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A Simple Convergence Question

  1. Nov 27, 2013 #1
    Hi,

    I have a basic question about convergence.

    I have two sequences, x1, x2, ... and y1, y2, ..., where yn = f(xn) for some function f : ℝN → ℝ.

    I have shown that the sequence, y1, y2, .... converges. What conditions do I need on the function, f, to ensure that the sequence x1, x2, ... also converges?

    Thanks in advance
     
  2. jcsd
  3. Nov 27, 2013 #2

    lurflurf

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    You need the function f to be continuous.
     
  4. Nov 27, 2013 #3

    Office_Shredder

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    In fact this is an equivalent definition of being continuous: f is continuous if and only if whenever [itex] x_n \to x [/itex], [itex] f(x_n) \to f(x) [/itex].
     
  5. Nov 27, 2013 #4
    But the poster is asking the converse. He's asking if ##f(x_n)\rightarrow x## implies ##x_n\rightarrow x##. So continuity is irrelevant here.
     
  6. Nov 28, 2013 #5
    Indeed, R136a1 is correct. I want the convergence of the xn → x given the convergence of f(xn) → f(x).

    In this case, continuity of f is not enough. For example, f(x) = x2 is continuous, but if we define the two series as follows

    x1 = √2, x2 = -√2, x3 = √2, x4 = -√2, ...

    and

    y1 = 2, y2 = 2, y3 = 2, y4 = 2, ...

    then yn is convergent and yn = f(xn), but xn is not convergent.

    If the function, f, were invertible and the inverse was continuous, then that would be enough. Right?

    The problem is that in my case I don't think that f is invertible.

    I don't necessarily need the result to hold globally, i.e. for all x. A local result might suffice.

    Any syggestions would be appreciated.
     
  7. Nov 28, 2013 #6
    You don't even need fully invertible. Just a continuous left-inverse suffices.

    I kind of doubt there is a simple condition. Could you perhaps give some more information about your specific situation?
     
  8. Nov 28, 2013 #7
    Hi,

    Thanks for the response.

    The problem is related to an algorithm that I've constructed for optimising Markov decision processes. I'm trying to prove the global convergence of the algorithm. I don't want to go into any unnecessary detail, so I will try to explain as briefly as I can.

    I have an objective function, f, of which the x's are the parameters I wish to optimise.

    My algorithm generates a series of parameter vectors, x1, x2, ...

    I've managed to show that the objective is strictly monotonically increasing with respect to this series, i.e. that

    f(x1) < f(x2) < ...

    I know that f is bounded from above, so I used the monotone convergence theorem to conclude that the series

    f(x1), f(x2), ...

    converges.

    Generally, f is not injective and so doesn't have an inverse.
     
  9. Nov 28, 2013 #8
    Thanks for your response. But what I really wanted to know is whether ##f## has some special properties that we could use. We know it's not injective, but is there perhaps something else? Obviously, if ##f## can be any continuous function, then what you're trying to do is false. So there must be something special going on.
     
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