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Math Help - how can u prove this argument

  1. #1
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    how can u prove this argument

    f_{n},g_{n} are sequences of functions :  X\rightarrow R which converge uniformly to f,g respectively.

    If  \lambda\in R , prove  \lambda f_{n} converges uniformly to  \lambda f

    and

    prove :  f_{n}g_{n} converges to  fg given f,g are bounded.
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    f_{n},g_{n} are sequences of functions :  X\rightarrow R which converge uniformly to f,g respectively.

    If  \lambda\in R , prove  \lambda f_{n} converges uniformly to  \lambda f

    and

    prove :  f_{n}g_{n} converges to  fg given f,g are bounded.
    You sent me a PM.
    I responded with a question: “What do you do for yourself?”
    Now I ask you the same question publicly.
    It seems to me that you simply expect to be given the answers.
    Am I wrong?
    If so, explain yourself!
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  3. #3
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    No. I have done part of this question.

    No. I done part of this question. I am not sure if my proof is right. and I am not sure my link between two arguments is right.

    i post my proof for the first part,


    when \lambda=0, question is trivial:
    \epsilon_{1}>0, \ \forall x\in X, \ |\lambda f_{n}(x)-\lambda f(x)|=0<\epsilon_{1}
    So assume \lambda\neq0
    f_{n} converges uniformly to f:
    \forall \epsilon_{1}>0, \ \exists N, \ \forall x\in X, \forall n\geq N, |f_{n}(x)-f(x)|<\epsilon_{1}
    by writing \epsilon_{2}=\frac{\epsilon_{1}}{|\lambda|}:
    \forall \epsilon_{2}>0, \ \exists N, \ \forall x\in X, \forall n\geq N, |\lambda f_{n}(x)-\lambda f(x)|<\epsilon_{2} \ \ \square.
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  4. #4
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    and then

    and then i just said, by setting:
    \lambda=g_{n}(x) \forall x\in X
    and use lemma just proved,
    I got:
    \lambda f_{n}(x) converges uniformly to  <br />
\lambda f
    and then set  \lambda=f(x) , use lemma again,
     g_{n}(x) f(x) converges pointwise to  fg
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  5. #5
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    ok

    ok. i posted all the proof i can get. so, could you please give me any correction of my proof if there is any problem.
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