Limsup = Liminf = L implies Lim = L, using neighborhoods

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SUMMARY

The discussion centers on proving the equivalence of limits using the definitions of limit superior (limsup) and limit inferior (liminf) for a function f defined on an interval I in ℝ. Specifically, it establishes that \(\lim_{x\rightarrow t}{f(x)} = L\) if and only if \(\limsup{f(t)} = \liminf{f(t)} = L\). The user attempts to manipulate the definitions of limsup and liminf but expresses confusion regarding the application of these concepts and their relationship to continuity. The discussion highlights the importance of understanding neighborhoods and the implications of continuity in limit proofs.

PREREQUISITES
  • Understanding of limits in calculus, specifically \(\lim_{x\rightarrow t}{f(x)}\)
  • Familiarity with the concepts of limit superior (limsup) and limit inferior (liminf)
  • Knowledge of neighborhoods in the context of real analysis
  • Basic understanding of continuity and its implications for limits
NEXT STEPS
  • Study the definitions and properties of limit superior and limit inferior in detail
  • Explore the relationship between continuity and limits in real analysis
  • Review theorems related to limits of sequences and their connection to functions
  • Practice problems involving neighborhoods and their role in limit proofs
USEFUL FOR

This discussion is beneficial for students of real analysis, particularly those grappling with the concepts of limits, continuity, and the definitions of limsup and liminf. It is especially relevant for those preparing for advanced calculus or analysis courses.

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Homework Statement



I am to prove that

\lim_{x\rightarrow t}{f(x)} = L \Longleftrightarrow \limsup{f(t)} = \liminf{f(t)} = L
where I is an interval in ℝ, and when f:I → ℝ, then \forall t \in I we let N_t denote the set of all neighborhoods of t relative to I, and we define:

\limsup{f(t)} := \inf_{U \in N_t} \sup{\{f(s) | s \in U\}}

\liminf{f(t)} := \sup_{U \in N_t} \inf{\{f(s) | s \in U\}}

The Attempt at a Solution



I'm working on \lim_{x\rightarrow t}{f(x)} = L \Longleftarrow \limsup{f(t)} = \liminf{f(t)} = L at the moment.

So, I've assumed \limsup{f(t)} = \liminf{f(t)} = L.

That is,
\inf_{U \in N_t} \sup{\{f(s) | s \in U\}} := \inf_{U \in N_t} \sup{\{f(U)\}} = L and
\sup_{U \in N_t} \inf{\{f(s) | s \in U\}} := \sup_{U \in N_t} \inf{\{f(U)\}} = L

I know that sup(f(U)) ≥ L for any U in Nt, so there exists some U1 in Nt such that sup(f(U1)) - ε > L for any ε > 0.
==> sup(f(U1)) > L + ε for all ε > 0

Also, inf(f(U)) ≤ L for any U in Nt, so clearly inf(f(U1)) ≤ L < L + ε for any ε > 0.

sup(f(U1)) > L + ε
inf(f(U1)) < L + εI don't think that last step is actually getting me anywhere...I'm completely stuck. Any suggestions would be greatly appreciated. My prof recently introduced liminf, limsup, and neighborhoods, and the new terminology is really messing with my head.
 
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When I look at this problem, I just want to use the theorem that for some sequence (tn), lim (tn) exists if and only if limsup (tn) = liminf (tn), and if we know the value of one of these, we know the value of all of them. But I assume that this can't be the proof, considering the book went to the trouble of defining the limsup and liminf in such special ways. Unless the info given is what I need to construct a sequence (tn) that converges to L.

(I'm very confused...I've been working on this problem for over five hours now.)
 
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Maybe I'm going about it the wrong way. Is it possible to show that f is continuous at t, then I can say that limx-->t f(x) = f(t) = L? Then I can use the limsup = liminf => lim exists theorem for functions (instead of sequences, like I hinted at in my last post), and it seems that the if and only if statement that I'm trying to prove would follow immediately.
 
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