为Lipα中的收敛列．所以，Lipα是Banach空间．(3) 记lipα = { f∈Lipα| lim δ→ 0 δ–αωδ( f ) = 0 }．f, g∈lipα，?λ∈ ，我们有δ–αωδ( f + g ) ≤δ–α(ωδ( f ) + ωδ( g ) ) = δ–αωδ( f ) + δ–αωδ( g ) → 0 (δ→ 0)．δ–αωδ( λ f ) = | λ| ·δ–αωδ( f ) → 0 (δ→ 0)．故f + g, λ f∈lipα，因此，lipα是Lipα的线性子空间．设{ f n }是lipα中的序列，且f n→f∈Lipα(n→∞)．则{ f n }一致收敛于f．ε > 0，存在N∈ +，使得|| f N →f || < ε /2．故有supδ > 0{δ–αωδ( f N-f ) } < ε /2．因为lim δ→ 0 δ–αωδ( f N) = 0，所以，?? > 0，使得?δ∈(0, ?)，有δ–αωδ( f N) < ε /2．此时我们有δ–αωδ( f ) ≤δ–α(ωδ( f N) + ωδ( f -f N))= δ–αωδ( f N) + δ–αωδ( f -f N)< ε /2 + supδ > 0{δ–αωδ( f N-f ) } < ε．所以，lim δ→ 0 δ–αωδ( f ) = 0，即f∈lipα．所以lipα是Lipα的闭子空间．1.4.17 (商空间) 设X是线性赋范空间，X0是X的闭线性子空间，将X中的向量分类，凡是适合x’-x’’∈X0的两个向量x’, x’’归于同一类，称其为等价类，把一个等价类看成一个新的向量，这种向量的全体组成的集合为X/X0表示，并称其为商空间．下列是关于商空间的命题．(1) 设[ y ]∈X/X0，x∈X，求证：x∈[ y ]的充分必要条件是[ y ] = x + X0．证明：设x’, x’’∈X，若它们归于同一类，则记为x’～x’’．我们用[ x ]表示x所在的等价类（大家注意，题目形式已经作了相应的修改）．(?) 若x∈[ y ]，则x～y．u ∈[ y ]，u～y，故u～x，即u –x∈X0．因此u ∈x + X0．所以[ y ] ?x + X0．反过来，?u ∈x + X0，则u～x，故u～y．因此u ∈[ y ]．所以x + X0 ? [ y ]．所以[ y ] = x + X0．(?) 若[ y ] = x + X0，则y –x∈X0，即y～x．从而x∈[ y ]．(2) 在X/X0中定义加法与数乘如下：[ x ] + [ y ] = x + y + X0(?[ x ], [ y ] ∈X/X0 )λ[ x ] = λ x + X0(?[ x ]∈X/X0 , ?λ∈ )其中x和y分别表示属于等价类[ x ]和[ y ]的任一元素．又规定范数|| [ x ] ||0 = inf z∈[ x ] || z || ( ?[ x ]∈X/X0 )求证：(X/X0, || · ||0)是一个B*空间．证明：第(1)部分说明了[ x ] = x + X0．容易看出加法与乘法的定义是合理的．进一步可以证明X/X0 构成数域上的线性空间，且其零元为[ θ] = X0．下面证明|| · ||0是X/X0 上的范数．显然，?[ x ]∈X/X0，|| [ x ] ||0≥ 0．若[ x ] = [ θ] = X0，则|| [ x ] ||0 = 0．若|| [ x ] ||0 = 0，则inf z∈[ x ] || z || = 0．存在z n∈[ x ]使得|| z n || → 0，即z n→θ (n→∞)．那么，x-z n∈X0，x-z n→x (n→∞)，而X0是闭集，故x∈X0．所以x～θ，即[ x ] = X0．因此|| · ||0有正定性．[ x ]∈X/X0，?λ∈ ，|| λ[ x ]||0 = || [ λ x ] ||0 = inf y∈[ x ] || λ y || = inf y∈[ x ] | λ| · || y ||= | λ| · inf y∈[ x ] || y || = | λ| · ||[ x ]||0．因此|| · ||0有齐次性．[ x ], [ y ]∈X/X0，|| [ x ] + [ y ] ||0 = inf z∈[ x ] + [ y ] || z || = inf u∈[ x ], v∈[ y ] || u + v ||≤ inf u∈[ x ], v∈[ y ] { || u || + || v || } ≤ inf u∈[ x ] { inf v∈[ y ] { || u || + || v ||} }≤ inf u∈[ x ] { inf v∈[ y ] { || u || + || v ||} } = inf u∈[ x ] { || u || + inf v∈[ y ] || v || }= inf u∈[ x ] || u || + inf v∈[ y ] || v || = || [ x ] ||0 + || [ y ] ||0．因此|| · ||0的三角不等式成立．所以，(X/X0, || · ||0)是一个B*空间．(3) 设[ x ]∈X/X0, 求证对?y∈[ x ]有inf { || y -z || | z∈X0 } = ||[ x ] ||0．证明：|| [ x ] ||0 = inf u∈[ x ] || u || = inf u∈[ y ] || u || = inf { || u || | u∈y + X0 }= inf { || y + v || | v∈X0 } = inf { || y -z || | z∈X0 }．(4) 定义映射? : X →X/X0为? (x) = [ x ] = x + X0(?x∈X )．求证?是线性连续映射．证明：?x, y∈X，?α, β∈ ，( α x + β y ) = [α x + β y ] = [α x ] + [ β y ] = α [ x ] + β[ y ] = α? (x) + β? (y)．|| ? (x) -? (y) ||0 = || [ x ] - [ y ] ||0 = || [ x-y ] ||0 = in f z∈[ x-y ] || z || ≤ || x-y ||．所以，?是线性连续映射．(5) ?[ x ]∈X/X0，求证?y∈X，使得? (y) = [ x ]，且|| y || ≤ 2|| [ x ] ||0．证明：因为|| [ x ] ||0 = inf z∈[ x ] || z ||，若|| [ x ] ||0 = 0，则由|| · ||0的正定性，知[ x ] = X0，取y = θ即满足要求．若|| [ x ] ||0≠ 0，则inf z∈[ x ] || z || = || [ x ] ||0 < 2 || [ x ] ||0，存在?y∈[ x ]，使得|| y || ≤ 2|| [ x ] ||0．此时显然有? (y) = [ x ] = [ y ]．(6) 设(X, || · ||)完备，求证(X/X0, || · ||0)也是完备的．证明：设{ [ x ]n }是X/X0中的基本列．为证明它是收敛列，只需证明它存在收敛子列．由基本列性质，可选出子列{ [ x ]n(k)}使得|| [ x ]n(k) - [ x ]n(k+1) ||0 ≤ 1/2k．故∑k ≥ 1 || [ x ]n(k) - [ x ]n(k+1) ||0 收敛．根据(5)，?k∈ +，?y k∈[ x ]n(k+1) - [ x ]n(k)，使得|| y k || ≤ 2|| [ x ]n(k+1) - [ x ]n(k) ||0．那么，∑k ≥ 1|| y k ||收敛．由X的完备性，s k = ∑ 1 ≤j ≤k y j是X中的收敛列．设其极限为s．由(5)中?的连续性，在X/X0中，?(s k) →?(s) ( k→∞ )．而?(s k) = ?( ∑ 1 ≤j ≤k y j ) = ∑ 1 ≤j ≤k ?( y j )= ∑ 1 ≤j ≤k ( [ x ]n(j+1) - [ x ]n(j)) = [ x ]n(k+1) - [ x ]n(1)．故{[ x ]n(k+1) - [ x ]n(1)}收敛，因而{[ x ]n(k)}是收敛列．因此X/X0中的基本列{ [ x ]n }存在收敛子列{[ x ]n(k)}，所以，{ [ x ]n }是X/X0中的收敛列．因此，(X/X0, || · ||0)是完备的．(7) 设X = C[0, 1]，X0 = { f∈X | f (0) = 0 }，求证：X/X0 ? ，其中记号“?”表示等距同构．证明：显然，X0是C[0, 1]中的线性子空间．记X0所确定的等价关系为～，则f～g ? f (0) = g (0)．定义Φ : X/X0 → ，Φ([ f ]) = f (0)．显然定义是合理的．f, g∈X，?α, β∈ ，Φ(α[ f ] + β[ g ]) = Φ([αf + β g ]) = (αf + β g )(0)= αf (0)+ β g (0) = αΦ([ f ])+ βΦ([ g ])．因此Φ是线性映射．因Φ(X0) = 0，故Φ是单射．而?c∈ ，若记所对应的常值函数为h c∈C[0, 1]，则Φ( [ h c] ) = c．故Φ是满射．综上所述，Φ : X/X0 → 是线性同构．f∈X，|| [ f ]||0 = inf g∈[ f ] { || g || } ≥ inf g∈[ f ] { | g (0) | }= inf g∈[ f ] { | f (0) | } = | f (0) | = | Φ([ f ]) |．另一方面，因为常值函数h f (0)∈[ f ]，故|| [ f ]||0 = inf g∈[ f ] { || g || } ≤ || h f (0) || = | f (0) | = | Φ([ f ]) |．所以，?f∈X，都有|| [ f ]||0 = | Φ([ f ]) |，因此Φ : X/X0 → 是等距同构．[第4节完] 泛函分析题1_5凸集与不动点p521.5.1 设X是B*空间，E是以θ为内点的真凸子集，P是由E产生的Minkowski 泛函，求证：(1) x∈int(E) ?P(x) < 1；(2) cl(int(E)) = cl(E)．证明：(1) (?) 若x∈int(E)，存在δ > 0，使得Bδ(x) ?E．注意到x + x/n→x ( n→∞ )，故存在N ∈ +，使得x + x/N ∈Bδ(x) ?E．即x/( N/( 1 + N ) ) ∈E．因此P(x) ≤N/( 1 + N ) < 1．(?) 若P(x) < 1．则存在a > 1，使得y = a x∈E．因θ∈int(E)，故存在δ > 0，使得Bδ(θ) ?E．令η = δ(a - 1)/a，?z∈Bη(x)，令w = (a z-y )/(a - 1)，则|| w || = || (a z-y )/(a - 1) || = || a z-y ||/(a - 1)= || a z-a x ||/(a - 1) = a || z-x ||/(a - 1) < aη/(a - 1) = δ．故w∈Bδ(θ) ?E．故z = ((a - 1)w + y )/a ∈E，因此，Bη(x) ?E．所以x∈int(E)．(2) 因int(E) = E，故有cl(int(E)) ? cl(E)．下面证明相反的包含关系．若x∈cl(E)，则?ε > 0，存在y∈E，使得|| x -y || < ε/2．因ny/(n + 1) →y ( n →∞ )．故存在N ∈ +，使得|| Ny/(N + 1) -y || < ε/2．令z = Ny/(N + 1)，则z∈E，且P(z) ≤N/(N + 1) < 1，由(1)知z∈int(E)．而|| z -x || ≤ || z -y || + || y -x || < ε/2 + ε/2 = ε．故x∈cl(int(E))，因此cl(E) ? cl(int(E))所以cl(int(E)) = cl(E)．1.5.2 求证在B空间中，列紧集的凸包是列紧集．证明：设A是B空间X中的列紧集，?ε > 0，存在A的有限ε /3网B．设B = {b1, b2, ..., b n}，M = max j{ || b j || }，取δ > 0，使得n δ M < ε /3．设[0, 1]分划D为0 = t0 < t1 < t2 < ... < t m = 1，使得max 1 ≤j ≤m {| t j–t j–1|} < δ．设?x∈co(A)，设x= λ1 a1 + λ2 a2+ ... + λ k a k，其中a j∈A，λ j > 0，∑ j λ j = 1．对每个j ≤k，存在b i( j )∈B使得|| a j-b i( j ) || < ε /3；令y= λ1 b i(1) + λ2 b i(2)+ ... + λ k b i(k)，则|| x - y || = || λ1 (a1 -b i(1)) + λ2 (a2 -b i(2))+ ... + λ k (a k-b i(k))||，≤λ1 · || a1 -b i(1) || + λ2 · || a2 -b i(2) || + ... + λ k · || a k-b i(k) ||≤ ( λ1 + λ2 + ... + λ k ) · (ε /2) = ε /3．将y= λ1 b i(1) + λ2 b i(2)+ ... + λ k b i(k)中的那些含有相同b j 的项合并起来，于是，y可表示为y= μ1 b1 + μ2 b2+ ... + μ n b n，其中μj ≥ 0，且∑ j μj = 1．对每个l ≤n，存在t s( l )∈D，使得|| μl-t s( l ) || < δ；令z= t s(1) b1 + t s(2) b2+ ... + t s(n) b n，则|| y - z || = || (μ1 -t s(1))b1 + (μ2 -t s(2))b2+ ... + (μn -t s(n))b n ||≤∑ l | μl-t s( l ) | · max j{ || b j || } ≤n δ M < ε /3；令C = {t s(1) b1 + t s(2) b2+ ... + t s(n) b n | t s(i)∈D，1 ≤i≤n}，则C是有限集，且C是co(A)的有限ε网．因空间是完备的，故co(A)是列紧集．1.5.3 设C是B*空间X中的一个紧凸集，映射T : C →C连续，求证T在C上有一个不动点．证明：因为C是紧集，所以C是闭集．因为C是紧集，故C的任意子集都列紧．而T(C) ?C，故T(C)列紧．于是，由Schauder不动点定理，T在C上有一个不动点．。
2015年全国医学博士入学统一考试英语真题及答案解析
2015年全国医学博士入学统一考试英语真题及答案解析Part I: Listening comprehension(略)Part II: Vocabulary(10%)Section ADirection: In this section, all the sentences are incomplete. Four word or phrases marked A,B,C and D are given beneath each of them. You are to choose the word the word or phrase that best completes the sentence, then mark your answer on the ANSWER SHEET.31. Despite his doctor’s note of caution, he never____from dring and smorking.A. retainedB. dissuadedC. alleviatedD. abstained32. people with a history of recurrent infections are warned that the use of personal stereos with headsets is likely to____their hearing.A. rehabilitateB. jeopardizeC. tranquilizeD. supplement33. impartial observers had to acknowledge that lack of formal education did not seem to____larry in any way in his success.A. refuteB. ratifyC. facilitateD. impede34. when the supporting finds were reduced, they should have revised their plan______.A. accordinglyB. alternativelyC. considerablyD. relatively35. it is increasingly believed among the expectant parents that prenatal education of classical music can_____ future adults with appreciation of music.A. acquaintB. familiarizedC. endowD. amuse36. if the gain of profit is solely due to rising energy prices, then inflation should be subsided when energy prices_____A. level outB. stand outC. come offD. wear off37. heat stroke is a medical emergency that demands immediate_____ from qualified medical personnel.A. prescriptionB. palpationC. interventionD. interposition38. asbestos exposure results in Mesothelioma, asbestosis and internal organ cancers, and_____ of these diseases is often decades after the initial exposure.A. offsetB. intakeC. outletD. onset39. ebola, which spreads through body fluid or secretions such as urine,______ and semen, can kill up to 90% of those infected.A. salineB. salivaC. scabiesD. scrabs40. the newly designed system is ____ to genetic transfections, and enables an incubation period for studying various genes.A. comparableB. transmissibleC. translatableD. amenable Section BDirections: each of the following sentences has a word or phrase underlined. There are four words or phrases beneath each sentence. Choose the word or phase which can best keep the meaning of the original sentence if it issubstituted for the underlined part. Mark your answer on the ANSWER SHEET.41. every year more than 1000 patients in Britain die on transplant waiting lists, prompting scientists to consider other ways to produce organs.A. propellingB. prolongingC. puzzlingD. promising42. improved treatment has changed the outlook of HIV patients, but there is still a serious stigma attached to AIDS.A. disgraceB. discriminationC. harassmentD. segregation43. surviviors of the shipwreck were finally rescued after their courage of persistence lowered to zero by their physical lassitude.A. depletionB. dehydrationC. exhaustionD. handicap44. scientists have invented a 3D scan technology to read the otherwise illegible wood-carved stone, a method that may apply to other areas such as medicine.A. negativeB. confusingC. eloquentD. indistinct45. top athletes scrutinize both success and failure with their coach to extract lessons from them, but they are never distracted from long-term goals.A. anticipateB. clarifyC. examineD. verify46. his imperative tone of voice reveals his arrogance and arbitrariness.A. challengingB. solemnC. hostileD. demanding47. the discussion on the economic collaboration between the United States and the European Union may be eclipsed by the recent growing trade friction.A. erasedB. triggeredC. shadowedD. suspended48. faster increases in prices foster the belief that the future increases will be also stronger, so that higher prices fuel demand rather than quench it.A. nurtureB. eliminateC. assimilateD. puncture49. some recent developments in photography allow animals to be studied in previously inaccessible places and in unprecedented detail.A. unpredictableB. unconventionalC. unparalleledD. unexpected50. a veteran negotiation specialist should be skillful at manipulating touchy situation.A. estimatingB. handlingC. rectifyingD. anticipatingPart III Cloze(10%)Direction: in this section there is a passage with ten numbered blanks. For each blank, there are four choices marked A, B, C, and D on the right side. Choose the best answer and mark the letter of your choice on the ANSWER SHEET.A mother who is suffering from cancer can pass on the disease to her unborn child in extremely rare cases 51 a new case report published in PNAS this week.According to researchers in Japan and at the Institute for Cancer Research in Sutton, UK, a Japanese mother had been diagnosed with leukemia a few weeks after giving birth 52 tumors were discovered in her daughter’s cheek and lung when she was 11 months old. Genetic analysis showed that the baby’s cancer cells had the same mutation as the cancer cellsof the mother. But the cancer cells contained no DNA whatsoever from the father 53 would be expected if she had inherited the cancer from conception. That suggests the cancer cell made it into the unborn child’s body across the placental barrier.The Guardian claimed this to be the fires 54 case of cells crossing the placental barrier. But this is not the case----microchimerism 55 cells are exchanged between a mother and her unborn child, is thought to be quite common, with some cells thought to pass from fetus to mother in about 50 to 70 percent of cases and to go the other way about half,56.As the BBC pointed out, the greater 57 in cancer transmission from mother to fetus had been how cancer cells that have slipped through the placental barrier could survive in the fetus without being killed by its immune system. The answer, in this case at least, lies in a second mutation of the cancer cells, which led to the 58 of the specific features that would have allowed the fetal immune system to detect the cells as foreign. As a result, no attack against the invaders was launched.59, according to the researchers there is little reason for concern of “cancer danger”. Only 17 probable cases have been reported worldwide and the combined 60 of cancer cells both passing the placental barrier and having the right mutation to evade the baby’s immune system is extremely low.51. A. suggests B. suggesting C. having suggested D. suggested52. A. since B. although C. whereas D. when53. A. what B. whom C. who D. as54. A. predicted B. notorious C. proven D. detailed55. A. where B. when C. if D. whatever56. A. as many B. as much C. as well D. as often57. A. threat B. puzzle C. obstacle D. dilemma58. A. detection B. deletion C. amplification D. addition59. A. therefore B. furthermore C. nevertheless D. conclusively60. A. likelihood B. function C. influence D. flexibilityPart IV Reading Comprehension(30%)Directions: in this part there are six passages, each of which is followed by five questions. For each question there are four possible answers marked A, B, C, and D. choose the best answer and mark the letter of your choice on the ANSWER SHEET.Passage OneThe American Society of Clinical Oncology wrapped its annual conference this week, going through the usual motions of presenting a lot of drugs that offer some added quality or extension of life to those suffering from a variety of as-yet incurable diseases. But buried deep in an AP story are a couple of promising headlines that seems worthy of more thorough review, including one treatment study where 100 percent of patients saw their cancer diminish byhalf.First of all, it seems pharmaceutical companies are moving away from the main cost-effective one-size-first-all approach to drug development and embracing the long cancer treatments, engineering drugs that only work for a small percentage of patients but work very effectively within that group.Pfizer announced that one such drug it’s pushing into late-stage testing is target for 4% of lung cancer patients. But more than 90% of that tiny cohort responded to the drug initial tests, and 9 out of ten is getting pretty close to the ideal ten out of ten. By gearing toward more boutique treatments rather than broad umbrella pharmaceuticals that try to fit for everyone it seems cancer researchers are making some headway. But how can we close the gap on that remaining ten percent?Ask Takeda Pharmaceutical and Celgene, two drug makers who put aside competitive interests to test a novel combination of their treatments. In a test of 66 patients with the blood disease multiple myeloma, a full 100 percent response to a cancer drug(or in this case a drug cocktail) is more or less unheard of. Moreover, this combination never would’ve been two competing companies hadn’t sat down and put their heads together.Are there more potentially effective drug combos out there separated by competitive interest and proprietary information? Who’s to say, but it seems like with the amount of money and research being pumped into cancer drug development, the outcome pretty good. And if researchers can start pushing more of their response numbers toward 100 percent, we can more easily start talking about oncology’s favorite four-letter word: cure.61. which of the following can be the best title for the passage?A. Competition and CooperationB.Two Competing Pharmaceutical CompaniesC. The promising Future of PharmaceuticalsD. Encouraging News: a 100% Response to a Cancer Drug62. in cancer drug development, according to the passage, the pharmaceuticals now ____A. are adopting the cost-effective one-size-fits-all approachB. are moving towards individualized and targeted treatmentsC. are investing the lion’s shares of their moneyD. care only about their profits63. from the encouraging advance by the two companies, we can infer that____A. the development can be ascribed to their joint efforts and collaborationB. it was their competition that resulted in the accomplishmentC. other pharmaceuticals will join them in the researchD. the future cancer treatment can be nothing but cocktail therapy64. from the last paragraph it can be inferred that the answer to the question___A. is nowhere to be foundB. can drive one crazyC. can be multipleD. is conditional65. the tone of the author of this passage seems to be_____A. neutralB. criticalC. negativeD. potimistPassage TwoLiver disease is the 12th leading cause of death in the US, chiefly because once it’s determined that a patient needs a new liver it’s difficult to get one. Even in case where a suitable donor match is found, there’s guarantee a transplant will be successful. But researchers Massachusetts General Hospital have taken a huge step toward building functioning livers in the lab, successfully transplanting culture-grown livers into rats.The livers aren’t grown from scratch, but rather within the infrastructure of a donor liver. The liver cells in the donor organ are washed out with a detergent that gently strips away the liver cells, leaving behind a biological scaffold of proteins and extracellular architecture that is very hard to duplicate synthetically.With all of that complicated infrastructure already in place, the researchers then seeded the scaffold(支架) with liver cells isolated from health livers, as well as some special endothelial cells to line the bold vessels. Once repopulated with healthy cells, these livers lived in culture for 10 days.The team also translated some two-day-old recellularized livers back into rats, where they continued to thrive for eight hours while connected into the rat’s vascular systems. However, the current method isn’t perfect and can not seem to repopulate the blood vessels quite densely enough and the transplanted livers can’t keep functioning for more than about 24 hours(hence the eight-hour maximum for the rat thansplant).But the initial successes are promising, and the team thinks they can overcome the blood vessel problem and get fully functioning livers into rats within two years. It still might be a decade before the tech hits the clinic, but if nothing goes horribly wrong—and especially if stem-cell research established a reliable way to create health liver cells from the every patients who need transplants-lab-generated livers that are perfect matches for their recipients could become a reality.66. it can be inferred from the passage that the animal model was mainly intended to____A. investigate the possibility of growing blood vessels in the labB. explore the unknown functions of the human liverC. reduce the incidence of liver disease in the US.D. address the source of liver transplants67. what does the author mean when he says that the livers aren’t grown from scratch?A. the making of a biological scaffold of proteins and extracellular architectureB. a huge step toward building functioning livers in the labC. the building of the infrastructure of a donor liverD. growing liver cells in the donor organ68. the biological scaffold was not put into the culture in the lab until____A. duplicated syntheticallyB. isolated from the healthy liverC. repopulated with the healthy cellsD. the addition of some man-made blood vessels69. what seems to be the problem in the planted liver?A. the rats as wrong recipientsB. the time point of the transplantationC. the short period of the recellularizationD. the insufficient repopulation of the blood vessels70. the research team holds high hopes of_____A. creating lab-generated livers for patients within two yearsB. the timetable for generating human livers in the labC. stem-cell research as the future of medicineD. building a fully functioning liver into ratsPassage ThreePatients whose eyes have suffered heat or chemical bums typically experience severe damage to the cornea—the thin, transparent front of the eye that refracts light and contributes most of the eye’s focusing ability. In a long-term study, Italian researchers use stem cells taken from the limbus, the border between the cornea and the white of the eye, to cultivate a graft of healthy cells in a lab to help restore vision in eyes. During the 10-years study, the researchers implanted the healthy stem cells into the damaged cornea in 113 eyes of 112 patients. The treatment was fully successful in more than 75 percent of the patients, and partially successful in 13 percent. Moreover, the restored vision remained stable over 10 years. Success was defined as an absence of all symptoms and permanent restoration of the cornea.Treatment outcome was initially assessed at one year, with up to 10 years of follow-up evaluations. The procedure was even successful on several patients whose bum injuries had occurred years earlier and who had already undergone surgery.Current treatment for burned eyes involves taking stem cells from a patient’s healthy eye, or from the eyes of another person, and transferring them to the burned eye. The new procedure, however, stimulates the limbal stem cells from the patient’s own eye to reproduce in a lab culture. Several types of treatments using stem cells have proven successful in restoring blindness, but the long-term effectiveness shown here is significant. The treatment is only for blindness caused by damage to the cornea; it is not effective for repairing damaged retinas or optic nerves.Chemical eye burns often occur in the workplace, but can also happen due to mishaps involving household cleaning products and automobile batteries.The result of the study, based at Italy’s University of Modena and Reggio Emilia, were published in the June 23 online issue of the New England Journalof Medicine.71. what is the main idea of this passage?A. stem cells can help restore vision in the eyes blinded by bums.B. the vision in the eyes blinded by bums for 10 years can be restoredC. the restored vision of the burned eyes treated with stem cells can last for10 yearsD. the burned eyes can only be treated with stem cells from other healthy persons72. the Italian technique reported in this passage_____A. can repair damaged retinasB. is able to treat damaged optic nervesC. is especially effective for burn injuries in the eyes already treated surgicallyD. shows a long-term effectiveness for blindness in vision caused by damage to cornea73. which of the following is NOT mentioned about eye bums?A. the places in which people workB. the accidents that involve using household cleaning productsC. the mishaps that involved vehicles batteriesD. the disasters caused by battery explosion at home74. what is one of the requirements for the current approach?A. the stem cells taken from a healthy eyeB. the patient physically healthyC. the damaged eye with partial visionD. the blindness due to damaged optic nerves75. which of the following words can best describe the author’s attitude towards the new method?A. sarcasticB. indifferentC. criticalD. positivePassage FourHere is a charming statistic: divide the us by race, sex and county of residence, and differences in average life expectancy across the various groups can exceed 30 years. The most disadvantaged look like denizens of a poor African country: a boy born on a Native American reservation in Jackson County, South Dakota, for example, will be lucky to reach his 60th birthday, a typical child in Senegal can expect to live longer than that.America is not alone in this respect. While the picture is extreme in other rich nations, health inequalities based on race, sex and class exist in most societies—and are only party explained by access to healthcare.But fresh insights and solutions may soon be at hand. An innovative project in Chicago to unite sociology and biology is blazing the trail(开创), after discovering that social isolation and fear of crime can help to explain the alarmingly high death rate from breast cancer among the city’s black women. Living in these conditions seems to make tumors more aggressive by changing gene activity, so that cancer cells can use nutrients more effectively.We are already familiar with the lethal effect of stress on people clinging to the bottom rungs of the societal ladder, thanks to pioneering studies of British civil servants conducted by Michael Marmot of University College London. What’s exciting about the Chicago project is that it both probes the mechanisms involved in a specific disease and suggests precise remedies that it both probes the mechanisms invlilved in a specific disease and suggests precise remedies. There are drugs that may stave tumors of nutrients and community coordinators could be employed to help reduce social isolation. Encouraged by the US National Institutes of Health , similar projects are springing up to study other pockets of poor health, in populations ranging from urban black men to while poor women in rural Appalachia.To realize the full potential of such projects, biologists and sociologists will have to start treating one other with a new respect and learn how to collaborate outside their comfort zones. Too many biomedical researchers still take the arrogant view that sociology is a “soft science” with little that’s serious to say about health. And too many sociologists reject any biological angle—fearing that their expertise will be swept aside and that this approach will be used to bolster discredited theories of eugenics, or crude race-based medicine.It’s time to drop these outdated attitudes and work together for the good of society’s most deprived members. More important, it’s time to use this fusion of biology and sociology to inform public policy. This endeavor has huge implications, not least in cutting the wide health gaps between blacks and whites, rich and poor.76. as shown in the 1st paragraph, the shaming statistic reflects______.A. injustice everywhereB. racial discriminationC. a growing life spanD. health inequalities77. which of the following can have a negative impact on health according to the Chicago-based project?A. where to liveB. which race to belong toC. how to adjust environmentallyD. what medical problem to suffer78. the Chicago-based project focuses its management on_____A. a particular medical problem and its related social issueB. racial discrimination and its related social problemsC. the social ladder and its related medical conditionsD. a specific disease and its medical treatment79. which of the following can most probably neglected by sociologists?A. the racial perspectiveB. the environmental aspectC. the biological dimensionD. the psychological angel80. the author is a big fan of______A. the combination of a traditional and new way of thinking in promoting healthB. the integration of biologists and sociologists to reduce health inequalitiesC. the mutual understanding and respect between racesD. public education and health promotionPassage FiveAmerican researchers are working on three antibodies that many mark a new step on the path toward an HIV vaccine, according to a report published online Thursday, July 8,2010, in the journal Science.One of the antibodies suppresses 91 percent of HIV strains, more than any AIDS antibody ever discovered, according to a report on the findings published in the Wall Street Journal. The antibodies were discovered in the cells of a 60-year-old African-American gay man whose body produced them naturally. One antibody in particular is substantially different from its precursors, the Science study says.The antibodies could be tried as a treatment for people already infected with HIV, the WSJ reports. At the very least, they might boost the efficacy of current antiretroviral drugs.It is welcome news for the 33 million people the United Nations estimated were living with AIDS at the end of 2008.The WSJ outlines the painstaking method the team used to find the antibody amid the cells of the African—American man, known as Donor 45. First they designed a probe that looks just like a spot on a particular molecule on the cells that HIV infects. They used the probe to attract only the antibodies that efficiently attack that spot. They screened 25 million of Donor 45’s cell to find just 12 cells that produced the antibodies.Scientists have already discovered plenty of antibodies that either don’t work at all or only work on a couple of HIV strains. Last year marked the first time that researchers found ”broadly neutralizing antibodies”, which knock out many HIV strains. But none of those antibodies neutralized more than about 40 percent of them, the WSJ says. The newest antibody, at 91 percent neutralization , is a marked improvement.Still, more work needs to be done to ensure the antibodies would activate the immune system to produce natural defenses against AIDS, the study authors say. They suggest there test methods that blend the three new antibodies together—in raw form to prevent transmission of the virus, such as from mother to child; in a microbicide gel that women or gay men could use before sex to prevent infection; or as a treatment for HIV/AIDS, combined with antiretroviral drug.If the scientists can find the right way to stimulate production of the antibodies, they think most people could produce then, the WSJ says.81. we can learn from the beginning of the passage that_______A. a newly discovered antibody defeats 91% of the HIV strainsB. a new antiretroviral drug has just come on the marketC. American researchers have developed a new vaccine for HIVD. the African—American gay man was cured of this HIV infection82. what is the implication of the antibodies discovered in the cells of the African—American gay man?A. they can cure the 33 million AIDS patients in the worldB. they may strengthen the effects of the existing antiretroviral drugsC. they will kill all the HIV virusesD. they will help make a quick diagnosis of an HIV infection83. the newest antibody found in Donor 45 reflects a dramatic advance in terms of_____.A. pathologyB. pharmacologyC. HIV neutralizationD. HIV epidemiology84. according to the study authors, the three test methods are intended to____.A. advance the technology in condom production to prevent HIV infectionB. facilitate the natural immune defense against AIDSC. develop more effective antiretroviral drugs85. the passage is most likely_____.A. a news reportB. a paper in ScienceC. an excerpt from an Immunology TextbookD. an episode in a science fiction novel.Passage SixWhitening the world's roofs would offset the emissions of the world's cars for 20 years, according to a new study from Lawrence Berkeley National Laboratory.Overall, installing lighter-colored roofs and pavement can cancel the heat effect of two years of global carbon dioxide emissions, Berkeley Lab says. It's the first roof-cooling study to use a global model to examine the issue.Lightening-up roofs and pavement can offset 57 billion metric tons of carbon dioxide, about double the amount the world emitted in 2006, the study found. It was published in the journalEnvironmental Research Letters.Researchers used a conservative estimate of increased albedo, or solar reflection, suggesting that purely white roofs would be even better. They increased the albedo of all roofs by 0.25 and pavement by 0.15. That means a black roof, which has an albedo of zero, would only need to be replaced by a roof of a cooler color -- which might be more feasible to implement than a snowy white roof, Berkeley Lab says.The researchers extrapolated a roof's CO2 offset over its average lifespan. If all roofs were converted to white or cool colors, they would offset about 24 gigatons (24 billion metric tons) of CO2, but only once. But assuming roofs last about 20 years, the researchers came up with 1.2 gigatons per year. That equates to offsetting the emissions of roughly 300 million cars, all the cars in the world, for 20 years.Pavement and roofs cover 50 to 65 percent of urban areas, and cause a heat-island effect because they absorb so much heat. That's why cities aresignificantly warmer than their surrounding rural areas. This effect makes it harder -- and therefore more expensive -- to keep buildings cool in the summer. Winds also move the heat into the atmosphere, causing a regional warming effect.Energy Secretary Steven Chu, a Nobel laureate in physics (and former Berkeley Lab director), has advocated white roofs for years. He put his words into action Monday by directing all Energy Department offices to install white roofs. All newly installed roofs will be white, and black roofs might be replaced when it is cost-effective over the lifetime of the roof."Cool roofs are one of the quickest and lowest-cost ways we can reduce our global carbon emissions and begin the hard work of slowing climate change," he said in a statement.86. which of the following can be the best title for the passage?A. a Decline in Car EmissionsB. white Roofs or Black PavementsC. the Effect of Linghting-up RoofsD. climate Change and Extreme Weathers87. a indicated by the passage, black roofs______A. are better than snowy white onesB. reflect not heat from the sunC. are more expensive to build in the urban areasD. are supposed to be placed by snowy white ones88. if they are converted to white or cooler colors, all roofs in the world in their lifetime_____A. can absorb 1.2 gigattons of CO2 a yearB. could serve as 300 million cars in terms of emissionC. would offset the emissions from 300 million carsD. would offset about 24 gigatons of CO2 as emitted from the cars89. according to the passage, it is hard and expensive to keep the urban buildings cool because of______A. the heat-island effectB. the lack of seasonal windsC. the local unique weatherD. the fast urban shrinkage90. energy Secretary Steven Chu implies that_____A. nothing could be more effective in cooling global warming than method he has advocatedB. the method in question still needs to be justified in the futureC. our global carbon emissions can be reduced by half if cool roofs are installedD. weather change and global warming can be addressed in no timePart V Writing(20%)Directions: in this part there is an essay in Chinese. Read it carefully and then write a summary of 200 words in English on the ANSWER SHEET. Make sure that your summary covers the major points of the passage.什么是健康？人的健康包括身体健康和心理健康两个方面。
2015考博真题
一、单选1×50
上下尖牙区别
monson球面的半径
下颌神经前支中的感觉神经
前牙切割运动的杠杆运动形式
单囊性成釉细胞瘤处理方式
腺淋巴瘤病理特点
舌下腺结构
放射性骨髓炎病理表现
翼下颌间隙内容
下颌运动特点
下颌体骨化中心
颞下颌关节手术时切口方式
牙受垂直向力时牙龈主纤维中不受力的是
（以后想起来再补充）
二、名解2×10
近唇线角
pterygoid process
Terra dentition index
mento-cervical angle
taste threshold
alveolar bone proper
candidiasis
chronic gingivitis
branchial cleft cyst
lymphoepithelial carcinoma
三、简答5×6
解剖
1.根管系统在根部侧面开口的系统名称，并从解剖角度解释牙周病和牙髓病的相互影响。
2.口颌系统肌链的组成与功能？
3.临床上面神经的解剖方法，面神经主干的解剖标志点？
病理
1.口腔黏膜鳞癌有很多亚型，请举3例口腔黏膜鳞癌亚型，并叙述其镜下特点及生物学行为？
2.根据牙骨质组织结构学特性，叙述牙骨质龋特点？
3.肌上皮细胞来源的唾液腺良恶性肿瘤各举两例，及其镜下鉴别要点。
泛函分析试卷
泛函分析期末考试试卷（总分100分） 一、选择题（每个3分，共15分）1、设X 是赋范线性空间，X y x ∈,，T 是X 到X 中的压缩映射，则下列哪个式子成立（ ）.A ．10<<-≤-αα， y x Ty Tx B.1≥-≤-αα， y x Ty Tx C.10<<-≥-αα， y x Ty Tx D.1≥-≥-αα， y x Ty Tx 2、设X 是线性空间，X y x ∈,，实数x 称为x 的范数,下列哪个条件不是应满足的条件：（ ）.A. 0等价于0且,0==≥x x xB.()数复为任意实,αααx x =C. y x y x +≤+D. y x xy +≤ 3、下列关于度量空间中的点列的说法哪个是错误的（ ）. A ．收敛点列的极限是唯一的 B. 基本点列是收敛点列 C ．基本点列是有界点列 D.收敛点列是有界点列4、巴拿赫空间X 的子集空间Y 为完备的充要条件是（ ）. A ．集X 是开的 B.集Y 是开的 C.集X 是闭的 D.集Y 是闭的5、设(1)p l p <<+∞的共轭空间为q l ，则有11p q+的值为（ ）.A. 1-B.12 C. 1 D. 12- 二、填空题（每个3分，共15分）1、度量空间中的每一个收敛点列都是（ ）。
2、任何赋范线性空间的共轭空间是（ ）。
3、1l 的共轭空间是（ ）。
4、设X按内积空间<x,y>成为内积空间，则对于X中任意向量x,y 成立不等式（）当且仅当x与y线性相关时不等式等号成立。
5、设T为复希尔伯特空间X上有界线性算子，则T为自伴算子的充要条件是（）。
三、判断题（每个3分，共15分）1、设X是线性赋范空间，X中的单位球是列紧集，则X必为有限维。
( )2、距离空间中的列紧集都是可分的。
( )3、若范数满足平行四边形法则，范数可以诱导内积。
( )4、任何一个Hilbert空间都有正交基。
2015南京师范大学考博英语真题阅读理解精练
2015南京师范大学考博英语真题阅读理解精练Every living thing has an inner biological clock that controls behavior.The clock works all the time;even when there are no outside signs to mark the passing of time.The biological clock tells plants when to form flowers and when the flowers should open.It tells insects when to leave the protective cocoon and fly a way.And it tells animals when to eat，sleep and wake.It controls body temperature，the release of some hormones and even dreams.These natural daily events are circadian rhythms.Man has known about them for thousands of years.But the first scientific observation of circadian rhythms was not made until1729. In that year French astronomer，Jean-Jacques d“Ortous de Mairan，noted that one of his plants opened it s leaves at the same time every morning，and closed them at the same time every night.The plant did this even when he kept it in a dark place all the ter scientists wondered about circadian rhythms in humans.They learned that man”s biological clock actually keeps time with a day of a little less than 25hours instead of the24hours on a man-made clock.About four years ago an American doctor，Eliot Weitzman，established a laboratory to study how our biological clock works.The people in his experiments are shut off from the outside world.They are free to listen to and live by their circadian rhythms.Dr.Weitzman hopes his research will lead to effective treatments for common sleep problems and sleep disorders caused by ageing and mental illness.The laboratory is inthe Monteflore Hospital in New York City.It has two living areas with three small rooms in each.The windows are covered，so no sunlight o r moonlight comes in.There are no radios or television receivers. There is a control room between the living areas.It contains computers，one-way cameras and other electronic devices for observing the person in the living area.The instruments measure heartbeat，body temperature，hormones in the blood，other substances in the urine and brain waves during sleep.A doctor or medical technician is on duty in the control room24hours a day during an experiment.They do not work the same time each day and are not permitted to wear watches，so the person in the experiment has no idea what time it is.In the first four years of research，Dr Weitzman and his assistant have observed16men between the ages of21and80. The men remained in the laboratory for as long as six st month，a science reporter for“The New York Times”newspaper，Dava Sobol，became the first woman to take part in the experiment.She entered the laboratory on June13th and stayed for25days.Miss Sobol wrote reports about the experiment during that time，which were published in the newspaper.（PS:The way to contact yumingkaobo TEL:si ling ling-liu liu ba-l iu jiu qi ba QQ:si jiu san san qi yi liu er liu）1、The biological clock is believed to play an essential role in ____.A、the regulation of body temperatureB、the secretion of hormonesC、animal reproductionD、many aspects of plant and animal physiology2、In his observation，the French scientist noticed that the leaves of a certain plant maintained its opening-and-closing cycles ____.A、even when it was kept in a murky place all dayB、even if it was placed in the moonlightC、even when he was observing it from a dark placeD、even during the night time3、The sentence“They are free to listen to and live by their circadian rhythms.”(In Paragraph4)probably means____.A、They can lead their daily lives according to their biological clocks，without referring to a man-made clock.B、They can listen to the wonderful rhythms of the biological clock and live close to them.C、They can live by regulating their own circadian rhythms.D、They are free from the annoying rhythms of everyday life.4、In the experiment conducted by Mr.Weitzman，the doctor who is on duty does not work the same time each day____.A、insgroupsto observe the abnormal behavior of the people at different timesB、so as not to be recognized by the peopleC、so as to avoid indicating to the people what time it is whenhe starts workD、so as to leave the people“s circadian rhythms in disorder5、Miss Sobol left the laboratory____.A、on June13thB、on June25thC、at the end of JuneD、on July7thKeys to PassageD A A C D本文由“育明考博”整理编辑。
【2024版】2015年南京大学新闻传播学院620传播史论考研真题及详解【圣才出品】
可编辑修改精选全文完整版2015年南京大学新闻传播学院620传播史论考研真题及详解【圣才出品】2015年南京大学新闻传播学院620传播史论考研真题及详解科目一：传播史论一、名词解释（每题5分，共20分）1．卡尔·霍夫兰2．社会顺从理论3．KDKA4．时间偏向的媒介二、简答题（每题10分，共50分）1．简述议程设置理论的作用机制。
2．简述芝加哥学派和哥伦比亚学派观点的异同。
3．简述书面文化的社会影响。
4．经验性研究方法的主要特征和原则。
5．简述法兰克福学派的主要观点。
三、论述题（每题20分，共80分）1．谈谈第三人效果的学术意义及现实意义。
2．有人认为，是广播而不是报纸和电视开启了大众传播时代并促成了大众传播学的诞生。
谈谈你的看法。
3．试论全球化与身份认同之间的关系，以及如何理解与媒介的关系。
4．20世纪70年代，受斯图亚特·霍尔的影响，英国的文化研究学派理论出现转型，你如何理解这一转型，哪些理论促使此转型。
参考答案：一、名词解释（每题5分，共20分）1．卡尔·霍夫兰答：卡尔·霍夫兰是美国实验心理学家，传播学四大先驱之一。
霍夫兰于二战期间担任美国陆战心理实验室主任，研究宣传电影对士兵士气的影响，通过对传播者、传播技巧、传播内容、受传者等进行实验研究，提出影响传播效果的一系列因素。
二战期间和战后，霍夫兰和一批心理学家进行了大量实验，对态度与说服进行了细致研究，形成了具有影响力的“耶鲁学派”。
其传播研究的成果集结于著作《传播与说服》一书中。
霍夫兰对传播学的贡献主要表现在三个方面：①他在彼得森、瑟斯顿等学者的基础上，首次较完善地把心理学控制试验的方法用于传播效果的研究。
②他对军事教育电影的研究不仅从一个方面证明了坎特里尔的观点，而且为打破“魔弹论”的神话提供了更有价值的证据。
③他注意到影响说服效果的多种因素，尤其是说服者及其发出的信息者两个因素，并提出了改善说服效果的一系列有价值的建议，为“可说服性”这个当代传播学的重要课题奠定了基础。
泛函分析9§1-5,习题选讲与答案
第九章 内积空间和希尔伯特空间例题选讲例1． Hilbert 是X 可分的充分必要条件X 存在一个可数的完全规范正交系{}n e证明：若X 是可分的，设{}n x 是X 的一个可数稠密子集。
不妨设{}n x 是线性无关的。
用Gram Schmidt -方法，存在可数的完全规范正交系{}n e ，使span{}1,,n e e L = {}1,,n span x x L 。
这样。
因此{}n e 是完全的。
反之，若{}n e 是X 的一个完全规范正交系，则span {}n e 在X 中稠密。
()01,,1,2,3,n k k k k k k X a ib e a b Q N =⎧⎫=+∈=⎨⎬⎩⎭∑L 是X 中的可数稠密子集，因此 X 是可分的。
证毕例2．求证：P 是Hilbert 空间X 上的投影算子的充分必要条件是：2P P =且*P P =证明：设P 是X 中相对应与闭线性子空间Y 的投影算子。
对任意x ∈X ，存在1x Y∈，2x ∈Y ⊥，使12x x x =+，1Px x =。
对于1x ，1x =10x +，其中1x Y ∈，0Y ⊥∈。
因此11Px x =，即21P x Px Px ==，因此2P P = 设,x y X ∈，12x x x =+，12y y y =+。
其中11,x y Y ∈，22,x y Y ⊥∈。
这样()()()()()1121112,,,,,Px y x y y x y x x Py x Py =+==+=。
这就证明了*P P =。
反之，若P 满足*P P =，*P P =。
令{}Y x Px x ==，则Y 是X 中的线性子空间。
Y 还是闭的。
事实上，若n x Y ∈，0lim n n x x →∞=，则00lim lim n n n n Px Px x x →∞→∞===。
故0x Y ∈，因此Y 是闭的线性子空间，我们要证明P 是Y 上的投影算子。
设x X ∈，则()x Px x Px =+-。
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f ( x ) g ( x )dx M
g ( x ) dx q ，
q
1
1 1 1 ，证明 f Lp () 并且 ‖f‖ M. Lp ( ) p q
第1页
试题编号
共 2 页
5． （15 分）设 X 为紧的度量空间，证明在 X 上存在 Borel 测度 使得对 X 上的任何非负连 续函数 f ，并且 f 0 ，有
Ax, x x ，
其中 , 表示 H 中的内积，证明对任何 y H ，方程
2
Ax y
有唯一的解. 4． （15 分）设 为 n 中的有界开集， f 为 上的 Lebesgue 可测函数，并且存在 M 0 ， 使得对 上的任何有界连续函数 g ，有
其中 1 q ，
X
f d 0 .
6． （15 分）设 (, , ) 为正测度空间，如果存在一列可测子集 {En } 使得当 n m 时，
En Em ，并且 0 ( En ) ，证明 Banach 空间 L1 (, ) 不是自反的.
第2页
p
证明：⑴ f Lp (, ) ； ⑵ lim
p
x | f n ( x )| M
f n ( x ) dx .
p
n
f n ( x ) f ( x ) d 0 .
3． （20 分）设 H 为 Hilbert 空间， A : H H 为有界线性算子，并且存在 0 使得对任 何 x H ，有
2． （20 分）设 (, ) 为正测度空间， () ，再设 { f n } L (, ), 1 p ，满 足如下条件： （i）存在 上的可测函数 f 使得 { f n } 在 上几乎处处收敛于 f ； （ii）对任意 0 ，存在 M 0 ，使得对任何 n 有
南京大学 2015 年博士学 数 学 满分： 分
注意：①所有答案必须写在答题纸或答题卡上，写在本试题纸或草稿纸上均无效； ②本科目不允许使用计算器。 1． （15 分）求
lim (1 cos n x ) x 3dx .
n