Exercises2

Exercises

1. Copy the statements of five definitions and five theorems from one of your

math textbooks. Identify the use of the defined words in the statements of the

theorems. Give examples that illustrate the theorems. Show how the conclusions

of the theorems don’t necessarily hold if the hypotheses are not satisfied.

Definition1 (definition 2.1)

Let X be a vector space over F .A norm on X is a function .:X→R such that for

all x,y ,X andαF,

(1)0x;

(2)0x if and only if x=0,

(3)xx,

(4)yxyx.

EXAMPLE1

The function .:Fn→R defined by 212121)(),...,,(njjnxxxx is a norm on nF.

(1)212121)(),...,,(njjnxxxx0

(2)212121)(),...,,(njjnxxxx=000Xxj

(3)2121212121)()(),...,,(njjnjjnxaaxxxxa

(4)21212121212121212121)()()()(),...,,(),...,,(njjnjjnjjnjjjnnyxxyxyyyxxx

It is a norm.

Definition2 (definition 3.3)

Let X be a complex vector space .an inner product on X is a function (., .):CXXsuch that for all x,y,zX,C,,

(1) (x,x)R,and (x,x)0;

(2) (x,x)=0 if and only if x=0;

(3) ),(),(),(zyzxzyx

(4) ),(),(xyyx

Example

If )(,2XLgf and the function (.,.):FXLXL)()(22 defined by Xdgfgf),(

is an inner product on )(2XL. This inner product will be called the standard inner product on

)(2XL.

Proof

Let )(,2XLgf. Then by holder's inequality ,with p=q=2 (theorem 1.55)and the definition

of)(2XL,212212)()(dgdfdgfXXX,so )(1XLgf

(a)0),(2dfffX

(b)00),(2fdfffX

(c)),(),(),(hghfhghfdhgfhgfXXX

(d)),(),(fgdfgdgfgfXX

so it is an inner product

Definition 3 ( definition 3.26)

Let X be an inner product space and let A be a subset of X . The orthogonal complement of A

is the set 0,:axXxAfor all Aa.

Example (exercise 3.15)

If 0:22nnxlxAfor all Nn,find A.

Solution :

Let 0:122nnxlxSfor allNn.if Sxand Ay then 0),(1nnnyxyx,

thus Ax, and soAS. Conversely, let Axsuppose 012mx for some Nm. The vector12me in the standard

orthonormal basis in2l belongs to A ,so 1212,mmxex, which is a contradiction . Thus

012mx for all Nm, so Sx .hence SA. And so SA

Definition 4 ( definition 4.16)

let X and Y be normed linear spaces and let ),(YXBT. The norm of T is defined

by

1:)(supxxTT.

Example (example 4.18)

If FCTF1,0: is the bounded linear operator defined by T(f)=f(0).then 1T

Solution :

Because ffT)( for all 1,0FCf,hence xkxTkT)(:inf 1for all

Xx. On the other hand , if Cg1,0: is defined by g(x)=1, for all Xxthen

1,0CCg with 11,0:)(supxxggand 1)0()(ggT.hence

TgTgT)(1 . Therefore 1T.

Definition 5 (definition 4.20)

Let X and Y be normed linear spaces and let ),(YXLT. If xxT)( for all

Xxthen T is called an isometry .

Example (example 4.22)

(a) if 221,...),(lxxx then 221,...),,0(lxxy.

(b) The linear transformation 22:llS defined by ,...),,,0(,...),,(321321xxxxxxS

Proof :

(a) since 221,...),(lxxx, ......0222122212xxxx

And so 2ly 22221222122......0)()(xxxxxxSb

Theorem 1(theorem 3.15)

Let X be an inner product space with inner product (.,.) and induced norm .. Then for all

x,yX:

)(22222yxyxyx

Example (3.7)show that the non-standard norm knnxx11on the space kR is not induced

by an inner product.

Proof

From the definition of the norm we have 8222221212121eeee,

4)11(2)(2212211ee.thus the parallelogram rule does not hold and so the norm

cannot be induced by an inner product.

Theorem 2 ()

2. Examine the proofs of three or four mathematical theorems. What is the

structure of these proofs? Identify where the hypotheses of the theorems are

used in the proofs.

Theorem2.25(Riesz' Lemmma )

Suppose that X is a normed vector space ,Y is a closed linear subspace of X such

that XYand ɑ is a real number such that 0

such that

1x andzx for all yY.

Proof

First , begin with the hypotheses of Y is a closed linear subspace of X such that XY.

We have 0:infYzzxd,where x X/Y by part (d) of theorem 1.25

(0:,infAyyxdAx)

With the hypotheses of 0

that

1dzx.

Next ,we should make 1x,so let zxzxx.then 1x

Last ,we show zx

ddyzxzxzxzxyzxzxzzxxyzxzxyx)()(11

It is proved .

Theorem 3.34

Let Y be a closed linear subspace of a Hilbert space H. For any xH, there exists