A lemma about upper derivatives (upper, non right)

  • Context: Graduate 
  • Thread starter Thread starter Castilla
  • Start date Start date
  • Tags Tags
    Derivatives
Click For Summary
SUMMARY

The discussion centers on a lemma regarding upper derivatives, specifically the behavior of the function h(y) defined as {g(y) - g(a)} / (y - a) as y approaches a. The lemma asserts that for any ε > 0 and δ > 0, there exists a y such that 0 < |y - a| < δ and h(y) > L - ε, where L is the limit superior of h(y) as y approaches a. The proof utilizes the definition of limit superior and supremum to establish the existence of such a y, correcting an earlier misstatement regarding the function g(y).

PREREQUISITES
  • Understanding of limit superior in calculus
  • Familiarity with the concept of upper derivatives
  • Knowledge of epsilon-delta definitions in mathematical analysis
  • Basic proficiency in mathematical proofs and notation
NEXT STEPS
  • Study the properties of limit superior in real analysis
  • Explore the definitions and applications of upper derivatives
  • Review epsilon-delta proofs in calculus
  • Learn about the implications of supremum in mathematical functions
USEFUL FOR

Mathematicians, students of advanced calculus, and researchers interested in real analysis and the behavior of functions near points of discontinuity.

Castilla
Messages
241
Reaction score
0
Let be

lim sup {g(y) - g(a)} / (y - a) = L.
y-> a

It seems that this lemma exists: For all epsilon e > 0 and delta d > 0, there is an y such that 0 < l y - a l < d and g(y) > L - e.

Can someone give me some hint to proof this?

Thank you.
 
Physics news on Phys.org
If someone is interested, there was a mistake in my statement of the lemma.

Le h(y) be = { g(y) - g(a) } / (y - a). What the lemma says is that for any pair of e and d, there is an "y" such that
0 < | y - a | < d and h(y) > L - e (erroneously I put g(y) instead of h(y) in the original statement).

The proof is this (excuse the clumsiness, I can't work on Latex, not for lazyness but for some technical reason beyond my powers):

Lets take and e and d. The known fact is that

limsup h(y) = L.
y->a

So, by definition

L =

lim sup {h(y) / 0 < | y - a | < k }
k->0

So for the selected e there exists a "k" such that k < d and
| sup {h(y) / 0 < |y-a|< k } - L | < e, and therefore
sup {h(y) / 0 < |y-a|< k} > L - e.

But this (and taking on account the definition of supremum) implies that there must be an y such that 0<|y-a|<k (hence 0<|y-a|<d) and h(y) > L - e.

That is all.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K