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Just a quick question on completeness

  1. Sep 13, 2005 #1
    Just a quick question on completeness.

    The space of all continuous complex values functions [itex]\mathcal{C}([0,1])[/itex] with the norm

    [tex]\|f\| = \sup\{|f(x)| \,:\, x \in [0,1]\}[/tex]

    is a Banach space because every Cauchy sequence in [itex]\mathcal{C}([0,1])[/itex] converges to a limit point in [itex]\mathcal{C}([0,1])[/itex].

    My question is this: What if the norm on [itex]\mathcal{C}([0,1])[/itex] is instead defined as

    [tex]\|f\| = \int_0^1 |f(x)|dx[/tex]

    This is a perfectly good norm (as it satisfies the 3 axioms), but does it cause [itex]\mathcal{C}([0,1])[/itex] to be complete?

    My guess is no. But Im not sure if my solution is very good. Here it is.

    Take a Cauchy sequence [itex]\{f_n\}[/itex] in [itex]\mathcal{C}([0,1])[/itex] and write

    [tex]f_n = \{f_1,f_2,\dots\}[/tex].

    So [itex]\{f_n\}[/itex] is just a sequence of continuous complex valued functions. I want to show that this converges in [itex]\mathcal{C}([0,1])[/itex], that is, the distance (as defined by the norm) between successive terms approaches zero the further along the sequence you go. More precisely:

    Given any [itex]\epsilon > 0[/itex], we need to find an [itex]N[/itex] such that for all [itex]m,n > N[/itex]

    [tex]\|f_n - f_m\| < \epsilon[/tex]

    This implies

    [tex]\int_0^1 |f_n(x) - f_m(x)|dx < \epsilon[/tex]
    [tex]\int_0^1 |f_n(x)|dx - \int_0^1|f_m(x)|dx < \epsilon[/tex]
    [tex]\|f_n\| - \|f_m\| < \epsilon[/tex]

    This is where I am struggling. I have the notion that

    [tex]\|f_n\| - \|f_m\| \rightarrow 0[/tex]

    Does not necessarily mean that [itex]\|f_n\| \rightarrow 0[/itex]. In other words, I dont see how this norm makes [itex]\mathcal{C}([0,1])[/itex] a Banach space because the distance between successiver terms may approach zero regardless of whether [itex]m,n \rightarrow \infty[/itex].

    This is a fairly straight-forward exercise in completeness, but I am struggling to come to a clear and concise proof. Any help is appreciated.
     
  2. jcsd
  3. Sep 13, 2005 #2

    matt grime

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    Find a sequence of functions that converges "graphically", but not to a continuous function. note here that the distance between the functions is simply the area between the graphs. so, can you think of any graphs of functions that to something not conintuous? there are infinitely many such things. might help to think "spikey"
     
  4. Sep 13, 2005 #3
    Well, off the top of my head, the function

    [tex]f(x) = \frac{1}{x^3}[/tex]

    is not continuous on [itex][0,1][/itex] (but it is continuous on (0,1))but there is obviously a sequence of continuous functions on [itex][0,1][/itex] which converge to this [itex]f(x)[/itex], (not sure how to prove this but though).

    Is this what you mean by being able to find a sequence of continuous functions on some closed interval which converge to a discontinuous function on the same interval - implying that the difference in the integral of the functions converges (ie the area under the graph between two continuous functions approaches zero) even though the functions themselves converge to a discontinuous function. Therefore Cauchy sequences do not necessarily converge to points in the space - hence the space is not a Banach space (with respect to the p-norm - I think that is what it is called?).
     
    Last edited: Sep 13, 2005
  5. Sep 13, 2005 #4

    matt grime

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    that isn't a fucntion defeined on [0,1] is it? and i doubt you'll get a sequence converging to it

    but es you should think of a discontinuous function on [0,1] with a sequence of continuous functions whose difference in area with this function tends to zero.

    hint, try the fucntion f(x)=0 of 0<=x<1 and 1 when x=1. I can you think of a sequence of fucntions through f_n(x) with say f_n(0)=0 and f_n(1)=1 that converge in area to zero?
     
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