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Math Help - linear operator??!

  1. #1
    Junior Member miss_lolitta's Avatar
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    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
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  2. #2
    Super Member Rebesques's Avatar
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    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
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  3. #3
    Super Member Rebesques's Avatar
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    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.
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  4. #4
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    Quote Originally Posted by Rebesques View Post
    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?
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  5. #5
    Super Member Rebesques's Avatar
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    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
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  6. #6
    Super Member Rebesques's Avatar
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    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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