linear operator??!

**miss_lolitta** · December 7th 2006, 12:14 PM

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

**Rebesques** · June 14th 2007, 12:43 PM

The spaces here should have a topology, or continuity is pointless.

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 $T:X\rightarrow Y$ be
additive (i.e. $T(x+x')=T(x)+T(x'), \ x,x'\in X$ ) and continuous.

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

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

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

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

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

**Rebesques** · June 14th 2007, 02:48 PM

It would also be kind of instructive, if someone proves the convergence $\lambda_n T(x)\rightarrow \lambda T(x)$ by using the topology of the underlying space, instead of the cheesy way followed here.

**Plato** · June 14th 2007, 03:32 PM

Originally Posted by Rebesques

It would also be kind of instructive, if someone proves the convergence $\lambda_n T(x)\rightarrow \lambda T(x)$ by using the topology of the underlying space, instead of the cheesy way followed here.

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 $\lambda f(x) \not= f\left( {\lambda x} \right)$ that means: $0 < \left\| {\lambda f(x) - f\left( {\lambda x} \right)} \right\| \le \left\| {\lambda f(x) - \lambda _n f\left( x \right)} \right\| + \left\| {f\left( {\lambda _n x} \right) - f\left( {\lambda x} \right)} \right\|$ .

Do you see the contradiction?

**Rebesques** · June 14th 2007, 03:49 PM

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

Oops. In the books I used to read, linear space just went for vector space.

if you were sure of that last sentence.

Which one? That $\lambda_n T(x)\rightarrow \lambda T(x)<br />$ looks good in a TVS?

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

In a TVS, the fun begins

**Rebesques** · June 16th 2007, 07:28 AM

And forgot to mention

Using the fact $<br /> T(\lambda x)\leftarrow T(\lambda_n x)=\lambda_n T(x)\rightarrow\lambda T(x)<br />$ to get $T(\lambda x)=\lambda T(x)$ , means that limits in this space must be unique. So it should be atleast a Hausdorff TVS.

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