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Proof of the collections of sequences are linear spaces or vector space.

by irmctn
Tags: linear, metric, norm, sequence, vector
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irmctn
#1
Dec24-08, 05:19 AM
P: 4
[b]1. The problem statement, all variables and given/known data[/b
Lets s denote the collection of all sequences in lR, let m denote the
collection of all bounded sequences in lR, let c denote the collection
of all convergent sequences in lR, and let Co denote the collection of
all sequences in lR which converge to zero.

(a)With the definition of sum given in (*) and the definition of product of
a sequence and real number given by a(xsubn)=(axsubn), show that each of
these collections has the properties of theorem(**). In each case the zero
element is the sequence teta=(0,0,...,0,..). (We sometimes say that these
collections are linear spaces or vector spaces.)



(b) If X=(xsubn) belongs to one of the collections m,c,csub0, define the norm
of X by lXl=sup{lxsubnl:nEN}. Show that thisnorm function has the properties
of (***). (For this reason, we sometimes say that these collections are
normed linear spaces.)




(*)definition:If X=(xsubn) and Y=(ysubn) are sequences in R to p, then we define their
sum to be the sequence X+Y=(xsubn + ysubn) in R to p, their difference
to be thesequence X-Y=(xsubn - xsubn), and their inner product to be the
sequence X.Y=(xsubn.ysubn) in R which is obtained by taking the inner
product of corresponding terms.
Similarly, if X=(xsubn) is a sequence in R
and if Y=(ysubn) is a sequence in R to p, we define the product of X and Y
to be the sequence in R to p denoted by XY=(xsubnysubn).
Finally,if Y=(ysubn) is a sequence in R with ysubn is not equal to 0, we
can define the quotient of a sequence X=(xsubn) in R to p by Y to be the
sequence X/Y=(xsubn/ysubn).


2. Relevant equations



3. The attempt at a solution
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HallsofIvy
#2
Dec24-08, 05:22 AM
Math
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PF Gold
P: 39,691
Okay, what have YOU done on this? I see no attempt at a solution. Also, you talk about "properties of theorem (**)" and "properties of (***) without saying what they are!
irmctn
#3
Dec24-08, 05:33 AM
P: 4
my project
Attached Files
File Type: doc project math type.doc (96.5 KB, 12 views)

irmctn
#4
Dec24-08, 07:35 AM
P: 4
Proof of the collections of sequences are linear spaces or vector space.

in theorem(**) there are properties about vector space
A1) x+y=y+x
A2)(x+y)+z=x+(y+z)
A3)0+x=x and x+0=x
A4) u=(-1)x satisfies x+u=0
M1)1x=x
M2)b(cx)=(bc)x
D)c(x+y)=cx+cy and (b+c)x=bx+cx
irmctn
#5
Dec24-08, 07:48 AM
P: 4
theorem(***) is the norm properties
(i)lxl>=0;
(ii)lxl=0 iff x=0
(iii)lcxl=lcl lxl
(iv)l lxl-lyl l<=lx+-yl<=lxl+lyl


my problem is i dont`n know how i can explain these properties for a sequence, for example m is collection of all bounded sequences, c is collection of all convergent sequences how i show difference between them?
thanks
HallsofIvy
#6
Dec24-08, 08:35 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,691
Quote Quote by irmctn View Post
in theorem(**) there are properties about vector space
A1) x+y=y+x
Here, x is a sequence {xn} and y is a sequence {xy}. According to your definitions, x+ y is the sequence {xn+ yn} and y+ x is the sequence {yn+ xn}. Are those two sequences equal?

A2)(x+y)+z=x+(y+z)
With x and y as above, z is the sequence {zn}. x+ y is the sequence {xn+ yn} as before so (x+y)+ z is the sequence {(xn+ Yn)+ zn}. For the same reasons, x+ (y+ z) is the sequence {xn+ (yn+ zn)}. Are those two sequences equal?

A3)0+x=x and x+0=x
There exist 0 such that... Suppose 0 is the sequence {0} (for all n). What is the sequence 0+ x? x+ 0?

A4) u=(-1)x satisfies x+u=0
If x is the sequence {xn}, what is (-1)x?

M1)1x=x
If x is the sequence {xn}, what is 1x?

M2)b(cx)=(bc)x
If x is the sequence {xn}, cx is the sequence {cxn} so b(cx) is the sequence {b(cxn)} and (bc)x is the sequence {(bc)xn}. Are those two sequences the same?

D)c(x+y)=cx+cy and (b+c)x=bx+cx
If x is the sequence {xn} and y is the sequence {yn}. What is c(x+y)? What is cx+ cy? Are they the same?
HallsofIvy
#7
Dec24-08, 08:45 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,691
Quote Quote by irmctn View Post
theorem(***) is the norm properties
You are told that if x is the sequence {xn}, |x| is the supremum (least upper bound) of the absolute value all numbers in that sequence. Since these are bounded sequences, that supremum exists.

(i)lxl>=0;
If x is the sequence {xn}, |xn| is never negative so its suprememum can't be negative.

(ii)lxl=0 iff x=0
The "if" part should be trivial. If |x|= 0, then 0 is an upper bound for the absolute values of the numbers in the sequence: |xn<= 0 for all n. But an absolute value cannot be negative

(iii)lcxl=lcl lxl
If x is the sequence {xn}, what is cx? What is |cx| then?

(iv)l lxl-lyl l<=lx+-yl<=lxl+lyl
Do you know that |x|- |y|<= |x+ y| for NUMBERS? And so for each term of the sequences {xn} and {yn}?


my problem is i dont`n know how i can explain these properties for a sequence, for example m is collection of all bounded sequences, c is collection of all convergent sequences how i show difference between them?
thanks
You don't need to show any difference between them.


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