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A lemma about upper derivatives (upper, non right)

  1. May 8, 2006 #1
    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. jcsd
  3. May 10, 2006 #2
    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.

    So, by definition

    L =

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

    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.
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