Re: [分析] Hardy–Littlewood maximal function

[ppia ((= =))]

※ 引述《tsaihohan (有人想一起下圍棋嗎)》之銘言：
:         Given a Lebesgue measurable subset A of R^n, we
:         denote by │A│ the Lebesgue measure of A. Given
:         a Lebesgue integrable function g on R^n, the Hardy
:         -Littlewood maximal function M_g of g is defined
:         on R^n by
:                M_g(x)=sup  ∫│g│ /│Br(x)│
:                       r>0  Br(x)
:         where Br(x) is the open ball with radius r centered
:         at x屬於R^n.
:         (a) Prove that for each t屬於R, the set {x屬於R^n:M_g(x)>t}
:         is open.

Fix some x with M_g(x) > t. Then for some R > 0

               ∫ │g│
               B_R(x)    defn    I
               ────  === ────  > t.
               |B_R(x)|       ωn R^n

                                                  I
Clearly, there exists some δ > 0 with       ─────  > t.
                                             ωn (R+δ)^n
For each y with |y-x| < δ,

        ∫  │g│   ≧  ∫  │g│ = I.
         B     (y)        B (x)
          R+|y-x|          R

Hence

                   ∫ │g│
                    B     (y)
                     R+|y-x|            I
     M (y)  ≧    ─────── ≧  ────     > t.
      g           │B     (y)│     ωn (R+δ)^n
                  │ R+|y-x| │


:         (b) It follows from (a) that M_g is a measurable function
:         on R^n. Prove that when ∫│g│>0, we have ∫│M_g│=∞.
:                                 R^n                R^n


If ∫│g│> 0, then ∫│g│= L > 0 for some R large enough.
                    B_R

For each y , M_g(y) ≧ L/[ωn (|y|+R)^n].
Hence
                             n                   n
                            d y                 d y
     ∫│M_g│ >  L ∫   ────      > ∫   ────
     |R^n          |R^n  (|y|+R)^n      |y|>R  (2|y|)^n

         ∞  r^(n-1)
    = L ∫  ──── nωn dr = ∞.
         R    (2r)^n

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推 tsaihohan   :感謝您^^
※ 編輯: ppia            來自: 125.231.219.112      (03/19 23:03)