Results 1 to 2 of 2

Math Help - sup of equicontinuous is continuous

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    Canada
    Posts
    1

    Smile sup of equicontinuous is continuous

    Let U={f(x)} be a subset of C_0 (the set of continuous functions f:[a,b]->R) that is equicontinuous and bounded. Prove that sup(f(x): feU) is a continuous function of x.

    Note: I am not sure if the question means U is itself bounded wrt the sup norm so that sup(|f(x)-g(x)| : xeR) < M for some M or if the question is just poorly worded and really means that each f(x) is point-wise bounded so that sup(f(x): feU) always exists. Then again, f(x) is continuous and [a,b] compact so boundedness is already a given property of the functions...

    I am one step short of a proof:


    • Define g(x)=sup(f(x): feU)
    • Fix x and choose any epsilon.
    • By the definition of supremum, there is some f_1 in U s.t. |f_1(x) - g(x)| < epsilon.
    • By equicontinuity, there exists a delta such that |x-y|<delta => |f_1(x) - f_1(y)| < epsilon.
    • By the definition of supremum, there is there is some f_2 in U s.t. |f_2(y)-g(y)| < epsilon (note that at this step, y determines which f_2 is selected from U).
    • Now, to show continuity of g at x, I need to show that f_1(x) is close to f_2(y).


    Problems:

    • I know that f_1(x) is close to f_1(y) and IF I could show that f_1(x) is close to f_2(x) [or that f_1(y) is close to f_2(y)] then I would be done by triangle inequality.


    • I know that f_1(y) shouldn't be far from f_2(y) because h = |f_1 - f_2| = 0 somewhere on (x-delta, x+delta) and h is continuous so it is bounded on (x-delta, x+delta) by some constant C but to make this C as small as epsilon I need to make my delta-neighborhood smaller which will change the f_2 that is selected from U and therefore change the point where h=0 and change the constant C.


    I would very much appreciate any help correcting this proof. I knew it was a bad idea to fix y in order to select an f_2 and I also haven't used boundedness
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    592
    Thanks
    25

    Re: sup of equicontinuous is continuous

    Since U is equicontinuous and bounded in C_0, the Arzela-Ascoli theorem implies that U is relatively compact.

    So, if g=supU, there exists a sequence (g_n)\subset U with g_n\rightarrow g pointwise.
    But then, (g_n) must possess a subsequence (denoted still by (g_n)) convergent in C_0.

    This means g_n\rightarrow g in C_0, which proves that g is the uniform limit of continuous functions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 18th 2011, 09:19 AM
  2. Replies: 3
    Last Post: April 18th 2011, 08:24 AM
  3. if family equicontinuous, prove lipschitz
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 4th 2010, 08:18 AM
  4. Replies: 1
    Last Post: April 20th 2010, 01:40 PM
  5. equicontinuous and uniformly equicontinuous
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 6th 2006, 03:25 PM

Search Tags


/mathhelpforum @mathhelpforum