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Homework Help: (C[0,1],|| ||2) is a complete metric space

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data


    I am required to show that (C[0,1], || ||2) is a complete metric space, or to disprove that it is

    2. Relevant equations

    C[0,1] is the set of continuous functions on the bounded interval 0,1

    3. The attempt at a solution

    I am immediately confused as I am told in my notes that if every cauchy sequence from X converges then (X,d) is a complete metric space, but I find it hard to see that there clould exist a divergent cauch sequencey since cauchy sequences are all convergent right?
     
  2. jcsd
  3. Mar 14, 2007 #2

    StatusX

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    Cauchy sequences converge, but not necessarily to something in the space (every metric space has a completion in which every cauchy sequence converges, and which the original metric space is a dense subset of). Can you find a cauchy sequence of continuous functions whose limit isn't continuous?
     
  4. Mar 14, 2007 #3
    would such an example be the sequence of spike functions? i.e the sequence of functions gn on [0,1] for which gn(x) is n2x when x is in [0,1/n], n(2-nx) when x is in [1/n,2/n] and zero elsewhere. The pointwise limit of gn(x) is always zero (x=0 included, since gn(0)=0 for any n). Yet, the integral of gn is always equal to 1,
     
  5. Mar 14, 2007 #4

    HallsofIvy

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    Cauchy sequences in a complete metric space converge- that's the definition of "complete"!

    Perhaps you are thinking of the fact that Cauchy sequences of real numbers converge- the real numbers with the "usual" metric is complete.

    Here, your metric space is functions continuous on [0,1] so your "Cauchy sequences" are sequences of continuous functions. You need to show that Cauchy sequence of continuous functions converge. What is the precise definition of || ||2 ?
     
  6. Mar 15, 2007 #5
    || ||2 is the 2-norm so ||f||2 =the square root of the integral from 0 to 1 of [f(x)^2]dx
     
  7. Mar 15, 2007 #6

    HallsofIvy

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    That finally dawned on me! I was thinking of the infinite case where continuous function may not be square integrable.
     
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