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匿名使用者 發問時間: 科學數學 · 9 年前

實分析(證明題)有關llinear vector space

Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

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  • Sam
    Lv 6
    9 年前
    最佳解答

    Equivalent norms

    Two norms |•|_1 and |•|_2 on a vector space V are called equivalent if there exist positive real numbers C and D such that

    C|x|_1 ≦ |x|_2≦ D|x|_1. for all x in V.

    Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

    [Proof]

    (=>) Let A be a set which is bounded in the first norm |*|_1, and the bound is M.

    Then |x|_2 ≦ (1/C)|x|_1 ≦ (1/C)M, for all x in A, i.e. A is bounded in the norm |*|_2 with bound (1/C)M.

    Let A be a set which is bounded in the second norm|*|_2, and the bound is M.

    Then |x|_1 ≦ D|x|_2 ≦ DM, for all x in A, i.e. A is bounded in the norm |*|_1 with bound DM.

    The part (=>) is done.

    (<=)

    Let US={x: |x|_1=1}, where US is the abbreviation of unit sphere , by the assumption there is M >0, such that

    |x|_2 ≦ M, for all x in US.

    For any x in V, x=|x|_1*(x/|x|_1), where (x/|x|_1) in US, and

    |x|_2=||x|_1*(x/|x|_1)|_2

    =|x|_1*|(x/|x|_1)|_2

    {since (x/|x|_1) in US}

    ≦ M|x|_1.

    Take D=M, we have

    |x|_2≦ D|x|_1.

    For the same reason as the above proof, we can easy obtain the other part of the inequality, C|x|_1≦|x|_2, for all x in V, so omit it and leave you as an exercise.

    Hence |*|_1 is equivalent to |*|_2.

    (<=) is done.

    Q.E.D.

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