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Spaces of continuous functions and Wronskians

  1. Jul 23, 2010 #1
    I'm struggling to understand continuous functions as subspaces of each other. I use ⊆ to mean subspace below, is this the correct notation? I also tried to write some symbols in superscript but couldn't manage. Anyway I know that;

    Pn ⊆ C∞(-∞,∞) ⊆ Cm(-∞,∞) ⊆ C1(-∞,∞) ⊆ C(-∞,∞) ⊆ F(-∞,∞)

    I know the axioms required to be a polynomial, but I am struggling to conceive the difference between, say, C(-∞,∞), C1(-∞,∞) and C2(-∞,∞) for example. Could someone possibly write out a few examples of functions that would, for example be vectors of C(-∞,∞) but not of C1(-∞,∞). I really appreciate any light that can be shed on this as I have been struggling to get it for a while.

    Also when it comes to the Wronskian, I know how to use it to show linear independency, but I don't actually understand why. Specifically, why are the derivatives of functions important in determining whether a homogenous linear combination of functions has only the nontrivial solution, because I was under the impression that because the system is linear, derivatives don't come into it . I always prefer to understand the maths I am applying, so please help me!

    I really appreciate any help.
     
  2. jcsd
  3. Jul 23, 2010 #2

    quasar987

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    Asking the difference between C and C^1 just boils down to finding a continuous function that is not differentiable. Surely you know of such a function. The most memorable example is probably the absolute value function x-->|x|.

    A function that's in the difference C^1 - C^2 is x^{3/2}. More generally, x^{(2k+1)/2} is in C^k but not C^{k+1}.
     
  4. Jul 23, 2010 #3

    quasar987

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  5. Jul 23, 2010 #4
    Sorry if I'm being slow, but how is x^(3/2) continuous for all values of x? Surely it is only continuous for (0,∞). Or does this matter? Furthermore I can find the first and second derivative of this with:
    f'(x) = (3/2)x^(1/2)
    and
    f''(x)=(3/4)x(-(1/2))
    and they are both continuous for (0,∞) but not for C(-∞,∞) (aren't they?)
     
  6. Jul 23, 2010 #5

    quasar987

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    You're right. Make that |x|^{3/2}.

    This one is well defined on all R and is differentiable at 0 because
    [tex]\lim_{h\rightarrow 0}\frac{|h|^{3/2}-|0|}{h}=\lim_{h\rightarrow 0}\sqrt{h}=0[/tex]
     
  7. Jul 23, 2010 #6
    OK, I think the penny might have dropped, but if I write out my understanding of things please can you reply letting me know if I have got the jist (this ties back into the wronskian I was mentioning earlier).

    Say I have f1=cosx and f2=sinx

    The wronskian tells us that the determinant of (forgive my inability to write out a proper matrix):

    cos x sinx
    -sinx cos x

    is 1. Therefore since this does not equal zero we know the two functions form a linearly independent set in C^1(-∞,∞). However, in truth this linearly independent set is also in C^∞(-∞,∞) since every fourth derivative of cosx = cosx, and the same goes for sinx therefore you can repeat the process of finding the derivatives an infinite number of times and they will always be continuous.

    Have I got this right?
     
  8. Jul 23, 2010 #7

    quasar987

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    Yes, completely right.
     
  9. Jul 23, 2010 #8
    Brilliant, thankyou!
     
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