Upper and lower derivatives


by Castilla
Tags: derivatives, upper
Castilla
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#1
Apr17-06, 06:00 PM
P: 240
Hello, does someone knows where I can find information about upper and lower derivatives??

For example, why they always exists and why if both have the same value L then the derivative exists and it is L ??

Thank you.
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HallsofIvy
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#2
Apr17-06, 06:57 PM
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PF Gold
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The "upper derivative" at x= a is just
[tex]lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}[/tex]
while the "lower derivative is
[tex]lim_{h\rightarrow 0^-}\frac{f(a+h)- f(a)}{h}[/tex]
and, of course, the derivative is
[tex]lim_{h\rightarrow 0}\frac{f(a+h)- f(a)}{h}[/tex]
By the time you get to derivatives you should have learned that the limit exists if and only if the two one-sided limits exist and are the same.
Castilla
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#3
Apr17-06, 07:28 PM
P: 240
Excuse me, but the upper and lower derivatives I refer to are not the "lateral derivatives" but

(upper derivative)

lim sup { f(x) - f(a) } / (x - a)
x->a

and the lower derivative defined with lim inf.

HallsofIvy
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#4
Apr18-06, 06:17 AM
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PF Gold
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Upper and lower derivatives


My apologies. When I read "they always exists" I thought it was just poor English for "if they exist".

The lim sup and lim inf will always exist if the function f is bounded since every bounded set of real numbers has a lim inf and lim sup. There are unbounded functions for which they do not exist. If lim sup and lim inf for any function exist and are the same then it follows that the limit exists and is that common value.
Castilla
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#5
Apr18-06, 09:11 AM
P: 240
That is my question. How the equality

Lim [supremum of { (f(x) - f(a) }/ (x-a) s.t. 0< lx-al <e } ] =
e->0

Lim [ infimum of { f(x) - f(a) } / (x-a) s.t. 0< lx-al < e } ]
e->0

implies the existence of f '(a).

Thanks.
Hurkyl
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#6
Apr18-06, 05:51 PM
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I think you're just making it too complicated.

What could you say if you had:

[tex]
\limsup_{x \rightarrow 0} g(x) = \liminf_{x \rightarrow 0} g(x)
[/tex]

?
Castilla
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#7
Apr18-06, 07:25 PM
P: 240
My problem is that I only know this definitions of lim sup and lim inf (those I put in my last post) and I dont see how to simplify things.
Hurkyl
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#8
Apr18-06, 07:47 PM
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Then your goal is to try and prove something when these two are equal!

(Or look in your book and find the relevant theorem)

If [itex]\limsup_{x \rightarrow 0} g(x) = L[/itex], what does that mean? (I bet you can guess the next question I'm going to ask)

So if they are both equal to L, then what?


This will, of course, require you to either use some theorems about sups and infs, or substitute in their definition too.
Castilla
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#9
Apr18-06, 08:52 PM
P: 240
I can only see this long way.

For some reason Latex forms can not be generated in PCs of this city (Lima).

Assume the function f is bounded in a (probably deleted) neighbourhood of "a".

Let b_n = supr. { f(x) / 0 < lx-al < 1/n }. Then the sequence (b_n) is decreasing, therefore convergent.

Let c_n = inf { f(x) / 0< lx-al < 1/n }. Then the sequence (c_n) is increasing, therefore convergent.

Let L be lim b_n. Then L = inf { b_n / n = 1,...}. Then for all e(psilon) there exists an n(e) / b_n - L < e.
That means that for all e there exists and n(e) / supr. {f(x) / 0<lx-al< 1/n} - L < e;
Then for all e there exists and n(e) / if x fullfills 0 < lx-al < 1/n -> f(x) - L < e. (1)

Suppose L is also lim c_n. Then L = sup { c_n / n = 1,...} Analogous reasoning concludes in this: for all e there exists and n(e) / if x fulfills 0 < lx-al < 1/n -> -e < f(x) - L. (2)

Joining (1) and (2) I got that for all e there exists an n(e) / if x fullfills 0 < lx-al < 1/n -> lf(x) - Ll < e. In other words, f(x) -> L (x->a).
Castilla
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#10
Apr20-06, 12:56 PM
P: 240
Hurkyl, sorry to bother, may be can you check the previous post? Thanks.
Nimz
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#11
Apr20-06, 03:38 PM
P: 81
For a hint that is a bit more direct, another name for sup is the least upper bound, and likewise, inf is the greatest lower bound. If your upper bound matches your lower bound, what does that tell you about the thing you are bounding?
Castilla
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#12
Apr26-06, 08:04 PM
P: 240
That helps. Thank you, Nimz.


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