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Continuous linear transformation

  1. Aug 20, 2007 #1
    T is a linear transformation from R^m->R^n, prove that T is continuous.

    I have proved that there's always a positive real number C that |T(x)|<=C|x|. How shall I proceed then?
  2. jcsd
  3. Aug 20, 2007 #2


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    May I ask how you have shown that?
  4. Aug 20, 2007 #3
    x is an element of R^m
    let A=max{|T(e1)|,|T(e2)|...|T(em)|)
    then |T(x)|<=A*sum_i(xi)<A/sqrt(m)*sqrt(sum_i(xi^2))=A/sqrt(m)*|x| (AM<RMS)

    sorry, i don't know how to use latex here:frown:
  5. Aug 20, 2007 #4


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    1) You forgot an absolute value sign around xi in line 2:


    2) How did you get

    sum_i|xi|<[1/sqrt(m)]*sqrt(sum_i(xi^2)) ?

    Surely this is not true. Take for instance the case when all but one of the x_i are non-vanishing. The inequality becomes

    |xi|<[1/sqrt(m)]*|xi| <==> sqrt(m)<1

    You had good ideas in your attempt but I suggest working directly with the definition of continuity. Given e>0 and x0 , try to find a d>0 such that |x-x0|<d ==> |T(x)-T(x0)|<e.
    Last edited: Aug 20, 2007
  6. Aug 20, 2007 #5
    Well, can you show that it is Lipschitz?
  7. Aug 21, 2007 #6
    1.Thanks for pointing that out, it should be the absolute value.:smile:
    2.sorry i made a mistake there, it should be
    sum_i|xi|<sqrt(m)*sqrt(sum_i(xi^2)) by AM<RMS
  8. Aug 21, 2007 #7
    Here is the standard way of doing it:

    For any linear transformation we define the norm of it as [itex]||T||=\sup_{|x|<1}|T(x)|[/itex]. Your arguments show that this norm always exists for any linear map.

    Then we have [itex]|T(x-y)|\le||T|||x-y|[/itex]. Therefore [itex]T[/itex] is Lipschitz, hence uniformly continuous. If you don't know anything about Lipschitz and uniform continuity you can just let [itex]|x-y|<\epsilon/||T||[/itex] and we have our desired result.
  9. Aug 21, 2007 #8


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    Oh, ok. I did not know this inequality, but it is true as long as you change that < for a <=, because obviously for m=1, we have an equality. What does AM and RMS stand for?

    What you've proven in showing |T(x)|<=C|x| is that T is Lipschitz continuous at 0. Usually, we say that a function f is Lipschitz at x0 if there exists numbers c and M such that |x-x0|<c ==> |f(x)-f(x0)| [itex]\leq[/itex] M|x-x0|. A function that is Lipschitz is necessarily continuous because given e>0, choose d=min{c,e/M}.

    So, can you extend your result and show that T is Lipschitz everywhere?
    Last edited: Aug 21, 2007
  10. Aug 21, 2007 #9


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    Shouldn't there be a sqrt{m} also on the RHS as in Andy's AM<RMS?
  11. Aug 21, 2007 #10
    oops, it should be <=, thanks.
    AM=arithmetic mean
    RMS=root mean square AM<=RMS (the equality holds when |x1|=|x2|....=|xm|)
  12. Aug 21, 2007 #11
    Thanks quasar987 and ZioX
    I forgot to use T(x)-T(y)=T(x-y) for the linear transformation...
    let d=e/C then, |x-y|<d --> |T(x)-T(y)|=|T(x-y)<cd=e
  13. Aug 27, 2007 #12
    I'll explain the logic a bit.

    Define [itex]||T||=\sup_{|x|<1}|T(x)|[/itex].

    Now the previous argument shows that there exists a C such that [itex]|T(x)| \le C|x|[/itex].

    Now, if we restrict [itex]|x| < 1[/itex] we get [itex]|T(x)| \le C|x| < C[/itex] so that |T(x)| has an upper bound, hence a supremum.

    Now, for continuity, it just becomes a matter of verifying that [itex]0 \le |T(x)-T(y)|=|T(x-y)|<\epsilon[/itex] whenever [itex]0 \le |x-y|< \min\{\epsilon / ||T||,1\}[/itex].
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