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A few questions on continuity.

  1. Dec 29, 2006 #1
    1) let f:[a,b]->R be a continuous function, for every a<=t<=b we define:
    M(t)=sup{f(x)|a<=x<=t} prove that M(t) is a continuous function on [a,b].
    2) let f be a continuous function on [a,b], and we define that A={x in [a,b]| f(x)>=0} where f(a)>0>f(b).
    i need to show that the supremum of A, s, that f(s)=0.
    3) i need to prove/disprove that f(x)=arctan(x) is a uniform continuous on R.

    for the first question, i think bacuase f is continuous then by wierstrauss theorem f achieves its max and min in the interval, so also M(t) is bounded, and the supremum of f(x) in a<=x<=t is also achieved in this interval and thus M(t) is continuous, i think this is basically the idea but i need to formalise this, anyone can help me on this?
    and the second one, i showed that A is bounded and non-empty so it has got a supremum, now we have two options, or s in A or s isnt in A.
    if s is in A, then f(s)>=0, so in oreder to show that f(s)=0, i tried to show ad absurdum that f(s)>0 leads to contradiction, but i didnt succeded in it, where did i go wrong?

    also any help on the third question will helpful.
     
  2. jcsd
  3. Dec 29, 2006 #2

    matt grime

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    1) so your definition of continuous is what precisely? All you've done so far is show that it doesn't have any singularities.

    2) seems easy enough, the sup is obvisously in A: f is continuous. We don't know where you went wrong since your attempted proof doesn't appear to be posted here. The method you cite will work - if f(s) is not zero there must be some t with f(t)>0 and t>s. Hint: t=s+delta/2, and f(t)=epsilon/2.
     
  4. Dec 29, 2006 #3

    StatusX

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    If f is continuous on a closed interval, then it's bounded on that interval. But the converse, that any such function is continuous, clearly does not hold (just take f(x)=sgn(x) on [-1,1]). Use the epsilon delta definition.

    For the third, it might be easier to prove the more general result that any differentiable function whose derivative is bounded is uniformly continuous, and then prove arctan is such a function.
     
  5. Dec 30, 2006 #4
    so do you have any advice?
    but why the sup is obviously in A?
    bacuse f is continuous means that the image of the interval [a,b] is also an interval, but how does this show that s is in A.
    s might be greater than b and b isnt in A, or smaller than b, but still not in A.
    i dont understand why it's obvious it's in A.
    and for the method, if f(s) isnt zero, i.e f(s)>0, but why there must be some t such that f(t)>0, and t>s, is it also because f is continuous?
     
  6. Dec 30, 2006 #5
    so you mean:
    Ae>0,Ed>0,Ax in [a,b], |x-x0|<d => |f(x)-f(x0)|<e
    (x0 is in [a,b])
    i need to show that for every e>0 there exists d'>0 such that for every t in [a,b] (t0 in [a,b]) |t-t0|<d' then |M(t)-M(t0)|<e
    but M(t)=sup{f(x)|a<=x<=t}
    so M(t0)=sup{f(x)|a<=x<=t0}
    because f is continuous: for every x in [a,t] which satisfies |x-x0|<d1
    then |f(x)-f(x0)|<e/2 (where x0 is in [a,t])
    also for every x in [a,t0] which satisfies |x-x0|<d2 (x0 is in [a,t0]) then |f(x)-f(x0)|<e/2
    then: lets take d=min{d1,d2}
    for |x-x0|<d
    we have |M(t)-M(t0)|<=|M(t)-f(x0)|+|M(t0)-f(x0)|<e
    is this correct?
     
  7. Dec 30, 2006 #6

    matt grime

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    Yes, and I gave it - start with the definition of continuity, since you haven't acutally done so. (There might be a better way of doing this, but I don't see it off the top of my head.)


    I suppose it depends on how much analysis you know

    That is impossible. A is a subset of [a,b], so its sup must be less than b.

    Because A is the inverse image of a closed set, hence closed. As I say, this might not be a result you know. However, let's do it from scratch - sup(A) is a limit point of A - there is a sequence x_n of elements in A tending to s by definition. f(x_n)=>0, so what can you say about f(s)?

    Draw a picture. Then prove what you've seen. Have you met the Intermediate value theorem? I'm just using the epsilon delta definition of continuity
     
  8. Dec 30, 2006 #7
    yes i met already the intermediate theorem.
    so lets see if i understand correctly.
    if f(s)>0 let t=s+d/2 for d>0, then: for |s-t|=d/2<d then |f(s)-f(t)|<e/2
    let f(s)>e/2 then f(s)<f(t)+e/2 then 0<f(s)-e/2<f(t) so we have t in A, and t>s a contradiction to that of s being the supremum of A.
    is this correct?
     
  9. Dec 30, 2006 #8

    matt grime

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    It appears to be written backwards. You should start with f(s) as being the epsilon. You're just plucking d from nowhere: don't. In particular d should not appear before the e: d depends on e.

    The point is that if f(x) is strictly bigger than zero, there is a neighbourhood around x where f is always strictly bigger than zero. I'm sure this idea has come up many times in your studies so far.
     
  10. Dec 30, 2006 #9
    yes the idea has come up.
    so you mean i should write it as:
    let e>0, there exists d>0, let t=s+d/2 then |s-t|=d/2<d because f is continuous, let's choose f(s)=e>0, then |f(s)-f(t)|<e
    0=f(s)-e<f(t) so f(t) is bigger than 0 and so is in A.

    i want just to clear something out, what i wrote in post 7, is just the idea, obviously when i write it on paper it will include the quantifiers and the order of epsilon and delta would be adequate.

    btw, what about my appraoch to question 1, in post 5, where should i make any corrections if needed in my proof?
     
  11. Dec 30, 2006 #10
    let me know if my proof is correct.
    let f be a differnetiable function in (a,b) and suppose that the derivative f'(x) is bounded in that interval, then
    |f'(x)|<=M, |lim(x->x0)[f(x)-f(x0)]/x-x0|<=M, then for some neibourhood (x0-d,x0+d) |f(x)-f(x0)|<=M|x-x0| bacuse f is differnetiable then it's also continuous then for every |x-x0|<d for some d (which is connected to e>0) we can choose d=e/M, and we get that f is uniformly continuous.

    and, now for arctan(x), i need to show that it's derivative is bounded, and its derivative is 0<1/(1+x^2)<1.
    is this correct?
     
  12. Dec 30, 2006 #11

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    This doesn't quite follow. Just because the limit of a function as x->x0 is <=M doesn't mean any neighborhood of x0 has values <=M. For example, the limit as x->0 of f(x)=x is 0, which is <=0, but every neighborhood of 0 contains positive numbers. Instead, integrate the function, ie, start with:

    [tex] |f(x)-f(x_0)|=\left| \int_{x_0}^x f'(x) dx \right| \leq \int_{x_0}^x |f'(x)| dx[/tex]
     
  13. Dec 30, 2006 #12
    did i say any neighbourhood, i meant there exists a neighbourhood where the quetient is smaller than M.
    anyway, i cant use integrals.
     
  14. Dec 30, 2006 #13
    and i didnt undersatnd your example, if f(x)=x then f'(x)=1 which isnt smaller then zero
     
  15. Dec 30, 2006 #14

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    When I said it doesn't follow that any neighborhood has that property, I meant it doesn't follow that there exists a single one with the property. And I was just talking about limits in general, but if you need to use a derivative as an example, then take f(x)=x^2. Even though the derivative of x^2 is 0 at x=0, and so in particular is <=0 there, there is no neighborhood of 0 on which the derivative is <=0. Of course the result that |f(x)-f(x_0)|<=M|x-x0| is true, I'm just saying you haven't proven it yet.

    If you can't use integrals, you'll need to at least use the fact that the derivative is <=M on an entire neighborhood of x0, not just at x0. Can you use the mean value theorem?
     
  16. Dec 30, 2006 #15

    matt grime

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    Can I suggest a different method for part 3?

    Pick e>0. Now, there exists a constant K such that

    0<pi/2 - arctan(x) < e for all x with |x| > K. Arctan is uniformly continuous on [-K,K], the rest you can fill in.
     
  17. Dec 30, 2006 #16
    yes i can.
    but how to use it here?
     
  18. Dec 30, 2006 #17

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    If |f(x)-f(x0)|>M|x-x0|, by the MVT, you can show there'd be some point c in (x,x0) with |f'(c)|>M, contradicting M as a bound for |f'|.
     
  19. Dec 30, 2006 #18
    oh, i see you mean lagrange theorem, yes i learned it, and it will do just fine here.
     
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