Results 1 to 6 of 6

Math Help - Metric spaces - continuity / compact

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    2

    Metric spaces - continuity / compact

    Hi, I'm doing the following assignment:

    We consider the metric space M = C([0,1], R) of continuous, real functions on [0,1]. M has the usual uniform metric:
    d_M(f,g)=sup{lf(x)-g(x)l l x E [0,1]}

    (a) Show that I:M-->R given by:
    I(f) = integral (from 0 to 1) f(x)*dx
    is continuous.

    (b) Show that I from (a) attains a max and a min on the closure of the unit ball:
    Closure of K(0,1) = {f E M l d_M(0,f) =< 1}
    Here 0EM is the nil-function and I don't have to show that the closure of the unit ball is as suggested.

    (c) Show that f_n E M given by
    f_n(x) = x^n, x E [0,1]
    for n=1,2, ... defines a series (f_n) in the closure of K(0,1) that does not have a convergent sub-series in M and conclude that the closure of K(0,1) is not compact.

    In (a) I think that an argument regarding I^(-1)(f) is sufficient. But I guess just saying that I^(-1)(f) = M, which is a closed set is too superficial?

    In (b) the problem is that the closure of K(0,1) is not compact and hence I cannot use the usual theorems...

    Any suggestions will be deeply appreciated... :-)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    For (a) use what my analysis professor would call "the world's most trivial integral inequality". That is if |f|\le M on [a,b] then \int_a^bf(x)dx\le M(b-a). Combine this with the regular definition of continuity and the given metric to so what I is continuous.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2009
    Posts
    2
    Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

    In (b), we look at a situation where f(x) <=0.

    The integral from

    1
    /
    l f(x) dx
    /
    0

    attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

    I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

    Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

    In (c) I'm still lost... I mean, for
    x<1: f_n(x) --> 0 for n --> infinity.
    x=1: f_n(x)=1 --> 1 for n--> infinity.
    And as {0,1} is a part of M i don't understand why we get into trouble at all :-)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by christina182 View Post
    Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

    In (b), we look at a situation where f(x) <=0.

    The integral from

    1
    /
    l f(x) dx
    /
    0

    attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

    I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

    Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

    In (c) I'm still lost... I mean, for
    x<1: f_n(x) --> 0 for n --> infinity.
    x=1: f_n(x)=1 --> 1 for n--> infinity.
    And as {0,1} is a part of M i don't understand why we get into trouble at all :-)
    In (a), how did you establish continuity by putting δ=ε, show some work if you like

    In (b) your definition is not very clear

    In (c),do you mean sequence and not series??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2009
    Posts
    113
    Quote Originally Posted by christina182 View Post
    Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

    In (b), we look at a situation where f(x) <=0.

    The integral from

    1
    /
    l f(x) dx
    /
    0

    attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

    I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

    Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

    In (c) I'm still lost... I mean, for
    x<1: f_n(x) --> 0 for n --> infinity.
    x=1: f_n(x)=1 --> 1 for n--> infinity.
    And as {0,1} is a part of M i don't understand why we get into trouble at all :-)
    Hi Christina182.

    Your argument for c) is correct!! Just observe that the limit (for ANY subsequence) function f(x)=0 0<=x<1 and f(1)=0 is NOT continuous!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    For (b): Using the mean value theorem for integrals we get that I(f)=f(\zeta) for some \zeta \in [0,1] and so we get \vert \int_0^1 f(x)dx \vert = \vert f(\zeta) \vert \leq 1 for all f \in B:= \{ f \in M : \Vert f \Vert _{\infty} \leq 1 \}
    so I(B) is bounded by 1. Let m:= \sup_{f \in B} I(f) and so m \leq 1. Now just let g(x)=m then g \in B and I(G)=m so I attains it's maximum in B. The case for the minimum is the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Any metric spaces can be viewed as a subset of normed spaces
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: December 15th 2011, 03:00 PM
  2. Continuity in metric spaces
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 23rd 2011, 04:41 PM
  3. Non-compact metric spaces
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: June 29th 2011, 08:50 PM
  4. Compact Metric Spaces
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 15th 2008, 07:15 PM
  5. Compact metric
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 14th 2008, 01:05 AM

Search Tags


/mathhelpforum @mathhelpforum