Results 1 to 6 of 6

Thread: linear operator??!

  1. #1
    Junior Member miss_lolitta's Avatar
    Joined
    Jan 2006
    From
    in math books
    Posts
    47

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Quote Originally Posted by Rebesques View Post
    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: $\displaystyle 0 < \left\| {\lambda f(x) - f\left( {\lambda x} \right)} \right\| \le \left\| {\lambda f(x) - \lambda _n f\left( x \right)} \right\| + \left\| {f\left( {\lambda _n x} \right) - f\left( {\lambda x} \right)} \right\|$.

    Do you see the contradiction?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear operator
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: Jun 5th 2011, 10:41 PM
  2. Linear operator
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 30th 2011, 09:43 PM
  3. Linear Operator
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 25th 2011, 11:59 PM
  4. Is this Linear Operator?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 14th 2010, 01:59 PM
  5. linear operator
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 16th 2008, 05:12 PM

Search Tags


/mathhelpforum @mathhelpforum