Help with a continuous function lemma

  • #1
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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
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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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?
 
  • #3
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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
 
  • #4
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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

Because by assumption, ##f(x)-f(a)=0## and thus ##f(x)=f(a)##.
 
  • #5
190
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Wow I am slow :)

Thanks for pointing that out
 
  • #6
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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?
 
  • #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.
 
  • #8
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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.

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)?
 
  • #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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