Re: Using the definition to prove convexity
- From: Paul Smith <phhs80@xxxxxxxxx>
- Date: Tue, 11 Sep 2007 10:59:25 -0000
On Sep 11, 11:04 am, Ray Vickson <C6...@xxxxxxx> wrote:
It is well-known that
f(x) := x^2, x in R,
is a convex function. However, I would like to prove that through the
definition of convex function. I have tried it, but with no success.
Perhaps, not being a professional mathematician, I am missing some
well-known inequality that would solve the problem. Some help would be
very appreciated.
Let u < v < w. On the graph of x^2, slope(u->v) = (v^2-u^2)/(v-u) = v
+u, slope(v->w) = v+w and slope(u->w) = u+w. Since v+u < v+w < u+w we
have, for f(x) = x^2, that [f(v)-f(u)/(v-u) < [f(w)-f(u)]/(w-u) <
[f(w)-f(v)]/(w-v). The first inequality says f(v) < f(u) + (v-u)[f(w)-
f(u)]/(w-u) = {1 - [(v-u)/(w-u)]} f(u) + [(v-u)/(w-u)] f(w). Letting r
= [(v-u)/(w-u)] (which is > 0 and < 1) we have v = u + r(w-u) = (1-r)u
+ rw. We have just finished getting f(v) < (1-r)f(u) + rf(w), so f
satisfies the definition of strict convexity.
Thanks, Ray, but I may be missing something. You start off by assuming
that u < v < w, and afterwards you claim that v+w < u+w. Since u < v
by assumption, the inequality v+w < u+w is not true.
Paul
.
- References:
- Using the definition to prove convexity
- From: Paul Smith
- Using the definition to prove convexity
- Prev by Date: Many Solutions Manuals and Ebooks in Electronic (PDF)Format!
- Next by Date: Re: JSH: SF Algorithm
- Previous by thread: Using the definition to prove convexity
- Next by thread: Re: Using the definition to prove convexity
- Index(es):
Relevant Pages
|
