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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- Dec 7th 2006, 12:14 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 - Jun 14th 2007, 12:43 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 $\displaystyle T:X\rightarrow Y$ be

additive (i.e. $\displaystyle T(x+x')=T(x)+T(x'), \ x,x'\in X$) and continuous.

To show $\displaystyle T$ is linear, we show that for $\displaystyle \lambda\in \mathbb{R}$, we have $\displaystyle T(\lambda x)=\lambda T(x)$(%).

We easily see that $\displaystyle T(nx)=T(x+x+\ldots+x)=T(x)+T(x)+\ldots+T(x)=nT(x)$, for $\displaystyle n\in\mathbb{N}$(*). Also, we notice that by additivity $\displaystyle 0=T(0)=T(x-x)=T(x)+T(-x)$, so $\displaystyle T(-x)=-T(x)$. So (*) also holds for negative n.

For $\displaystyle \lambda=\frac{1}{m} \in\mathbb{Q}$, we have $\displaystyle T(x)=T(\frac{m}{m}x)=mT(\frac{1}{m}x) \ \mbox{or} \ T(\frac{1}{m}x)=\frac{1}{m}T(x)$. Now (*) grants us (%) for the rationals.

For $\displaystyle \lambda\in\mathbb{R}-\mathbb{Q}$, consider a sequence of rationals $\displaystyle \lambda_n\rightarrow\lambda.$ We have

$\displaystyle T(\lambda_n x)=T(\lambda_n x-\lambda x+\lambda x)=T((\lambda_n-\lambda)x)+T(\lambda x)$. Now $\displaystyle T(\lambda_n x)\rightarrow T(\lambda x)$ and $\displaystyle T((\lambda_n -\lambda)x)\rightarrow T(0)=0$, thus

$\displaystyle T(\lambda x)\leftarrow T(\lambda_n x)=\lambda_n T(x)\rightarrow\lambda T(x)$. qed - Jun 14th 2007, 02:48 PMRebesques
It would also be kind of instructive, if someone proves the convergence $\displaystyle \lambda_n T(x)\rightarrow \lambda T(x)$ by using the topology of the underlying space, instead of the cheesy way followed here.

- Jun 14th 2007, 03:32 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 $\displaystyle \lambda f(x) \not= f\left( {\lambda x} \right)$ that means: $\displaystyle 0 < \left\| {\lambda f(x) - f\left( {\lambda x} \right)} \right\| \le \left\| {\lambda f(x) - \lambda _n f\left( x \right)} \right\| + \left\| {f\left( {\lambda _n x} \right) - f\left( {\lambda x} \right)} \right\|$.

Do you see the contradiction? - Jun 14th 2007, 03:49 PMRebesquesQuote:

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

Quote:

if you were sure of that last sentence.

$ looks good in a TVS? :confused:

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

In a TVS, the fun begins :o - Jun 16th 2007, 07:28 AMRebesques
And forgot to mention :o

Using the fact $\displaystyle

T(\lambda x)\leftarrow T(\lambda_n x)=\lambda_n T(x)\rightarrow\lambda T(x)

$ to get $\displaystyle T(\lambda x)=\lambda T(x)$, means that limits in this space must be unique. So it should be atleast a Hausdorff TVS.