# A lemma about upper derivatives (upper, non right)

1. May 8, 2006

### Castilla

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.

2. May 10, 2006

### Castilla

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 cant 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.