Prove the left hand limit and right hand limit exist and are finit

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Homework Statement



let f: I -> R, where I is an open interval and f is a monotone function. Assume f is discontinuous at c. show that lim x→c- f(x) and lim x→c+ f(x) exist and are finite

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The Attempt at a Solution



Well since f is discontinuous at c we know there exists a ε0, for all δ > 0 such that |x-c| < δ and |f(x)-f(c)| ≥ ε0.

So from that we know,

f(c) - ε0 ≥ f(x) ≥ f(c) + ε0

so it would seem that f(x) is bounded. If it is bounded and, WLOG since it is monotone, increasing it will converge to the sup. And i suppose that sup is the limit? This is where i get confused and where i get lost in how to show all of this using symbols and what not
 

Answers and Replies

  • #2
jbunniii
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So from that we know,

f(c) - ε0 ≥ f(x) ≥ f(c) + ε0
This inequality doesn't make any sense. The leftmost expression is smaller than the rightmost expression, so there can't be any ##f(x)## for which the inequality is true.

Let's consider the limit on the left side: ##\lim_{x \rightarrow c^-} f(x)##. The set ##\{f(x) : x < c\}## is bounded above by ##f(c)## (why?), so it has a least upper bound, call it ##M##. Can you show that ##\lim_{x \rightarrow c^-} f(x) = M##?
 
  • #3
jbunniii
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By the way, notice that your proof won't need to use the fact that ##f## is discontinuous at ##c##. The left and right hand limits certainly exist if ##f## is continuous at ##c##, so the theorem is indicating that they actually exist whether or not ##f## is continuous at ##c##. Therefore you just need a general proof that the left and right limits always exist if ##f## is monotone, regardless of continuity at ##c##.
 
  • #4
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This inequality doesn't make any sense. The leftmost expression is smaller than the rightmost expression, so there can't be any ##f(x)## for which the inequality is true.

Let's consider the limit on the left side: ##\lim_{x \rightarrow c^-} f(x)##. The set ##\{f(x) : x < c\}## is bounded above by ##f(c)## (why?), so it has a least upper bound, call it ##M##. Can you show that ##\lim_{x \rightarrow c^-} f(x) = M##?

Ok so ##\{f(x) : x < c\}## is bounded above by ##f(c)## because x < c and f is monotone (decreasing). Since it is bounded below it has a inf (by axiom of completeness) call it M. And by monotone convergence theorem f(x) converges to its inf? thus ##\lim_{x \rightarrow c^-} f(x) = M##?
 
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jbunniii
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Ok so ##\{f(x) : x < c\}## is bounded above by ##f(c)## because x < c and f is monotone (decreasing).
Well no, it's bounded above by ##f(c)## if ##f## is increasing. (We can handle the decreasing case later, by considering ##-f##.)

Since it is bounded below it has a inf (by axiom of completeness) call it M.
Let's stick with one or the other - either it is bounded above (increasing case) or below (decreasing case). If we assume ##f## is increasing, then ##\{f(x) : x < c\}## is bounded above by ##f(c)##, so it has a least upper bound, call it ##M##.

Now you should be able to make a simple epsilon-delta argument that ##\lim_{x \rightarrow c^-} f(x) = M##.
 
  • #6
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Well no, it's bounded above by ##f(c)## if ##f## is increasing. (We can handle the decreasing case later, by considering ##-f##.)


Let's stick with one or the other - either it is bounded above (increasing case) or below (decreasing case). If we assume ##f## is increasing, then ##\{f(x) : x < c\}## is bounded above by ##f(c)##, so it has a least upper bound, call it ##M##.

Now you should be able to make a simple epsilon-delta argument that ##\lim_{x \rightarrow c^-} f(x) = M##.

I cannot figure out what to choose δ to be in order to get |f(x) - M| < ε
 
  • #7
jbunniii
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If we put ##S = \{f(x) : x < c\}##, then we have defined ##M = \sup(S)##.

Therefore for any ##\epsilon > 0##, there must be an element ##y \in S## such that ##M - \epsilon < y \leq M##.

But every element of ##S## is of the form ##f(x)## for some ##x < c##. So ##y = f(x)## for some ##x < c##. Thus there is some ##x < c## such that ##M - \epsilon < f(x) \leq M##.

But we can conclude more: use the fact that ##f## is increasing.
 
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  • #8
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If we put ##S = \{f(x) : x < c\}##, then we have defined ##M = \sup(S)##.

Therefore for any ##\epsilon > 0##, there must be an element ##y \in S## such that ##y > M - \epsilon##.

But every element of ##S## is of the form ##f(x)## for some ##x < c##. So ##y = f(x)## for some ##x < c##. Thus there is some ##x < c## such that ##f(x) > M - \epsilon##.

But we can conclude more: use the fact that ##f## is increasing.

well since we have for some x0 we have f(x0) > M - ε that would imply that there exists an x ≥ x0 where (since f is increasing)
f(x0) ≤ f(x).

Since M is the sup we then have

M - ε < f(x0) ≤ f(x) ≤ M < M + ε

thus |f(x) - M| < ε.

Isn't this just the proof of the monotone convergence theorem, but applied to function? If so, would it have been valid to say that since f(x) is bounded above and monotone increasing it converges to its sup?
 
  • #9
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Also for the case where we approach from the right of c do we do the same thing, but only this time treat the function as decreasing and having an inf since we assumed that f was increasing and now we are going in reverse?
 
  • #10
jbunniii
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well since we have for some x0 we have f(x0) > M - ε that would imply that there exists an x ≥ x0 where (since f is increasing)
f(x0) ≤ f(x).

Since M is the sup we then have

M - ε < f(x0) ≤ f(x) ≤ M < M + ε

thus |f(x) - M| < ε.
Actually it's more than "there exists an ##x##" - it is true, for ALL ##x## satisfying ##x_0 \leq x < c##, that ##M - \epsilon < f(x_0) \leq f(x) \leq M##. And this is exactly what you need in order to conclude that ##\lim_{x \rightarrow c^-}f(x) = M##.
Isn't this just the proof of the monotone convergence theorem, but applied to function? If so, would it have been valid to say that since f(x) is bounded above and monotone increasing it converges to its sup?
It's the same general idea, but if you only have the theorem for sequences (a monotonically increasing sequence that is bounded above converges to its supremum), then you have to prove that it works for monotone functions too. That is essentially what the above proof has done.
 

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