Re: Convex Functions
- From: Maury Barbato <mauriziobarbato@xxxxxxxx>
- Date: Tue, 11 Sep 2007 10:58:57 EDT
quasi wrote:
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.
But it seems quite hard to prove or disprove this
statement.
Ok, here's a proof -- I think this one actually works
...
I plan to prove that f is continuous. Since f is
midpoint-convex,
continuity of f would suffice to imply that f is
convex.
Let p be a point of R^n.
Denote the i'th coordinate of p by p[i]. Thus, p =
(p[1], ..., p[n]).
For any nonzero vector v in R^n with nonnegative
entries, let L(p,v)
be the line through p, parallel to v.
Restricted to L(p,v), f is midpoint-convex and weakly
increasing,
hence f is continuous on L(p,v).
Let u be the unit vector in the direction of <1, ...,
1>. Then f is
continuous on L(p,u).
Let p_1, p_2, p_3, ... be an arbitrary sequence of
points approaching
p. To show that f is continuous at p, it suffices to
show that the
sequence
f(p_1), f(p_2), f(p_3), ...
approaches f(p).
Define a sequence of nonnegative reals c_1, c_2, c_3,
... by
c_m = max( abs(p_m[i] - p[i]), i = 1, ..., n )
Since p_1, p_2, p_3, ... converges to p, it follows
that c_1, c_2,
c_3, ... converges to 0 (from above).
Since f is (weakly) increasing on R^n,
f(p - c_m*u) <= f(p_m) <= f(p + c_m*u)
Letting m approach infinity, continuity of f on
L(p,u) implies
f(p - c_m*u) approaches f(p)
and
f(p + c_m*u) approaches f(p)
Thus, by the squeeze theorem, as m approaches
infinity, f(p_m)
approaches f(p).
Therefore f is continuous at p. Hence, since p was
arbitrary, f is
continuous on R^n, as claimed.
It follows that f is convex.
quasi
Very very good work, quasi!! A simple, elegant proof!
Thank you a lot for having devoted your attention to
my problem.
My Best Regards,
Maury
.
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- Re: Convex Functions
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