Hi for all,,

Can someone prove that:

If A is an additive operator and continuous from a linear space X into a linear space Y;then A is a linear operator ??!

thanks for all respondings

regards..

Miss_Lolitta

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- December 7th 2006, 01:14 PMmiss_lolittalinear operator??!
Hi for all,,

Can someone prove that:

If A is an additive operator and continuous from a linear space X into a linear space Y;then A is a linear operator ??!

thanks for all respondings

regards..

Miss_Lolitta - June 14th 2007, 01:43 PMRebesques
The spaces here should have a topology, or continuity is pointless. :o

So, consider them to be real topological vector spaces (which is consistent with the fact that we need to have a linear map here).

Let be

additive (i.e. ) and continuous.

To show is linear, we show that for , we have (%).

We easily see that , for (*). Also, we notice that by additivity , so . So (*) also holds for negative n.

For , we have . Now (*) grants us (%) for the rationals.

For , consider a sequence of rationals We have

. Now and , thus

. qed - June 14th 2007, 03:48 PMRebesques
It would also be kind of instructive, if someone proves the convergence by using the topology of the underlying space, instead of the cheesy way followed here.

- June 14th 2007, 04:32 PMPlato
BTW. In my experience the term

*linear space*means a__normed linear space__.

Thus the norm induces a metric space.

You did a good job is showing the standard properties.

Having graded this sort of proof for years, I did wonder if you were sure of that last sentence.

Suppose that that means: .

Do you see the contradiction? - June 14th 2007, 04:49 PMRebesquesQuote:

In my experience the term linear space means a normed linear space.

Quote:

if you were sure of that last sentence.

In a metric space, your thingie is much more neat.

In a TVS, the fun begins :o - June 16th 2007, 08:28 AMRebesques
And forgot to mention :o

Using the fact to get , means that limits in this space must be unique. So it should be atleast a Hausdorff TVS.