[Peterdlax reference for functional analysis exercises] Chapter 5th norm linear space

 

 

1. (a) proof (6) defines the norm.

(B) prove that they are equivalent in the (5) shift.

 

Proof: $ \ Bex | (z, u) | '\ Leq | (z, u) | \ Leq 2 | (z, u) | ', \ quad | (z, u) | ''\ Leq | (z, u) | \ Leq \ SQRT {2} | (z, u) | ''. \ EEx $

 

2. proof theorem 2.

 

Proof: For $ Y_1, Y_2 \ In \ bar y $, $ \ Bex \ exists \ Y \ ni Y _ {1N} \ To Y_1, \ quad Y \ ni Y _ {2n} \ To Y_2, \ EEx $ and $ \ Bex Y \ ni Ky _ {1N} + Y _ {2n} \ To ky_1 + Y_2. \ EEx $ ky_1 + Y_2 \ In \ bar y $.

 

3. Proof: If $ x $ is a list space and $ y $ is a closed subspace of $ x $, the commercial space $ x/y $ is complete.

 

Proof: If $ [x_n] $ is the Cauchy column in $ x/y $, then $ \ Bex \ forall \ ve> 0, \ exists \ n, \ m> n \ geq n \ rA | [x_m-x_n] | = | [x_m-x_n] | <\ ve. \ EEx $ defined by the norms of $ [x] $, $ \ Bex \ exists \ q_m, Q_n, \ st q_m-x_m \ In Y, \ q_n-x_n \ In y, \ st | q_m-q_n | <2 \ ve. \ EEx $ complete by $ x $, $ \ Bex \ exists \ Q, \ st Q_n \ to Q \ quad \ sex {n \ To \ infty }. \ EEx $ and $ \ beex \ Bea | [x_n]-[Q] | & = | [Q_n]-[Q] | \ & = | [q_n-q] | \ & \ Leq | q_n-q | \ to 0 \ quad \ sex {n \ To \ infty }. \ EEA \ eeex $

 

4. It is proved that every finite-dimensional sub-space in a norm linear space is closed.

 

Proof: Set $ x $ to a norm linear space. $ Y = \ span \ sed {e_1, \ cdots, e_n} $ is its $ N $ dimension linear space. original Certificate $ \ bee \ label {5_4_equiv} C_1 \ sex {\ sum _ {k = 1} ^ n y_k ^ 2} ^ \ frac {1} {2} \ Leq \ Sen {y} \ Leq C_2 \ sex {\ sum _ {k = 1} ^ n y_k ^ 2} ^ \ frac {1} {2 }. \ EEE $ in fact, $ \ bee \ label {5_4_continu} \ Bea \ Sen {y} & =\ Sen {\ sum _ {k = 1} ^ n y_ke_k} \ & \ Leq \ sum _ {k = 1} ^ n | y_k | \ cdot \ Sen {e_k} \ & \ Leq \ sex {\ sum _ {k = 1} ^ n y_k ^ 2} ^ \ frac {1} {2} \ cdot \ sex {\ sum _ {k = 1} ^ n \ Sen {e_k} ^ 2} ^ \ frac {1 }{ 2 }. \ EEA \ EEE $ in turn, consider $ \ Bex f (y) =\sen {y}, \ quad Y \ in S =\sed {Y \ In Y; \ sum _ {k = 1} ^ n y_k ^ 2 = 1 }. \ EEx $ is known by \ eqref {5_4_continu} $ F $ consecutively in the tightly set $ S $, and can be retrieved to the bottom confirmation $ M $. this $ m> 0 $ (otherwise $ \ exists \ Y \ in S, \ ST \ Sen {y} = 0 $ ). therefore, $ \ Bex f (y) \ geq M, \ quad \ sum _ {k = 1} ^ ny_k ^ 2 = 1, \ EEx $ \ Bex f (y) \ geq m \ sex {\ sum _ {k = 1} ^ n y_k ^ 2} ^ \ frac {1} {2}, \ quad \ forall \ Y \ in Y. \ EEx $ since \ eqref {5_4_equiv}, $ \ sex {Y, \ Sen {\ cdot} $ is the same as $ \ BBR ^ N $, it is also complete, and it is the closed sub-space of $ x $.

 

5. It is proved that the upper definite norm in examples (A), examples (C), examples (d), and examples (e) are not strictly times added.

 

Proof: Taking (a) as an example, take $ \ Bex x = (, \ cdots), \ quad y = (, \ cdots ), \ EEx $ then $ \ Bex \ Sen {x + y} = 2 =\sen {x} + \ Sen {y}, \ EEx $ but $ X, Y $ linear independence.

 

6. It is proved that the norm in example (B) and example (f) is not strictly added when $ p = 1 $.

 

Proof: Taking (B) as an example, take $ \ Bex x = (, \ cdots), \ quad y = (, \ cdots ), \ EEx $ \ Bex \ Sen {x + y} = 2 = \ Sen {x} + \ Sen {y}, \ EEx $ but $ X, Y $ linear independence.

 

7. The $ {\ BF m} $ introduced by (41) is linear.

 

Proof: $ \ bee \ label {5_7_linear} \ Bea 2Z '= x' + y' \ rA 2 {\ BF m} \ cfrac {x + y} {2 }= {\ BF m} X + {\ BF m} y. \ EEA \ EEE $ get $ y = 0 $ \ bee \ label {5_7_two} 2 {\ BF m} \ cfrac {x} {2} = {\ BF m} X. \ EEE $ prove by mathematical induction $ \ Bex {\ BF m} (kx) = K {\ BF m} X, \ Quad (k = 1, 2, \ cdots ). \ EEx $ actually, $ \ beex \ Bea {\ BF m} (kx) & ={\ BF m} (x + (k-1) X) \ & =\ cfrac {1} {2} {\ BF m} (2x) + \ cfrac {1} {2} {\ BF m} (2 (k-1) x) \ quad \ sex {\ eqref {5_7_linear }\\\&={\ BF m} X + {\ BF m} (k-1) X) \ quad \ sex {\ eqref {5_7_two }\\\& ={\ BF m} X + (k-1) {\ BF m} X \ quad \ sex {\ mbox {inductive hypothesis }\\\& = K {\ BF m} X. \ EEA \ eeex $ get $ y =-x $ in \ eqref {5_7_linear}, then $ \ Bex {\ BF m} (-x) =-{\ BF m} X, \ EEx $, and $ \ beex \ Bea {\ BF m} (kx) & ={ \ BF m} (-k) (-x) \\& = (-k) {\ BF m} (-x) \ & = (-k) (-{\ BF m} X) \ & = K {\ BF m} X \ quad \ sex {k =-1, -2, \ cdots }. \ EEA \ eeex $ and then by $ \ Bex {\ BF m} X ={\ BF m} \ sex {M \ cdot \ cfrac {1} {m} x} = m \ cdot {\ BF m} \ sex {\ cfrac {1} {m} x} \ EEx $ $ \ Bex {\ BF m} \ sex {\ cfrac {1} {m} X }=\ cfrac {1} {m} {\ BF m} X, \ quad m \ In \ bbz \ BS \ sed {0 }; \ EEx $ \ Bex {\ BF m} \ sex {\ cfrac {k} {m} x} = K {\ BF m} \ sex {\ cfrac {1} {m} X }=\ cfrac {k} {m} {\ BF m} X, \ quad k \ In \ bbz, \ m \ In \ bbz \ BS \ sed {0 }. \ EEx $ by $ {\ BF m} $ is an equidistance between $ {\ BF m} $ consecutive, with $ \ Bex {\ BF m} (\ Al x) = \ Al \ cdot {\ BF m} X, \ quad \ forall \ Al. \ EEx $ finally, by \ eqref {5_7_linear} And \ eqref {5_7_two}, $ \ Bex {\ BF m} (x + y) =\ cfrac {1} {2} {\ BF m} (2x) + \ cfrac {1} {2} {\ BF m} (2y) = {\ BF m} X + {\ BF m} y. \ EEx $

 

8. Prove that $ x $ is complete.

 

Proof: only proof that $ x $ is closed. set $ \ Bex x \ Ni x ^ k \ to X, \ EEx $ \ Bex \ max_n | a ^ k_n-a_n | \ to 0 \ quad \ sex {k \ To \ infty }. \ EEx $ and $ \ Bex \ forall \ ve> 0, \ exists \ K, \ ST \ sup_n | a ^ k_n-a_n | <\ cfrac {\ ve} {2 }. \ EEx $ for this $ K $, known by $ \ DPS {\ vlm {n} a ^ k_n = 0} $ \ Bex \ exists \ n, \ n \ geq n \ rA | a_n ^ k | <\ cfrac {\ ve} {2 }. \ EEx $ hence $ \ Bex n \ geq n \ rA | a_n | \ Leq | a_n-a ^ K_n | + | a ^ K_n | <\ ve. \ EEx $ this indicates $ \ DPS {\ vlm {n} a_n = 0 }$, and $ x \ In x $.

 

 

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