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
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
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?
Oops. In the books I used to read, linear space just went for vector space.In my experience the term linear space means a normed linear space.
Which one? That $\displaystyle \lambda_n T(x)\rightarrow \lambda T(x)if you were sure of that last sentence.
$ looks good in a TVS?
In a metric space, your thingie is much more neat.
In a TVS, the fun begins
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.