Re: Convex Functions

Robert Israel wrote:

quasi <quasi@xxxxxxxx> writes:

On Sat, 08 Sep 2007 18:23:00 EDT, Maury Barbato
<mauriziobarbato@xxxxxxxx> wrote:


Remembering that a function f:R^n->R is said
to be increasing if for every x, y in R^n, with
x_i >= y_i for every i in {1,...,n}, we have
f(x)>=f(y), we could have the following

Let f:R^n->R be an increasing function such that
for every x,y in R^n we have

f((x+y)/2)<=[f(x)+f(y)]/2.

then f is convex.

In fact, any function that is locally bounded above
and midpoint-convex
is convex.

Proof: suppose f is not convex. There exist x, y,
and t with
0 < t < 1 and f(t x + (1-t) y) > t f(x) + (1-t) f(y).
For convenience
I'll take t x + (1-t) y = 0. Now f(s x + (1-s) y) <=
s f(x) + (1-s) f(y)
for all dyadic rationals s = j/2^n, j = 0 .. 2^n. So
there is
a sequence u_n -> 0 with u_n <= b < f(0). Then for
any positive integer m,
f(0) <= (1 - 2^(-m)) f(u_n) + 2^(-m) f((1-2^m) u_n)
so f((1-2^m) u_n) >= (2^m - 1) (f(0) - f(u_n)) + f(0)

>= (2^m - 1) (f(0) - b) + f(0)
For any N and epsilon > 0, we can take m large enough
so
(2^m - 1) (f(0) - b) + f(0) > N,
and then n large enough so || (1 - 2^m) u_n || <
epsilon. This shows
that f is unbounded above in every neighbourhood of
0.
--
Robert Israel
israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada

Ohh, what a wonderful proof!!! I had a job trying to
understand it, but is is worth the trouble! Superb!
My Best Regards,
Maury
.

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