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Function Space C[a,b]

  1. Jun 26, 2010 #1
    Hi i'm using Kreyszig's Introductory Functional Analysis with Applications and he proves that the set of continuous functions on an interval [a,b] under the metric d(x,y) = max|x(t)-y(t)| is complete. Standard proof nothing hard over there.

    But isn't the sequence of x^n s on [0,1] a Cauchy sequence (under this metric) which does Not converge to a continuous function?
  2. jcsd
  3. Jun 26, 2010 #2
    What is your reasoning for why it is Cauchy?
  4. Jun 27, 2010 #3
    Well intuitively it seems so, but later on i tried checking this out as follows:

    i) compute the derivative of x^(n-1) - x^n.

    ii)Equating this to zero will give the point x in [0,1] at which this is highest.

    iii)Substituting this value back into x^(n-1) - x^n will give the disatance (w.r.t the metric d) between x^(n-1) and x^n.

    The attachment has the distances. Thanks! :)

    Attached Files:

  5. Jun 27, 2010 #4


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    You should try to give a proof of your claim.

    For all n,m: if x=0 then x^n=0, if x=1 then x^m=1. Hence the distance between x^n and x^m will always be at least 1 in the sup-norm:

    [tex]\max_{x\in[0,1]}|x^n-x^m|\geq 1.[/tex]

    No way this is a Cauchy sequence, or a convergent sequence. The sequence converges pointwise to the discintinuous function [itex]f(x)=\delta_{x,1},[/itex] but not uniform.
  6. Jun 27, 2010 #5
    Hi Landau, i know that the sequence converges pointwise and the limit function is discontinuous on [0,1]. However in the metric space we're treating the functions as points; have You looked at the metric we're using on [0,1]? It's d(x,y) = max|x(t)-y(t)| for all t in [0,1].

    The convergence isn't the usual one of a sequence of functions on a subset of R as in calculus, but in a metric space with respect to the metric d.

    So the distance between x and x^2 is 0.25, the dist. between x^2 and x^3 is 0.148148 and so on... (see the attachment in my previous post)
  7. Jun 27, 2010 #6


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    Hi Bolzano, I am aware that we are considering the metric d, this is also called the supremum (or uniform) norm. Convergence with respect to this metric is precisely the usual "uniform convergence".

    Bu I now see I made a mistake in my last post, arguing that the distance would be at least 1 for all n,m. Instead of [itex]\max_{x\in[0,1]}|x^n-x^m|[/itex] I considered [itex]\max_{x,y\in[0,1]}|x^n-y^m|[/itex].

    Still, you should try to make your proof rigorous (because it won't be possible). Given e>0, try to find N such that d(x^n,x^m)<e for all n,m>N.
    Last edited: Jun 27, 2010
  8. Jun 27, 2010 #7
    Yes precisely, the sequence I'm considering obviously isn't Cauchy, otherwise the space won't be complete. In fact the theorem tells us that the sequence can't be Cauchy since it converges to a function "outside" the space of Cont. Functions.

    However i really felt that this sequence was Cauchy wrt to the metric but it obviously isn't, proving it rigorously is the only way to make sure obviously :)
  9. Jun 27, 2010 #8


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    Perhaps easier is to show that the sequence does not converge uniformly to [itex]f(x):=\delta_{x,1}[/itex] (=1 if x=1 and =0 otherwise), i.e. it does not converge with respect to d to this function.

    |x^n-f(x)|=|1-1|=0 if x=1
    |x^n-f(x)|=|x^n-0|=|x|^n if x<1.

    Hence [itex]d(x^n,f)=\sup_{0\leq x<1}|x|^n.[/itex]
    E.g. take epsilon=1/2.
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