1、张恭庆泛函分析上册答案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x int(E),存在 0,使得B (x) E注意到x + x/n x ( n ),故存在N +,使得x + x/N B (x) E即x/( N/( 1 + N ) ) E因此P(x) N/( 1 + N ) 1( ) 若P(x) 1,使得y = a x E因 int(E),故存在 0,使得B ( ) E令 =
2、(a 1)/a, z B (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y E,使得| x y | /2因ny/(n + 1) y ( n )故存在N +,使得| Ny/(N + 1) y | /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 | 0,故Ax的各分量也非负但不全为零 x C,设f (x) = (A
3、x)/( 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x0 C即f (x0) = x0,也就是(Ax0)/( 1 i n (Ax0)i ) = x0令 = 1 i n (Ax0)i,则有Ax0 = x01.5.6证明:设B = u C0, 1 | 0, 1 u(x) dx = 1,u(x) 0 ,则B是C0, 1中闭凸集设max (x, y) 0, 1 0, 1 K(x, y) = M,min (x, y) 0, 1 0, 1 K(x, y) = m, 0, 1 ( 0, 1 K(x, y) dy) dx = N,max
4、 x 0, 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 u B因此S是从B到B的映射 u, v B,| 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 x 0, 1 | 0, 1 K(x, y) (u(y) v(y) dy | M | u v |;因此映射u 0, 1 K
5、(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 u B,| S u | = | ( 0, 1 K(x, y) u(y) dy )/( 0, 1 ( 0, 1 K(x, y) u(y) dy) dx )| = max x 0, 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 ) =
6、M/m,故S(B)是一致有界集其次,证明S(B)等度连续 u B, t1, t2 0, 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 /(m 0, 1 ( 0, 1 u(y) dy) dx ) (1/m) max y 0, 1 | K(t1, y) K(t2, y) |由K(x, y)在0, 1 0, 1上的一致连续性, 0,存在 0,使得 (x1,
7、 y1), (x2, y2) 0, 1,只要| (x1, y1) (x2, y2) | ,就有| K(x1, y1) K(x2, y2) | m 故只要| t1 t2 | 时,y 0, 1,都有| K(t1, y) K(t2, y) | m 此时,| (S u)(t1) (S u)(t2) | (1/m) max y 0, 1 | K(t1, y) K(t2, y) | (1/m) m = 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 = ( 0, 1 ( 0, 1 K(x, y) u0(y) dy) dx则(S u0)(x) = ( 0,
8、 1 K(x, y) u0(y) dy)/ = (T u0)(x)/ 因此(T u0)(x)/ = u0(x),T u0 = u0显然上述的 和u0满足题目的要求1.6.1 (极化恒等式)证明: x, y X,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(
9、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中数| |是可由某积( , )诱导出的,则数| |应满足平行四边形等式而事实上,Ca, b中数| |是不满足平行四边形等式的,因此,不能引进积( , )使其适合上述关系数| |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 | =
10、| f g | = 1,显然不满足平行四边形等式1.6.3证明: x L20, T,若| x | = 1,由Cauchy-Schwarz不等式,有| 0, T e ( T ) x( ) d |2 ( 0, T (e ( T )2 d ) ( 0, T ( x( )2 d )= 0, T (e ( T )2 d = e 2T 0, T e 2 d = (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 ,使得x( ) = e ( T )再注意| x | = 1,就有 0, T ( e ( T )2 d = 1解出 = (1
11、e 2T )/2) 1/2故当单位球面上的点x( ) = (1 e 2T )/2) 1/2 e ( 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x N ,则 y N,(x, y) = 0而M N,故 y M,也有(x, y) = 0因此x M 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 f E ,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, c ,S = e 2 i n x | n 是L2c, c + 1中的一个正交集再将其标准化,得到一个规正交集S1 = n(x) = dn e 2 i n x | n 其中的dn = | e 2 i n x | (n ),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 = 当b a 1时,若u L2a, b,且u S ,我们将u延拓成a, a + 1上
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