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Priestley step functions

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    I have to show that f: R->R is a step function if and only if:

    1) f is continuous except at finitely many points of R
    2) f takes only finitely many distinct values
    3) f(x) -> 0 as |x| -> infinity

    2. Relevant equations



    3. The attempt at a solution

    I think I have shown that assuming f is a step function then 1, 2 and 3 hold.

    However, I'm not sure about going the other way around, if 1,2 and 3 hold then f is a step function. The question advises using another theorem that I should have learnt last term, but it doesn't specify which and I can't figure out which it means. I was studying differentiation in analysis last term.

    Using 3, I can show there exists an a0 such that f=0 for x<a0 and an such that f=0 for x>an
    EDIT: I'm not so sure that this is as simple as I initially thought. With just 3 on it's own it may never reach 0.

    Does 2 imply that f must be constant over a finite number of intervals? This doesn't seem very rigorous. I suppose this also uses 1 that f is continuous?

    Thank you :)
     
    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2
  4. May 2, 2010 #3
    Sorry, my question was just about priestley step functions.
     
  5. May 2, 2010 #4
    OK sorry - I never was good at reading.
     
  6. May 2, 2010 #5
    I failed to find Priestley step function on Google, but from your other conditions it looks like just a step function with zero values beyond some bounds (presumably so that they always have a finite integral).

    Look at the function in the gaps between consecutive points of discontinuity (or [itex]\pm\infty[/itex]) what do you think you could prove about the function in one of these intervals?
     
  7. May 2, 2010 #6
    By the way it's bit of a heavy hint but the theorem you mention was probably called the "intermediate value theorem" - does that ring a bell?
     
  8. May 2, 2010 #7
    You don't say that the functions are differentiable, so I don't think you will be able to use much connected with differentiation, but the intermediate value theorem may well have been included in the course. If you didn't do it then the proof is still not too awkward.
     
  9. May 2, 2010 #8
    Off to bed. Please someone pick this up if Kate2010 comes back. (You don't need to prove IVT if it's not been covered - you can use (2) instead.)
     
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