# From decreasing sequences to decreasing functions

1. May 12, 2006

### Castilla

Hello.

I know this:

If (a_n) is a bounded below decreasing sequence, then

lim (a_n) = inf { a_n / n = 1,... }
n->oo

How to translate this to real functions ?

I mean, I have read that:

lim (sup { f(x) / 0< |x-a|< e}) =
e->0

inf { sup {f(x) / 0< |x-a|< e} / e > 0}

and i suppose it has something to do with the previous but I fail to see how.

Can you help me?

2. May 13, 2006

### Jimmy Snyder

For starters, consider sup { f(x) / 0< |x-a|< e} as a function of e. It is defined only for e > 0. It is bounded from below by f(a). Also, it is decreasing as $$e \rightarrow 0$$. Let me know if this is enough help.

Last edited: May 13, 2006
3. May 13, 2006

### Castilla

Lets say g: R+ -> R: g(e) = sup { f(x) / 0< |x-a| < e}.

But I fail to see why f(a) is an inferior bound of { g(e) / e > 0}.

"a" is not even in the dominion of f.

4. May 13, 2006

### Jimmy Snyder

That may or may not be so. However you bring up a valid point, f(a) is not necessarily a lower bound for g(e) even if it exists. My solution needs the following modification.

If g(e) is not bounded from below in any neighborhood of 0, then lim sup = inf sup = $$-\infty$$. If g(e) is bounded from below in some neighborhood of 0, then use that bound instead of f(a) in my previous suggestion. Either way lim sup = inf sup.

5. May 15, 2006

### Castilla

Let see...

We have g(e) = sup { f(x) / 0 < |x-a| < e }.

Let's suppose that there exist E and M such that if e belongs to the open interval (0, E) then g(e) < M. This fact, plus the increasing condition of function g, implies that M is a lower bound of A = { g(e) / e > 0}. Then A has an infimum, "w".

Let be d>0. By definition of infimum, there exists and e_1 which belongs to A and g(e_1) < w + d.
Being g an increasing function, we have, for all e of the interval (0, e_1), g(e_1) >= g(e). So for all e of (0, e_1):

w + d > g(e_1) >= g(e) >= w > w - d, therefore | g(e) - w | < d.

This means that

limit g(e) = w = inf A = inf {g(e) / e > 0}.
e->0+

Is this okay, jimmysnyder??

6. May 15, 2006

### Jimmy Snyder

I'm sorry, I did not read past this part of your message. In my message, I was assuming that there exist E and m such that if e belongs to the open interval (0, E) then g(e) > m.

I want to take advantage of the fact that as e approaches 0, g(e) is nonincreasing. That is, if 0 < e' < e, then
{ f(x) / 0 < |x-a| < e' } is a subset of { f(x) / 0 < |x-a| < e }.
Therefor g(e') is less than or equal to g(e).

7. May 15, 2006

### Castilla

I did not mention that the function g was an increasing one (or better, nondecreasing) but its definition implies this, because g(e) = sup { f(x) / 0 < |x-a| < e}.

Jimmysnyder, could you read my previous post? I think it is all right but may be you can confirm this...

There is a mistake at mi previous post!! It must say: M < g(e).

Last edited: May 15, 2006
8. May 15, 2006

### Jimmy Snyder

Well I read a little more but it still has problems.
M is a lower bound of A = {g(e) / 0 < e < E}.

However, I don't think any of this is necessary. Just use the fact that g(e) is bounded from below in (0,E) and nonincreasing as e approaches 0. Apply the same reasoning as in the sequential version. Q.E.D.

9. May 15, 2006

### Castilla

That is ok, thanks for your help!