Ph.D. Qualifying Exam – Spring 2004

We consider now two cases, depending on |z|:• If |z| 1: Just take N = 1 in (2) in order to get∣∣ g(z) ‖f‖∞ + K|z| ( ‖f‖ ∞ + K )|z|2 2• If |z| < 1: Choose N to be the only positive integer such thatthat isand see what happens in (2):1 1< |z| 2N+1 N ln 1/|z|ln 22 N< N + 1∣ g(z)∣ ∣ 4‖f‖∞ |z| + K|z|2 ln2 ln 1|z|Letting c be some bigass constant, for example(Kc = Max 4‖f‖ ∞ ,2 ln2 , ‖f‖ ∞ + K )2we get an inequality that combines both cases:∣ ∣ ∣ (∀x, z ∈ R ∣ g(z) = ∣f(x + z) − f(x) c 1 + ln + 1 )|z||z|18