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Conclusion about the dimension of C°(R)?

  1. Jul 12, 2017 #1
    [mentor note: thread moved from Linear Algebra to here hence no homework template]

    So, i was doing a Linear Algebra exercise on my book, and thought about this.

    We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
    Let's suppose that Aƒ= x0 ƒ(t)dt.

    By the Calculus Fundamental Theorem, d/dx(Aƒ) = ƒ, so we have a left inverse, which implies that ker(A)={0}.

    Supposing that dim(E) is finite, by the rank-nullity theorem we have that im(A)=E . As a result of that:
    (ƒ(x)=|x|) ∈ E ⇒ ƒ ∈ im(A)
    ⇒dƒ/dx ∈ E

    But we know that ƒ's derivative is not continuous. So, supposing that dim(E) is finite lead us to a contradition (dƒ/dx ∈ E ∧ dƒ/dx ∉ E) therefore dim(E) must be infinite.

    Is this argument valid? If not, could you guys point where does it fail? Thank you!
    Last edited by a moderator: Jul 12, 2017
  2. jcsd
  3. Jul 12, 2017 #2


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    Note that the right-hand side is ##(Af)(x)##, not ##Af##.

    The derivative is not even defined at ##x = 0##.

    Yes, it is valid, but it is quite convoluted. If you insist on doing it this way, I would prefer to say that you have found a linear operator ##A## that is injective but not surjective (the latter because ##A## maps into ##C^1(\mathbb{R})##). This already implies infinite dimensionality of ##E##, since on a finite dimensional space injectivity and surjectivity are equivalent for linear operators.

    Of course, if you merely care to show that ##E## is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.
  4. Jul 12, 2017 #3
    Thanks for the knowledge shared!
    I hadn't tought that, that's, indeed, much simpler.
    So the basis of the set of all polynomials would be enough?
  5. Jul 12, 2017 #4


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    The set of all polynomials is too large to be a basis for ##C_0##, but ##\{1,x,x^2,\dots\}## would do.
  6. Jul 12, 2017 #5


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    this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!
  7. Jul 13, 2017 #6


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    While I usually like the straightforward approach best (maybe this is one of the differences between a "pure" mathematician and an (aspiring) "applied" mathematician?), I gave it some thought and then came to the conclusion that I agree with you.

    Just as an exercise, you could try to complete the direct approach as well. (You are almost there, anyway.)
  8. Jul 13, 2017 #7
    Hm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C°(ℝ)?
    That's where i want to get someday. Thanks for the inspiration!
  9. Jul 13, 2017 #8


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    I think that I and post #4 misunderstood you when you wrote
    We (or I, at least) thought that here you asserted that the set of all polynomials itself is a basis for a linear subspace of ##C^0(\mathbb{R})##, maybe because you wrote "the basis". However, from what you wrote afterwards (quoted at the top of this post), I get that you meant any (Hamel) basis for the linear space of all polyniomials. (For example, you can indeed use the canonical basis mentioned in post #4. By the way, I think there the ##C_0## was written by small mistake.)

    Very good, as far as I can see, you are more than done.
    Last edited: Jul 13, 2017
  10. Jul 13, 2017 #9
    Yeah, that's what i meant. Sorry if i wasn't clear enough! English is not my mother language so, mostly when writing about math and science, my texts can get a little confusing. I will try to be more specific in the next time.

    Okay! Thanks for the support and knowledge shared!
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