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 ?

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- May 8th 2010, 04:09 PMneset44Proof 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 PMDrexel28
- May 9th 2010, 12:19 AMneset44
sequence in banach space

- May 9th 2010, 01:11 AMOpalg
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 AMHallsofIvy
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 AMneset44
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 AMOpalg
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 AMneset44
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 AMHallsofIvy
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 AMneset44
i asked and My teacher said . It is a scalar multiplication

- May 10th 2010, 03:25 AMHallsofIvy
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 PMneset44
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 PMJose27
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 AMmabruka
My guess is that he is talking about a Banach Algebra.