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Help with a continuous function lemma

  1. Oct 20, 2013 #1
    I am on page 97 of Spivak calculus and having trouble proving Theorem 3 of chapter 6 (which he says is a lemma for the next chapter). I don't know how to type the symbol delta so I am replacing delta with @, and replacing epsilon with &

    Theorem: Suppose f is continuous at a, and f(a)>0. Then there is a number @>0 such that f(x)>0 for all x satisfying |x-a|<@

    Proof: Consider the case f(a)>0 Since f is continuous at a, if &>0 there is a @>0 such that, for all x, if |x-a|<@, then |f(x)-f(a)<&

    Since f(a)>0, we can take f(a) as & Thus there is @>0 so that for all x, if |x-a|<@ then |f(x)-f(a)|<f(a)

    So far, I understand all of this. The part I don't understand is the very last line of the proof, which says:

    and this last inequality implies f(x)>0

    I don't get how if |x-a|<@ then |f(x)-f(a)|<f(a) implies that f(x)>0.

    INTUITIVELY I get it...because if |f(x)-f(a)|<f(a) that means the distance between f(x)-f(a) is smaller than the number f(a), and f(a)=f(a)-0, so if f(x) were negative, the distance between f(x) and f(a) would be greater than the distance between f(a) and 0. I get it...but I jus't cant prove it with rigor.

    Thank you in advance for your help



    Somebody posted this answer on yahoo answers:

    If |f(x) - f(a)| < f(a), then -f(a) < f(x) - f(a) < f(a), so in particular -f(a) < f(x) - f(a). Adding f(a) to both sides of this inequality gives 0 < f(x).

    But I dont see why the absolute value signs got dropped
     
  2. jcsd
  3. Oct 20, 2013 #2

    CompuChip

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    If you want, you can go back to the definition of the absolute value, which is
    $$|z| = \begin{cases} -z & \text{ if } z < 0; \\ 0 & \text{ if } z = 0; \\ z & \text{ if } z > 0. \end{cases}$$
    (actually you can cover the = 0 case in one of the inequalities, but let's be explicit).

    So let's work through them, setting z = f(x) - f(a):
    (1) The trivial case is f(x) - f(a) = 0, so |f(x) - f(a)| < f(a) immediately gives you 0 < f(a).
    (2) if f(x) - f(a) < 0 then you get |f(x) - f(a) = -(f(x) - f(a)) < f(a). [...]
    (3) if f(x) - f(a) > 0 then you get |f(x) - f(a)| = f(x) - f(a) < f(a). [...]
    Can you fill the dots by yourself now?
     
  4. Oct 25, 2013 #3
    Thanks for the reply, I have been busy all week so I just noticed it.

    Case 1 I see how you got 0 < f(a), but I don't see how that implies f(x)>0. Again I understand it intuitively but I still don't see how it is proven by this case
    Case 2 I filled in the dots, and it makes complete sense to me
    Case 3 I filled in the dots, and it makes complete sense to me
     
  5. Oct 25, 2013 #4
    Because by assumption, ##f(x)-f(a)=0## and thus ##f(x)=f(a)##.
     
  6. Oct 25, 2013 #5
    Wow I am slow :)

    Thanks for pointing that out
     
  7. Oct 25, 2013 #6
    One more question....so the epsilon-delta limit definition states that for ALL epsilon>0, there is some delta>0 such that for all x if 0<|x-a|<delta, then |f(x)-L|<epsilon. The key word being ALL.

    In the example spivak game he assumed Epsilon=f(a). What about for all other epsilon? How does proving that it works when epsilon=f(a) prove that it works for all epsilon?
     
  8. Oct 25, 2013 #7

    Office_Shredder

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    You're told that the function is continuous, you never have to prove that any limits exist at any point in time.
     
  9. Oct 25, 2013 #8
    I'm not sure I understand. if we don't have to prove any limits exist at any point in time why is the whole proof involving proving a limit?

    What's the point of proving it works only when Epsilon = f(a) if we cannot prove it works given any epsilon? What if epsilon does not equal f(a)?
     
  10. Oct 25, 2013 #9

    Office_Shredder

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    The function is continuous, so
    [tex] \lim_{x\to a} f(x) = a [/tex].

    You use the existence of the limit which is already known to prove things by picking a specific epsilon which is convenient, you aren't trying to prove that the limit exists or that things are true for arbitrary choices of epsilon.
     
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