# Proof for banach Space 2

• May 8th 2010, 04:09 PM
neset44
Proof for banach Space 2
Prove that, In a normed linear space Xn converges to X and Yn converges to Y then XnYn converges to XY .

Xn and Yn are sequence in banach space.

Anyone idea ?
• May 8th 2010, 07:15 PM
Drexel28
Quote:

Originally Posted by neset44
In a normed linear space Xn converges to X and Yn converges to Y then Xn.Yn converges to X.Y

Anyone idea ?

What is $\displaystyle X_n.Y_n$?
• May 9th 2010, 12:19 AM
neset44
sequence in banach space
• May 9th 2010, 01:11 AM
Opalg
Quote:

Originally Posted by Drexel28
What is $\displaystyle X_n.Y_n$?

Quote:

Originally Posted by neset44
sequence in banach space

What Drexel28 is asking is: What does the dot mean in $\displaystyle X_n.Y_n$? It looks as though it is supposed to be an inner product. But a Banach space doesn't have an inner product (unless it is a Hilbert space), so it's hard to know what this problem is supposed to be about.
• May 9th 2010, 01:54 AM
HallsofIvy
In fact, although you titled this "proof for Banach space 2", you only said "In a normed linear space Xn". Not all normed spaces are Banach spaces. Please state the problem in full.
• May 10th 2010, 01:37 AM
neset44
Xn and Yn are in Banach space. So in a normed linear space Xn converges to X and Yn converges to Y then prove that XnYn (multiply) converges to XY.

That is the full problem.
• May 10th 2010, 01:44 AM
Opalg
Quote:

Originally Posted by neset44
Xn and Yn are in Banach space. So in a normed linear space Xn converges to X and Yn converges to Y then prove that XnYn (multiply) converges to XY.

That is the full problem.

There is no operation of multiplication defined in a Banach space, so this question makes no sense.

In order for a product $\displaystyle X_n.Y_n$ to be defined, there must be some additional structure in the space. It has to be either a Hilbert space (with an inner product), or a Banach algebra (with a multiplicative structure).

Also, as HallsofIvy has pointed out, a Banach space is not the same as a normed linear space.
• May 10th 2010, 01:59 AM
neset44
i changed product with standart multiply as you see. but anyway;

This is the full problem .

It says on theorem :

In a normed linear space Xn coverges to X and Yn converges to Y => XnYn converges to XY . Prove it ?

Nothing more
• May 10th 2010, 03:14 AM
HallsofIvy
One more time and then we give up!

A "Banach space" is a vector space having a norm such that "Cauchy sequences" converge.

While "scalar multiplication" and "vector addition" are defined in any vector space, multiplication of vectors is NOT generally defined. There is no "standard multiply"! It is impossible to answer your question with knowing how you are defining "XnYn".
• May 10th 2010, 03:21 AM
neset44
i asked and My teacher said . It is a scalar multiplication
• May 10th 2010, 03:25 AM
HallsofIvy
Are you saying that {Xn} is a sequence of scalars that converge to X and {Yn} is a sequence of vectors in a Banach space that converge to Y? That was not what you said initially!
• May 10th 2010, 12:06 PM
neset44
as a result that what he given to me Xn converges to x and Yn converges to y then XnYn converges to xy ?

there is not any more information. then he said between Xn and Yn multiplication is scalar ...

that is :S(Angry)
• May 10th 2010, 07:22 PM
Jose27
Quote:

Originally Posted by neset44
as a result that what he given to me Xn converges to x and Yn converges to y then XnYn converges to xy ?

there is not any more information. then he said between Xn and Yn multiplication is scalar ...

that is :S(Angry)

You really need to improve you english writing skills.

That aside, if you meant $\displaystyle (x_n) \subset \mathbb{C}$ (or $\displaystyle \mathbb{R}$ it works the same) and $\displaystyle (y_n) \subset X$ (where this last one is a normed space over whichever field you pick from the above) such that $\displaystyle |x_n - x| \rightarrow 0$ and $\displaystyle \| y_n - y \| \rightarrow 0$ then

$\displaystyle \| x_ny_n - xy \| \leq \| x_ny_n -x_ny\| + \|x_ny-xy\| \leq M\| y_n-y\| + |x_n-x| \|y\| \rightarrow 0$

where $\displaystyle |x_n|\leq M$ for all $\displaystyle n$.
• May 11th 2010, 12:17 AM
mabruka
My guess is that he is talking about a Banach Algebra.