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Uniform Continuity- extentions of functions

  1. Dec 13, 2007 #1
    Hi guys. Final tomorrow and i had some last minute questions for proving/disproving a function is uniformly cont.

    Basically i want to know if the following proofs are acceptable

    Consider f(x)=1/x for x element (0, 2) = I

    Proof 1:

    f(x) does not converge uniformly on I. In order for a function to be uniformly continuous on an open interval, say (a,b) it must be able to be extended to a continuous function on [a,b].

    Since lim f(x) = Infinity , the extension function cannot be continuous, therefore f(x) is
    x-->0
    not uniformly continuous on I.//

    Proof 2:

    Thm: if f is uniformly continuous on S, and sn is a Cauchy sequence in S, then f(sn) is also a Cauchy sequence.

    consider the sequence sn= 1/n which is clearly Cauchy on I

    if f is uniformly continuous on I, then f(sn) will also be Cauchy. But f(sn) = n which is not a Cauchy sequence. thus, f is not uniformly continuous on I.//


    Are those 2 proofs acceptable? Any feed back would greatly be appreciated


    Also i wanna know if my epsilon delta proof is correct.

    Prove that X^2 is uniformly continuous on [0,3]
    (i know i can use that one theorem, that states if its continuous on a closed interval, its uniformly continuous but i wanna practice epsilon delta)

    Scratch work

    for each epsilon> 0 there exists delta>0 such that

    |x-y|< delta ==> |x^2 - y^2| < epsilon

    |x^2 - y^2| = |(x-y)(x+y)|< |(x-y)*9|<epsilon

    thus choose delta = epsilon/9

    Forward Proof

    for each epsilon>0 there exists a delta>0 s.t.

    |x-y|<delta= epsilon/9 implies |f(x)-f(y)|= |(x-y)(x+y)|< |(x-y)*9| < |(epsilon/9)*9| = epsilon //


    Is this done correctly? Am i forgetting to do something. Any help is appreciated. Sorry for such a long post... Please reply over!
     
  2. jcsd
  3. Dec 13, 2007 #2

    morphism

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    Everything looks good. Although I would reword your "Forward Proof" to "for each epsilon>0 choose [or let/take/set/etc] delta=epsilon/9 ..." but this is just a matter of taste.
     
    Last edited: Dec 13, 2007
  4. Dec 13, 2007 #3

    Dick

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    Don't confuse uniform continuity with uniform convergence. They are two different things. Your proofs of the first two are needlessly indirect. If you need practice with epsilon-delta proofs I would suggest you apply it to the first problem, it's much closer to the spirit of the concept. Show given an epsilon that there is no delta that will work. The second one is fine, but I'm a little confused why you chose 9, when if x and y are in [0,3] then (x+y)<=6. Not that 9 doesn't work :).
     
  5. Dec 13, 2007 #4

    morphism

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    I suspect the OP meant to say "f is not uniformly continuous on I" instead of "f(x) does not converge uniformly on I" - a Freudian slip perhaps!
     
  6. Dec 13, 2007 #5

    Dick

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    I thought so too. But it doesn't hurt to warn of coming of coming concepts with similar names. That's how freudian slips happen.
     
  7. Dec 13, 2007 #6
    ya sorry guys i meant uniformly continuous... oops

    and ya it should be epsilon over six, i forgot how to add...

    @ dick...

    Epsilon delta proof for the first problem...

    i actually dont know how to disprove something with an epsilon delta proof... could you help me out

    here is what i have so far...

    for some epsilon >0 there exits delta >0 such that
    |x-y|<delta implies |f(x)-f(y)|< epsilon

    |1/x - 1/y| = |(y-x)/xy|

    I know i have to find some epsilon that violates this, or that
    |x-y|< delta implies |f(x)-f(y)|> epsilon

    or maybe the other way around... but any way ya, i dont know where to go from there.
     
  8. Dec 14, 2007 #7

    Dick

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    f is unbounded. This means for NO epsilon is there a delta such that |x-y|<delta -> |f(x)-f(y)|<epsilon.
     
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