泛函分析上册答案.doc

1.1.5 1.1.61.1.71.2.21.2.31.2.41.3.31.3.41.3.51.3.71.3.81.3.91.4.1 1.4.5-61.4.91.4.111.4.121.4.131.4.141.4.151.4.171.5.1证明：(1) () 若xint(E)，存在d 0，使得Bd (x) E注意到x + x/n x ( n )，故存在N N+，使得x + x/N Bd (x) E即x/( N/( 1 + N ) ) E因此P(x) N/( 1 + N ) 1() 若P(x) 1，使得y = a xE因qint(E)，故存在d 0，使得Bd (q) E令h = d (a - 1)/a，zBh (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) 0，存在yE，使得| x - y | e/2因ny/(n + 1) y ( n )故存在N N+，使得| Ny/(N + 1) - y | e/2令z = Ny/(N + 1)，则zE，且P(z) N/(N + 1) 1，由(1)知z int(E)而| z - x | | z - y | + | y - x | 0，故Ax的各分量也非负但不全为零xC，设f (x) = (Ax)/( 1 i n (Ax)i )，则f (x)C容易验证f : C C还是连续的由Brouwer不动点定理，存在f的不动点x0C即f (x0) = x0，也就是(Ax0)/( 1 i n (Ax0)i ) = x0令l = 1 i n (Ax0)i，则有Ax0 = l x01.5.6证明：设B = uC0, 1 | 0, 1 u(x) dx = 1，u(x) 0 ，则B是C0, 1中闭凸集设max (x, y)0, 10, 1 K(x, y) = M，min (x, y)0, 10, 1 K(x, y) = m，0, 1 (0, 1 K(x, y) dy) dx = N，max x0, 1 | 0, 1 K(x, y) dy |= P令(S u)(x) = (0, 1 K(x, y) u(y) dy)/(0, 1 (0, 1 K(x, y) u(y) dy) dx )则0, 1 (S u)(x) dx = 1，u(x) 0；即S uB因此S是从B到B内的映射u, vB，| 0, 1 K(x, y) u(y) dy - 0, 1 K(x, y) v(y) dy |= | 0, 1 K(x, y) (u(y) - v(y) dy | = max x0, 1 | 0, 1 K(x, y) (u(y) - v(y) dy | M | u - v |；因此映射u # 0, 1 K(x, y) u(y) dy在B上连续类似地，映射u # 0, 1 (0, 1 K(x, y) u(y) dy) dx也在B上连续所以，S在B上连续下面证明S(B)列紧首先，证明S(B)是一致有界集uB，| S u | = | (0, 1 K(x, y) u(y) dy )/(0, 1 (0, 1 K(x, y) u(y) dy) dx )| = max x0, 1 | 0, 1 K(x, y) u(y) dy |/(0, 1 (0, 1 K(x, y) u(y) dy) dx ) (M 0, 1 u(y) dy |/(m 0, 1 (0, 1 u(y) dy) dx ) = M/m，故S(B)是一致有界集其次，证明S(B)等度连续uB，t1, t20, 1，| (S u)(t1) - (S u)(t2) | = | 0, 1 K(t1, y) u(y) dy - 0, 1 K(t2, y) u(y) dy |/(0, 1 (0, 1 K(x, y) u(y) dy) dx ) 0, 1 | K(t1, y) - K(t2, y) | u(y) dy /(m0, 1 (0, 1 u(y) dy) dx ) (1/m) max y0, 1 | K(t1, y) - K(t2, y) |由K(x, y)在0, 10, 1上的一致连续性，e 0，存在d 0，使得(x1, y1), (x2, y2)0, 1，只要| (x1, y1) - (x2, y2) | d，就有| K(x1, y1) - K(x2, y2) | m e故只要| t1 - t2 | d 时，y0, 1，都有| K(t1, y) - K(t2, y) | m e此时，| (S u)(t1) - (S u)(t2) | (1/m) max y0, 1 | K(t1, y) - K(t2, y) | (1/m) m e = e故S(B)是等度连续的所以，S(B)是列紧集根据Schauder不动点定理，S在C上有不动点u0令l = (0, 1 (0, 1 K(x, y) u0(y) dy) dx则(S u0)(x) = (0, 1 K(x, y) u0(y) dy)/l = (T u0)(x)/l因此(T u0)(x)/l = u0(x)，T u0 = l u0显然上述的l和u0满足题目的要求1.6.1 (极化恒等式)证明：x, yX，q(x + y) - q(x - y) = a(x + y, x + y) - a(x - y, x - y)= (a(x, x) + a(x, y) + a(y, x) + a(y, y) - (a(x, x) - a(x, y) - a(y, x) + a(y, y)= 2 (a(x, y) + a(y, x)，将i y代替上式中的y，有q(x + i y) - q(x - i y) = 2 (a(x, i y) + a(i y, x)= 2 (-i a(x, y) + i a( y, x)，将上式两边乘以i，得到i q(x + i y) - i q(x - i y) = 2 ( a(x, y) - a( y, x)，将它与第一式相加即可得到极化恒等式1.6.2证明：若Ca, b中范数| |是可由某内积( , )诱导出的，则范数| |应满足平行四边形等式而事实上，Ca, b中范数| |是不满足平行四边形等式的，因此，不能引进内积( , )使其适合上述关系范数| |是不满足平行四边形等式的具体例子如下：设f(x) = (x a)/(b a)，g(x) = (b x)/(b a)，则| f | = | g | = | f + g | = | f g | = 1，显然不满足平行四边形等式1.6.3证明：xL20, T，若| x | = 1，由Cauchy-Schwarz不等式，有| 0, T e - ( T - t ) x(t ) dt |2 (0, T (e - ( T - t )2 dt ) (0, T ( x(t )2 dt )= 0, T (e - ( T - t )2 dt = e - 2T 0, T e 2t dt = (1- e - 2T )/2因此，该函数的函数值不超过M = (1- e - 2T )/2)1/2前面的不等号成为等号的充要条件是存在lR，使得x(t ) = l e - ( T - t )再注意| x | = 1，就有0, T (l e - ( T - t )2 dt = 1解出l = (1- e - 2T )/2) - 1/2故当单位球面上的点x(t ) = (1- e - 2T )/2) - 1/2 e - ( T - t )时，该函数达到其在单位球面上的最大值(1- e - 2T )/2)1/21.6.4证明：若xN ，则yN，(x, y) = 0而M N，故yM，也有(x, y) = 0因此xM 所以，N M 1.6.51.6.6解：设偶函数集为E，奇函数集为O显然，每个奇函数都与正交E故奇函数集O E fE ，注意到f总可分解为f = g + h，其中g是奇函数，h是偶函数因此有0 = ( f, h) = ( g + h, h) = ( g, h) + ( h, h) = ( h, h)故h几乎处处为0即f = g是奇函数所以有 E O这样就证明了偶函数集E的正交补E 是奇函数集O1.6.7 证明：首先直接验证，cR，S = e 2p i n x | nZ 是L2c, c + 1中的一个正交集再将其标准化，得到一个规范正交集S1 = jn(x) = dn e 2p i n x | nZ 其中的dn = | e 2p i n x | (nZ)，并且只与n有关，与c的选择无关(1) 当b a =1时，根据实分析结论有S = q当b a 1时，若uL2a, b，且uS ，我们将u延拓成a, a + 1上的函数v，使得v(x) = 0 (x(b, a + 1)则vL2a, a + 1同时把S = e 2p i n x | nZ 也看成L2a, a + 1上的函数集那么，在L2a, a + 1中，有vS 根据前面的结论，v = q因此，在L2a, b中就有u = q故也有S = q；(2) 分成两个区间a, b 1)和b 1, b来看在a, b 1)上取定非零函数u(x) = 1 ( xa, b 1) )记pn = a, b 1) u(x)jn(x) dx我们再把u看成是b 2, b 1上的函数(u在b 2, a)上去值为0)那么pn就是u在L2b 2, b 1上关于正交集S1 = jn(x) | nZ 的Fourier系数由Bessel不等式，nZ | pn |2 m，则n - m - 1 0，从zn - m - 1而解析( zn/(2p)1/2, zm/(2p)1/2 ) = (1/i)| z | = 1 ( zn/(2p)1/2 (z*)m/(2p)1/2 )/z dz= (1/(2pi)| z | = 1 zn (z*)m/z dz = (1/(2pi)| z | = 1 zn - m - 1 dz = 0因此， zn/(2p)1/2 n 0是正交规范集1.6.91.6.10证明：容易验证en fn是正交规范集，下面只证明en fn是X的基xX，由正交分解定理，存在x关于X0的正交分解x = y + z，其中y X0，z X0因en, fn分别是X0和X0的正交规范基，故y = nN ( y, en ) en，z = nN ( z, fn ) fn 因z X0，故(x, en) = ( y + z, en) = ( y, en) + ( z, en) = ( y, en)因y X0，故(x, fn) = ( y + z, fn) = ( y, fn) + ( z, fn) = ( z, fn)故x = y + z = nN ( y, en ) en + nN ( z, fn ) fn= nN ( x, en ) en + nN ( x, fn ) fn因此en fn是X的正交规范基1.6.11证明：首先，令j k (z) = ( k +1 )/p)1/2 z k ( k 0 )，则 j k k 0是H 2(D)中的正交规范基那么，u(z)H 2(D)，设u(z) = k 0 a k z k，则kN，有(u, j k) = D u(z) j k(z)* dxdy = D ( j 0 a j z j) j k(z)* dxdy= j 0 a j (p/( j +1 )1/2D ( j +1 )/p)1/2 z j j k(z)* dxdy= j 0 a j (p/( j +1 )1/2D j j(z) j k(z)* dxdy= j 0 a j (p/( j +1 )1/2 (j j, j k)= a k (p/( k +1 )1/2即u(z)的关于正交规范基 j k k 0的Fourier系数为a k (p/( k +1 )1/2 ( k 0 )(1) 如果u(z)的Taylor展开式是u(z) = k 0 b k z k，则u(z)的Fourier系数为b k (p/( k +1 )1/2 ( k 0 )由Bessel不等式， k 0| b k (p/( k +1 )1/2 |2 | u | +，于是有 k 0| b k |2/( k +1 ) +(2) 设u(z), v(z)H 2(D)，并且u(z) = k 0 a k z k，v(z) = k 0 b k z k则u(z) = k 0 a k (p/( k +1 )1/2j k (z)，v(z) = j 0 b j (p/( j +1 )1/2j j (z)，(u, v) = ( k 0 a k (p/( k +1 )1/2j k (z), j 0 b j (p/( j +1 )1/2j j (z) )= k 0 j 0 (a k (p/( k +1 )1/2j k (z), b j (p/( j +1 )1/2j j (z) = k 0 j 0 (a k (p/( k +1 )1/2 b j*(p/( j +1 )1/2) (j k (z), j j (z)= k 0 (a k (p/( k +1 )1/2 b k* (p/( k +1 )1/2) = p k 0 (a k b k* )/( k +1 )(3) 设u(z)H 2(D)，且u(z) = k 0 a k z k因1/(1 - z) = k 0 z k，1/(1 - z)2 = k 0 (k +1) z k，其中| z | 1故当| z | 1时，有1/(1 - | z | )2 = k 0 (k +1) | z | k根据(2)，| u(z) |2 = p k 0 (a k a k* )/( k +1 ) = p k 0 | a k |2/( k +1 )| u |2/(1 - | z |)2 = (p k 0 | a k |2/( k +1 ) ( k 0 (k +1) | z | k ) (p k 0 | a k |2/( k +1 ) | z | k) ( k 0 (k +1) | z | k ) p ( k 0 ( | a k |/( k +1 )1/2 | z | k/2) (k +1)1/2 | z | k/2 )2 (Cauchy-Schwarz不等式)= p ( k 0 | a k | | z | k )2 p | k 0 a k z k |2 = p | u(z) |2 ，故| u(z) | | u |/(p1/2 ( 1 - | z | )(4) 先介绍复分析中的Weierstrass定理：若 fn 是区域U C上的解析函数列，且 fn 在U上内闭一致收敛到 f，则f在U上解析（见龚升简明复分析）回到本题设 un 是H 2(D)中的基本列则zD，由(3)知 un(z) 是C中的基本列，因此是收敛列设un(z) u(z)对C中任意闭集F D，存在0 r 0，存在NN+，使得m, n N，都有| un - um | e p1/2 ( 1 - r )再由(3)，zF，| un(z) - um(z) | | un - um |/(p1/2 ( 1 - | z | ) | un - um |/(p1/2 ( 1 - r ) -即f在X上有下界，因而f在C有下确界m = inf xC f (x)注意到a(x, y)实际上是X上的一个内积，记它所诱导的范数为| x |a = a(x, x)1/2，则| |a与| |是等价范数因此f (x) = a(x, x) - Re(u0, x) = | x |a2 - Re(u0, x)设C中的点列 xn 是一个极小化序列，满足m f (xn ) m + 1/n ( nN+ )则由平行四边形等式，| xn - xm |a2 = 2(| xn |a2 + | xm |a2 ) - 4| (xn + xm)/2 |a2 = 2( f (xn) + Re(u0, xn) + f (xm) + Re(u0, xm) ) - 4( f (xn + xm)/2) + Re(u0, (xn + xm)/2)= 2( f (xn) + f (xm) - 4 f (xn + xm)/2) + 2 Re( (u0, xn) + (u0, xm) - (u0, xn + xm) )= 2( f (xn) + f (xm) - 4 f (xn + xm)/2) 2( m + 1/n + m + 1/m ) - 4 m= 2(1/n + 1/m) 0 ( m, n )因此| xn - xm |2 (1/d) | xn - xm |a2 0 ( m, n )即 xn 为X中的基本列由于X完备，故 xn 收敛设xn x0 ( n )则| xn - x0 |a2 M | xn - x0 |2 0 ( m, n )而由内积a( , )，( , )的连续性，有a( xn , xn ) a( x0 , x0 )，且(u0, xn) (u0, x0)，( n )因此f (xn) = a(xn, xn) - Re(u0, xn) a(x0, x0) - Re(u0, x0) = f (x0)，( n )由极限的唯一性，f (x0) = m = inf xC f (x)至此，我们证明了f 在C上有最小值下面说明最小值点是唯一的若x0, y0都是最小值点，则交错的点列 x0, y0, x0, y0, x0, . 是极小化序列根据前面的证明，这个极小化序列必须是基本列，因此，必然有x0 = y0所以最小值点是唯一的最后我们要证明最小点x0C满足给出的不等式xC，t0, 1，有x0 + t ( x - x0)C，因此有f (x0 + t ( x - x0) f (x0)即| x0 + t ( x - x0) |a2 - Re(u0, x0 + t ( x - x0) | x0 |a2 - Re(u0, x0)展开并整理得到t Re ( 2a(x0, x - x0) - (u0, x - x0) ) - t2 | x - x0 |a2故当t(0, 1，有Re ( 2a(x0, x - x0) - (u0, x - x0) ) - t | x - x0 |a2令t 0就得到 Re ( 2a(x0, x - x0) - (u0, x - x0) ) 02.1.22.1.32.1.42.1.52.1.62.1.72.1.82.1.92.2.22.2.52.3.12.3.3-22.3.42.3.52.3.72.3.82.3.92.3.112.3.122.3.132.3.142.4.42.4.52.4.62.4.72.4.82.4.92.4.102.4.112.4.122.4.132.4.142.5.42.5.52.5.72.5.82.5.102.5.122.5.182.5.202.5.222.6.12.6.22.6.32.6.4