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Math Help - Equivalent Norms

  1. #1
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    Equivalent Norms

    Let ||.||_1 and ||.||_2 be norms on a vector space X

    such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

    If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

    show that convergence in X_1 implies convergence in X_2.

    and conversely show that there are positive numbers

    a and b such that for all x element of X,

    a||x||_1 <= ||x||_2 <= b ||x||_1.
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  2. #2
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    Quote Originally Posted by frater_cp View Post
    Let ||.||_1 and ||.||_2 be norms on a vector space X

    such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

    If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

    show that convergence in X_1 implies convergence in X_2.
    If x_1 converges to x in X_1, then x_1- X converges to 0. Therefore...

    and conversely show that there are positive numbers

    a and b such that for all x element of X,

    a||x||_1 <= ||x||_2 <= b ||x||_1.
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  3. #3
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    Hi Thank you for looking at this problem.

    the answer is not clear to me, can you please elaborate?
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  4. #4
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    Oh, c'mon! If x_n converges to x in ||.||_1, then x_n- x converges to 0 in ||.||_1. But your condition is "||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0" so x_n- x converges to 0 in ||.||2 from which x_n converges to x in ||.||_2.
    Last edited by HallsofIvy; June 28th 2009 at 07:09 AM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Oh, c'mon! If x_n converges to x in ||.||_1, then x_n- x converges to 0 in ||.||_2.

    You mean it converges to 0 in ||.||_1 and not ||.||_2
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  6. #6
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    Thanks. I had just noticed that myself and editted it. (So I could pretend I didn't make that mistake!)
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  7. #7
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    First part of the problem is perfectly explained by HallsofIvy.

    Second part consists of proving that || ||_1 and || ||_2 are equivalent norms. But the first part implies that the identity

    I: (X,|| ||_1)----------(X,|| ||_2)
    I(x)=x

    is continuous at zero, and hence continuous. Since both spaces are complete, Open Mapping Theorem implies that I is an isomorphism, and we are done.
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  8. #8
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    Quote Originally Posted by frater_cp View Post
    Let ||.||_1 and ||.||_2 be norms on a vector space X

    such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.

    If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,

    show that convergence in X_1 implies convergence in X_2.

    and conversely show that there are positive numbers

    a and b such that for all x element of X,

    a||x||_1 <= ||x||_2 <= b ||x||_1.

    Do you mean that ||x_n||_1------>0 means :

     lim_{n\rightarrow\infty}{||x_{n}||_{1}} =0??
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