1. ## linear 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

2. 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 $\displaystyle T:X\rightarrow Y$ 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

3. 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.

4. Originally Posted by Rebesques
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.
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: $\displaystyle 0 < \left\| {\lambda f(x) - f\left( {\lambda x} \right)} \right\| \le \left\| {\lambda f(x) - \lambda _n f\left( x \right)} \right\| + \left\| {f\left( {\lambda _n x} \right) - f\left( {\lambda x} \right)} \right\|$.

5. 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 $\displaystyle \lambda_n T(x)\rightarrow \lambda T(x)$ looks good in a TVS?

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

In a TVS, the fun begins

6. And forgot to mention

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.