高微,一題均勻收斂

f_n(x)=(1, cos(x)/ n^2) on R

這題是不是勻均收斂?

謝謝

2 個解答

評分
  • 1 0 年前
    最佳解答

    是均勻收斂!

    f_n(x)=(1, cos(x)/ n^2) -> f(x)=(1, 0)

    pf.

    For all ε>0, taking N= 1/√ε (N is indep. on x).

    If n>N, then | f_n(x) - f(x) | = |(0, cos(x)/n^2) | < 1/n^2 < 1/N^2 =ε

    By definition, f_n(x) converges to f(x)=(1, 0) uniformly.

  • 1 0 年前

    令 f_n(x)=(f_n'(x),f''_n(x)), Here f_n'(x)=1, f_n''(x)=cos(x)/n^2

    很名顯的 Both f_n'(x) and f_n''(x) are uniformly convergent

    For ε>0, there is N>0 such that |f_n'(x)-f_m'(x)|<ε/2

    and |f_n''(x)-f_m''(x)|<ε/2 whenever n,m>=N and for all x

    Hence for m,n>=N

    ||f_n(x)-f_m(x)||=√[(f_n'(x)-f_n'(x))^2+(f_n''(x)-f_m''(x))^2]

    <=|f_n'(x)-f_m'(x)|+|f_n''(x)-f_m''(x)|<ε/2+ε/2=ε

    for all x

    所以f_n(x) is nuiformly convergent

    2009-03-05 22:04:36 補充:

    f_n(x)€|R^2 要考慮它的norm

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