Re: [分析] 一題convex的問題

[ppia (papayaPaul)]

※ 引述《k6416337 (とある煞氣の光希)》之銘言：
: Suppose that f:|R^2->|R∪{∞} is convex with respect to each variable and
: f≠∞ at all.(f(x,y)=xy is convex w.r.t. x and y,but not convex w.r.t. (x,y))
: Let p=(a,b)屬於|R^2 and r>0 be given and fixed. Suppose that f(q)<∞ for all
: q屬於B_r(p).Show that there is a constant C>0 s.t.
:                     |f(q_1)-f(q_2)|≦C|q_1-q_2|
: for all q_1,q_2屬於B_r'(p) for some r' with 0<r'<r.
: 這題是有一本電子書可以參考 http://ppt.cc/GbfU 在P12~P13
: 不過他的前提是必須有上界，如果有上界就可以用它的方法證出
: 我有問過老師怎麼做，他說要利用凸性找出r'使得在裡面有上界，可是我沒辦法找出來
: 有請高手幫忙解答，謝謝


我直接證明這個命題:

   Let Q = [A,B]x[C,D] be a rectangle and R = [a,b]x[c,d] be another with

                    closure(R) ㄈ Interior(Q) ( RㄈㄈQ ).

Since f is continuous(<=convex) on [a,b]x{C}, [a,b]x{c}, [a,b]x{d}, [a,b]x{D},

             f(x,D)-f(x,d)                         f(x,C)-f(x,c)
       sup   ───────  := L    and    inf    ─────── := J
     a≦x≦b     D - d                    a≦x≦b      C - c

are both finite. This done, by the convexity of f along the y-direction,


        f(x,C) - f(x,c)       f(x,y2) - f(x,y1)       f(x,D) - f(x,d)
J ≦  ─────────  ≦  ───────── ≦ ───────── ≦ L,
             C - c                 y1 - y2                 D - d

for any x in [a,b], y1 < y2 in [c,d]. Similarly,

                            f(x2,y) - f(x1,y)
                     N  ≦  ───────── ≦ M,
                                 x1 - x2

for any y in [c,d], x1 < x2 in [a,b]. Put S = max{ |M|, |N|, |L|, |J| }.
Then for any q1=(x1,y1) and q2=(x2,y2) in R,

                         2                        2                     2
     │f(x2,y2)-f(x1,y1)│ ≦ │f(x2,y2)-f(x2,y1)│+│f(x2,y1)-f(x1,y1)│

                                2        2    2        2                 2
                           ≦  S │y2-y1│ + S │x2-x1│  = ( S│q2-q1│)

Note that S depends on both rectangles Q and R.


   Now for the case of B(r',p):= B' ㄈㄈ B(r,p):=B , it suffices to cover B' by
some finitely many interir(R)'s such that closure(Q)ㄈ B.

--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.32.4.99
推 seaxrrz     :你這樣選的C會不會跟x1,y1,x2,y2有關呢
→ seaxrrz     :喔  我好像看錯了
→ k6416337    :為什麼會連續?不是要在凸而且有上界才可以嗎?
"g:|R→|R;  g convex on (ξ,η) => g continuous on (ξ,η)"

Back to our case, since f is convex along any x-direction, f is continuous on
|R x {y} for any y, and therefore continuous on, say, [a,b]x{c}.
※ 編輯: ppia            來自: 114.32.4.99          (06/19 11:26)
→ k6416337    :能給一下連續的證明嗎?感謝
→ ppia        :是指我上面寫的 那個g的命題嗎?
→ k6416337    :恩恩
→ ppia        :http://ppt.cc/4X-m
推 k6416337    :the convexity of f implies that f is bounded by i
→ k6416337    :its values at two opposite faces of P
→ k6416337    :上面這句話不太懂
嗯.其實我也看不懂XD, 那我就直接證我舉的命題好了. 其實可以證明:

 g:(ξ,η)→|R  If g is convex, then g is Lipschitz continuous on any
                (a,b)ㄈㄈ(ξ,η).

這其實就是你本來問題的一維版本, 用它可以證明二維版, (然後可以一直推到n維這樣)


Pick some a' b' with ξ< a' < a < b < b'< η. By the convexity of g,

g(a') - g(a)     g(a) - g(y)      g(y) - g(z)
────── ≦  ────── ≦ ──────
   a' - a           a - y            y - z

                                             g(z) - g(b')      g(b') - g(b)
                                         ≦ ─────── ≦ ─────── ,
                                                z - b'            b' - b

for any y < z in (a,b). Hence f is Lipschitz continuous on (a,b).

※ 編輯: ppia            來自: 114.32.4.99          (06/20 12:43)