Results 1 to 6 of 6

Math Help - Cts functions on [0,1] tending ptwise to 0 - convergence uniform on an interval?

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    7

    Question Cts functions on [0,1] tending ptwise to 0 - convergence uniform on an interval?

    Hello everyone, first post! I've come across the following Analysis problem and given that it's one of the hardest questions on my problem sheet, I am having a few problems working my argument through.

    "Suppose that f_n:[0,1] -> Reals is a sequence of continuous functions tending pointwise to 0. Must there be an interval on which f_n -> 0 uniformly?"

    I have considered using the Weierstrass approximation theorem here, which states that we can find, for any continuous function [0,1] -> Reals, a uniform approximation by polynomials.

    Because of this, it seems to me - though I could be wrong - that these f_n -> 0 uniformly if this series of polynomials (each p_n approximating an f_n to a sufficient degree of accuracy) tends to 0 uniformly - in which case it suffices to prove the result for any series of polynomials.

    Even if this deduction -is- correct, which I'm not 100% confident about, I can't seem to follow through and show that there exists such an interval for a polynomial sequence. On the other hand, perhaps there is a counterexample and I'm going about this completely the wrong way! Could anyone lend a hand please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    I can see how the statement would be true if the family of functions was equicontinuous.

    First, there is a point x_0\in [0,1] such that \{f_n(x_0)\}\to 0. In other words, for any \varepsilon>0, there exists N such that n>N implies |f_n(x_0)|<\varepsilon. Since the f_n are equicontinuous, there is \delta>0 such that |x-x_0|<\delta implies |f_n(x)-f_n(x_0)|<\varepsilon for all f_n. This gives the inequality

    |f_n(x)|=|f_n(x)-f_n(x_0)+f_n(x_0)|\leq |f_n(x)-f_n(x_0)|+|f_n(x_0)|<\varepsilon + \varepsilon=2\varepsilon

    whenever x\in (x_0-\delta, x_0+\delta) and n>N. This proves that \{f_n\} is uniformly convergent to 0 on the interval (x_0-\delta, x_0+\delta).

    At the moment, I am not sure if equicontinuity can be dropped.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by mathmos8128 View Post
    Suppose that f_n:[0,1] -> Reals is a sequence of continuous functions tending pointwise to 0. Must there be an interval on which f_n -> 0 uniformly?
    The answer is No, but I found it quite tricky to construct a counterexample.

    Start with something easier. Define a sequence of functions f_n(x) on the unit interval as follows: on the interval from 0 to 2^{-n-1}, f_n(x) increases linearly from 0 to 1; on the interval from 2^{-n-1} to 2^{-n}, f_n(x) decreases linearly from 1 to 0; and on the remainder of the interval, f_n(x) = 0. Then it is well known that f_n(x)\to0 pointwise but not uniformly on the unit interval. In fact, if the binary expansion of x has a 1 in any of the first k places, then f_n(x) = 0 whenever n\geqslant k.

    Now extend f_n(x) by periodicity to a function on the whole real line, with period 1. Thus f_n(x) is a function with a narrow spike, of height 1, to the right of each integer point.

    Next, define a sequence of functions g_n(x) by g_n(x) = \sum_{k=1}^n 2^{-k}f_n(2^kx). Then g_n(x) is a continuous function. If n\geqslant k then g_n(x) \geqslant 2^{-k}f_n(2^kx). But 2^{-k}f_n(2^kx) attains the value 2^{-k} in every interval of length 2^{-k}. Therefore \max\{g_n(x):x\in J\}\geqslant 2^{-k} for all n\geqslant k in any interval J of length 2^{-k}. Therefore g_n(x) does not tend uniformly to 0 in any interval.

    However, g_n(x)\to0 pointwise. To see that, we have to consider two separate cases.

    Case 1: x has only finitely many 1's in its binary expansion. Suppose that the last 1 occurs in the r'th place in the binary expansion, and let n>r. Then for  k<r, 2^kx has a 1 somewhere in the first n places of its binary expansion, and so f_n(2^kx)=0. And for k\geqslant r, 2^kx is an integer, so again f_n(2^kx)=0. Thus g_n(x) = 0 for all n> r.

    Case 2: x has infinitely many 1's in its binary expansion. Then given \varepsilon>0, there exists r with 2^{-r} < \varepsilon/2 such that x has a 1 in its r'th binary place. If n>r then g_n(x) = \sum_{k=1}^r 2^{-k}f_n(2^kx) + \sum_{k=r+1}^n 2^{-k}f_n(2^kx). But the first of those sums is 0, and the second is less than \varepsilon. Therefore g_n(x) < \varepsilon.

    So in either of the two cases, g_n(x)\to0 as n\to\infty.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2010
    Posts
    7
    Quote Originally Posted by Opalg View Post
    The answer is No, but I found it quite tricky to construct a counterexample.

    Start with something easier. Define a sequence of functions f_n(x) on the unit interval as follows: on the interval from 0 to 2^{-n-1}, f_n(x) increases linearly from 0 to 1; on the interval from 2^{-n-1} to 2^{-n}, f_n(x) decreases linearly from 1 to 0; and on the remainder of the interval, f_n(x) = 0. Then it is well known that f_n(x)\to0 pointwise but not uniformly on the unit interval. In fact, if the binary expansion of x has a 1 in any of the first k places, then f_n(x) = 0 whenever n\geqslant k.

    Now extend f_n(x) by periodicity to a function on the whole real line, with period 1. Thus f_n(x) is a function with a narrow spike, of height 1, to the right of each integer point.

    Next, define a sequence of functions g_n(x) by g_n(x) = \sum_{k=1}^n 2^{-k}f_n(2^kx). Then g_n(x) is a continuous function. If n\geqslant k then g_n(x) \geqslant 2^{-k}f_n(2^kx). But 2^{-k}f_n(2^kx) attains the value 2^{-k} in every interval of length 2^{-k}. Therefore \max\{g_n(x):x\in J\}\geqslant 2^{-k} for all n\geqslant k in any interval J of length 2^{-k}. Therefore g_n(x) does not tend uniformly to 0 in any interval.

    However, g_n(x)\to0 pointwise. To see that, we have to consider two separate cases.

    Case 1: x has only finitely many 1's in its binary expansion. Suppose that the last 1 occurs in the r'th place in the binary expansion, and let n>r. Then for  k<r, 2^kx has a 1 somewhere in the first n places of its binary expansion, and so f_n(2^kx)=0. And for k\geqslant r, 2^kx is an integer, so again f_n(2^kx)=0. Thus g_n(x) = 0 for all n> r.

    Case 2: x has infinitely many 1's in its binary expansion. Then given \varepsilon>0, there exists r with 2^{-r} < \varepsilon/2 such that x has a 1 in its r'th binary place. If n>r then g_n(x) = \sum_{k=1}^r 2^{-k}f_n(2^kx) + \sum_{k=r+1}^n 2^{-k}f_n(2^kx). But the first of those sums is 0, and the second is less than \varepsilon. Therefore g_n(x) < \varepsilon.

    So in either of the two cases, g_n(x)\to0 as n\to\infty.
    Wow, that's genius! If you don't mind me asking, how did you come up with that counterexample? It's certainly not something I could just construct off the top of my head! Thankyou so much
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by mathmos8128 View Post
    If you don't mind me asking, how did you come up with that counterexample?
    The starting point was the Wikipedia article on Dini's theorem, which makes it pretty clear that pointwise convergence only implies uniform convergence if you have a strong additional condition, such as the sequence being monotonic. That made me confident enough to look for a counterexample rather than trying to prove a positive result. Finding a counterexample that actually worked was something that I had to think about overnight. The obvious strategy is to take a sequence that converges pointwise but not uniformly (usually because something blows up at a single point), and then find some way of spreading that pathology throughout an interval. But I had several false starts before getting the details right.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2010
    Posts
    7
    Quote Originally Posted by Opalg View Post
    The starting point was the Wikipedia article on Dini's theorem, which makes it pretty clear that pointwise convergence only implies uniform convergence if you have a strong additional condition, such as the sequence being monotonic. That made me confident enough to look for a counterexample rather than trying to prove a positive result. Finding a counterexample that actually worked was something that I had to think about overnight. The obvious strategy is to take a sequence that converges pointwise but not uniformly (usually because something blows up at a single point), and then find some way of spreading that pathology throughout an interval. But I had several false starts before getting the details right.
    Well that is extremely clever, I'd never heard of Dini's theorem before - I had actually looked at the idea of 'moving triangles' but I couldn't find a way to put them together to provide a solid counterexample - thankyou very much for all the effort! It's geniunely appreciated.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 19th 2011, 06:39 PM
  2. Replies: 1
    Last Post: October 31st 2010, 07:09 PM
  3. Uniform convergence of a sequence of functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 17th 2010, 12:49 PM
  4. Series of functions & uniform convergence
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 12th 2010, 02:26 PM
  5. uniform convergence of series of functions
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: December 3rd 2009, 03:59 AM

Search Tags


/mathhelpforum @mathhelpforum