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Strong differentiability condition

  1. Aug 21, 2013 #1
    Assume that a point [itex]x[/itex] is an interior point of domain of some function [itex]f:[a,b]\to\mathbb{R}[/itex], and assume that the limit

    [tex]
    \lim_{(\delta_1,\delta_2)\to (0,0)} \frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}
    [/tex]

    exists. What does this imply?

    Well I know it implies that [itex]f'(x)[/itex] exists, but does it imply more? Does it imply that the derivative of [itex]f[/itex] exists in some neighbourhood of [itex]x[/itex] and is continuous? Where's a counter example?

    To be more presice, we can define a set

    [tex]
    \mathcal{D}=\big\{(\delta_1,\delta_2)\;\big|\; |\delta_1|<D,\; |\delta_2|<D,\; \delta_1\neq \delta_2\big\}
    [/tex]

    with some small [itex]D[/itex], and then consider the given expression as a mapping

    [tex]
    \mathcal{D}\to\mathbb{R}
    [/tex]

    So the domain is a square with a thin diagonal removed. Then we seek its limit at the center point [itex](0,0)[/itex].
     
  2. jcsd
  3. Aug 22, 2013 #2

    Stephen Tashi

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    What about the function defined by
    [itex] f(x) = x^2 [/itex] for [itex] x \ne 0 [/itex]
    [itex] f(0) = 3 [/itex]

    What's [itex] f'(0) [/itex] ?
     
  4. Aug 22, 2013 #3
    I just managed to prove that if the mentioned limit exists, [itex]f[/itex] will necessarily be continuous on some interval [itex]]x-D,x+D[[/itex]. So not only will the function be continuous at the point [itex]x[/itex], but also in some interval around it.

    That was interesting. Next, I'm interested in the differentiability...

    With this function the mentioned limit does not exist. The question is that what are the consequences of the existence of the limit.
     
  5. Aug 22, 2013 #4
    I just understood that Tashi must have assumed that I had been demanding [itex]\delta_1\neq 0[/itex] and [itex]\delta_2\neq 0[/itex] during the limit. There is no such assumption. Only [itex]\delta_1\neq \delta_2[/itex]. See the explanation about the domain [itex]\mathcal{D}[/itex] in my opening post.
     
  6. Aug 22, 2013 #5

    HallsofIvy

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    Tashi was NOT "assuming" that- that is part of the definition of "limit". It is you that have right to say that [itex]\delta_1[/itex] and [itex]\delta_2[/itex] are or can be 0.

    Given that definition, of derivative, it is NOT true that f must be continuous or differentiable in the usual sense at the point.
     
  7. Aug 22, 2013 #6
    You are referring to some stupid high schoold definition that nearly nobody is interested about. This is the real definition: Assume [itex]X[/itex] is some topological space, [itex]x\in X[/itex] some point in it, and [itex]x_1,x_2,x_3,\ldots \in X[/itex] is some sequence in it. We say that the sequence converges to [itex]x[/itex] if for all open [itex]U\subset X[/itex] such that [itex]x\in U[/itex] there exist a [itex]N_U\in\mathbb{N}[/itex] with the property

    [tex]
    n\geq N_U\quad\implies\quad x_n\in U
    [/tex]

    Now the domain of parameters [itex](\delta_1,\delta_2)[/itex] is

    which is naturally a subset of [itex]\mathbb{R}^2[/itex], so we'll set [itex]X=\mathbb{R}^2[/itex] and use the ordinary topology there. It would make no sense to add some additional constraints such as [itex]\delta_1\neq 0[/itex] and [itex]\delta_2\neq 0[/itex] to the general definition.

    Just to be clear I want to ask one thing. I'm sure we agree that the sequence

    [tex]
    1,\frac{1}{2},\frac{1}{3},\ldots
    [/tex]

    converges to zero. What about the constant sequence

    [tex]
    0,0,0,\ldots ?
    [/tex]

    IMO this converges to the zero too. Do you insist that it does not converge to zero?
     
  8. Aug 22, 2013 #7
    I managed to prove one more thing. Assume that a function [itex]f:\;]a,b[\to\mathbb{R}[/itex] is differentiable in its domain. Fix some point [itex]x\in\;]a,b[[/itex]. Now the derivative [itex]f'[/itex] is continuous at point [itex]x[/itex] iff the limit
    [tex]
    \lim_{(\delta_1,\delta_2)\to (0,0)}\frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}
    [/tex]
    exists.

    Very interesting result! :cool: The easier direction comes with the intermediate value theorem. More difficult direction can be proven with an antithesis and some sequence [itex]\lambda_n\to x[/itex] that the antithesis provides.
     
  9. Aug 22, 2013 #8
    hmm.. ok I think it could be that my question was little provocative since it seems that limits

    [tex]
    \lim_{n\to\infty}x_n
    [/tex]

    and

    [tex]
    \lim_{x\to x_0} f(x)
    [/tex]

    could be given deliberately different kind of definitions. If you have some very authorative text that says that the latter limit "must be" taken under the constraint [itex]x\neq x_0[/itex], I believe you. Anyway, I have two comments: First, if a definition is unnatural, it is ok to defy it with more logical new definitions. Second, the definitions that you are referring to probably would imply the constraint [itex](\delta_1,\delta_2)\neq (0,0)[/itex], and not [itex]\delta_1\neq 0[/itex] and [itex]\delta_2\neq 0[/itex]? I made it clear in the beginning that we are assuming [itex]\delta_1\neq \delta_2[/itex] anyway.
     
  10. Aug 22, 2013 #9
    oh dear, I just realized I have still not made it 100% unambiquously clear what I meant with the limit then :redface:. Well here comes the definition. We say that a real number [itex]L[/itex] is the limit denoted as

    [tex]
    \lim_{(\delta_1,\delta_2)\to (0,0)}\frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}
    [/tex]

    if for all [itex]\epsilon>0[/itex] there exists an open set [itex]U_{\epsilon}\subset\mathbb{R}^2[/itex] such that [itex](0,0)\in U_{\epsilon}[/itex] and

    [tex]
    \Phi(U_{\epsilon}\cap\mathcal{D})\subset \;]L-\epsilon,L+\epsilon[
    [/tex]

    where the function [itex]\Phi[/itex] has been defined by the formula

    [tex]
    \Phi:\mathcal{D}\to\mathbb{R},\quad\quad \Phi(\delta_1,\delta_2) = \frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}
    [/tex]

    Do you believe me now?
     
  11. Aug 22, 2013 #10

    Stephen Tashi

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    You don't get any points for being humble, but at least you are objective enough to catch some of your own mistakes.

    It's ok to make new definitions, but you need to state them. I don't think you'll persuade the mathematical world that a new definition of limit is needed.

    You only made a statement that defined a mapping with the condition on it's domain that [itex] \delta_1 \ne \delta_2 [/itex] You didn't provide a complete statement that defined a new kind of limit.
     
  12. Aug 22, 2013 #11
    Check my definition in the post #9. Does it look like familiar or "a new kind of limit"? I would understand if it looks "new" to a high school student.
     
  13. Aug 22, 2013 #12

    Stephen Tashi

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    You clearly stated the definition of a new kind of limit. I don't see that the existence of this new kind of limit implies anything different that the existence of the (usual kind of) limit that defines the derivative f'(x). Do have an example where one limit exists and the other doesn't?
     
  14. Aug 22, 2013 #13
    I don't have a topology book myself, so I would need go to library to check what the real definitions are precisely. But anyway I have hard time believing that my definition could be new. It repeats the idea that small environments must be mapped into small environments. Can you tell what your definition is? It hasn't been made clear either in this thread.

    This works:

    [tex]
    f(x) = \left\{\begin{array}{ll}
    x^2\sin\Big(\frac{1}{x}\Big),\quad & x \neq 0 \\
    0,\quad & x= 0 \\
    \end{array}\right.
    [/tex]

    Now

    [tex]
    \lim_{\delta_2\to 0}\lim_{\delta_1\to 0 }\frac{f(\delta_2)-f(\delta_1)}{\delta_2-\delta_1}
    = \lim_{\delta_2\to 0} \delta_2\sin\Big(\frac{1}{\delta_2}\Big) = 0
    [/tex]

    so if the limit exists, it cannot be anything else than zero. Then define sequences [itex]\delta_1^1,\delta_1^2,\delta_1^3,\ldots[/itex] and [itex]\delta_2^1,\delta_2^2,\delta_2^3,\ldots[/itex] by setting

    [tex]
    \delta_1^n = \frac{1}{2\pi n+ \frac{\pi}{2}},\quad\quad \delta_2^n = \frac{1}{2\pi n - \frac{\pi}{2}}.
    [/tex]

    Now [itex](\delta_1^n,\delta_2^n)\to (0,0)[/itex] when [itex]n\to\infty[/itex], but on the other hand

    [tex]
    \frac{f(\delta_2^n)-f(\delta_1^n)}{\delta_2^n-\delta_1^n}
    = \frac{\frac{1}{(2\pi n - \frac{\pi}{2})^2}\sin\big(2\pi n - \frac{\pi}{2}\big)
    -\frac{1}{(2\pi n + \frac{\pi}{2})^2}\sin\big(2\pi n + \frac{\pi}{2}\big)}
    {\frac{1}{2\pi n - \frac{\pi}{2}} - \frac{1}{2\pi n + \frac{\pi}{2}}}
    = \cdots = -\frac{2}{\pi} + O\Big(\frac{1}{n^2}\Big)
    [/tex]

    so we see that if the limit exists, it cannot be anything else than [itex]-\frac{2}{\pi}[/itex]. According to these remarks, the limit does not exist.
     
  15. Aug 23, 2013 #14

    Stephen Tashi

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    For a real valued function [itex] f [/itex] defined on a metric space, [itex] \lim_{x \rightarrow a } f(x) = L [/itex] means for each real number [itex] \epsilon > 0 [/itex] there exists a real number [itex] \delta [/itex] such that [itex] 0 < | x - a| < \delta [/itex] implies [itex] | f(x) - L | < \epsilon [/itex] (where the [itex] |...| [/itex] denotes the norm function that defines the metric).

    If we want to know the definition for a topological space, we can consult http://en.wikipedia.org/wiki/Limit_of_a_function. Note it makes a statement about [itex] f(U \cap \Omega - \{p\}) [/itex], not about [itex] f(U \cap \Omega) [/itex].
     
  16. Aug 23, 2013 #15
    It is my opinion that the definitions are bad, because it makes no sense to artificially remove the limit point from the allowed domain through which the convergence happens, but it doesn't matter now because of this:

     
  17. Aug 23, 2013 #16

    Stephen Tashi

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    Returning to the question in the original post:

    What about
    [itex] f(x) = 0 [/itex] if [itex]x [/itex]is a rational number.
    [itex] f(x) = x^2 [/itex] if [itex] x [/itex] is an irrational number.
     
  18. Aug 24, 2013 #17
    First choose some sequence [itex]\delta_1^1,\delta_1^2,\delta_1^3,\ldots [/itex] such that all [itex]\delta_1^n[/itex] are irrational and [itex]0<\delta_1^n<\frac{1}{n}[/itex]. With fixed [itex]n[/itex] we can choose a rational [itex]\delta_2^n[/itex] to be as close to [itex]\delta_1^n[/itex] as we want, so we can choose a rational sequence [itex]\delta_2^1,\delta_2^2,\delta_2^3,\ldots [/itex] such that [itex]0<\delta_2^n<\frac{1}{n}[/itex] and

    [tex]
    \Big|\frac{(\delta_1^n)^2}{\delta_2^n - \delta_1^n}\Big| \geq n
    [/tex]
     
  19. Aug 24, 2013 #18

    pwsnafu

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    It makes perfect sense to remove p. Take the function f(0) = 1, and f(x) = 0 elsewhere. Suppose we wanted to take the limit at x=0. We know the limit is zero, so if V = (-1/4, 1/4) we must exclude ##f(p) = 1 \notin V##
     
  20. Aug 24, 2013 #19
    If the limit

    [tex]
    \lim_{x\to x_0}f(x)
    [/tex]

    exists, and if the limit

    [tex]
    \lim_{n\to\infty} x_n = x_0
    [/tex]

    holds, it would be reasonable to expect

    [tex]
    \lim_{n\to\infty} f(x_n) = \lim_{x\to x_0} f(x)
    [/tex]

    too. The standard definition implies that this result does not always hold, which means that the standard definition is bad.
     
  21. Aug 24, 2013 #20

    pwsnafu

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    What‽ For an arbitrary topological space? No, it would not be reasonable to expect that.
     
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