Hi Dime,
Thanks for your question. There are many different ways to prove this.
Here's one version:
By definition, a strictly convex function f has the second derivative
f'' > 0. This is the definition you might be familiar with. We'll go
with a different definition of convexity, which makes it a bit easier
to prove:
f is strictly convex if and only if
f(t*x + (1t) y) < t * f(x) + (1t) * f(y) for all x,y in R and t between 0 and 1.
Rearranging this, we get
f(y)  f(x) > [f(x + t (yx))  f(x)]/t = [f(x + t (yx))  f(x)] *
(yx) / (t * (yx))
Take the limit as t goes to 0 and you have the expression in terms of derivatives:
f(y)  f(x) > f'(x) * (yx)
So now we have the statement that f is strictly convex iff
f(y)  f(x) > f'(x) * (yx), for all x,y in R.
Pick x so that f'(x) = 0.
Then f(y) > f(x) for all y in R (as long as y is not equal to x).
=> f(x) will be the global minimum.
q.e.d.
Best wishes,
gwagnerga
Source: Simon, Carl P. and Lawrence Blume. Mathematics for Economists.
Norton. 1994. [an excellent book, by the way.] 
Request for Answer Clarification by
dime365ga
on
10 Nov 2003 18:00 PST
Hiya
Sorry i didnt state this, but im not allowed to assume
differentiability, and i have to prove that there is only one global
minimum
http://planetmath.org/encyclopedia/ExtremalValueOfConvexconcaveFunctions.html
http://en2.wikipedia.org/wiki/Global_minimum
i think these ones show?
im not quite sure though

Clarification of Answer by
gwagnerga
on
11 Nov 2003 05:13 PST
Hey Dime,
You are right. The first link
[http://planetmath.org/encyclopedia/ExtremalValueOfConvexconcaveFunctions.htm]
shows exactly what you need to prove. The only difference is that it
assumes a convex, not a strictly convex function. All you need to do,
though, is adjust the lessthanorequal signs to lessthan signs and
you should be all set.
Best,
gwagnerga

Request for Answer Clarification by
dime365ga
on
11 Nov 2003 06:47 PST
ah ok
but im a little confused, are they missing any steps?
it says:
it follows that f(x) < f(ty + (1t)x)
Since f is convex, we get:
1) f(x) < f(ty + (1t)x)
2) f(x) < tf(y) + (1t)f(x)
so 3) f(x) < f(y) as claimed
how did they go from 1 to 2 to 3??
thanks

Request for Answer Clarification by
dime365ga
on
11 Nov 2003 08:19 PST
would this be a good way to prove this aswell ?
http://web.mit.edu/6.291/wwwold/Lecture_5.pdf (page 4)

Clarification of Answer by
gwagnerga
on
11 Nov 2003 11:54 PST
To answer your previous question, you get from (1) to (2) by the fact
that f(ty + (1t)x) < tf(y) + (1t)f(x). To see that most easily,
refer to the graph provided on page 4 of the PDF from MIT
[http://web.mit.edu/6.291/wwwold/Lecture_5.pdf]. Take x as x_bar, y
as x* and set t = 1/2. A strictly convex function will have the
property that f(x) < f(0.5y + 0.5x) and also the property that f(0.5y
+ 0.5x) < 0.5f(y) + 0.5f(x).
f(0.5y + 0.5x) is the value of the function f at the point halfway
between x and y (or x_bar and x*). You get 0.5f(y) + 0.5f(x) by
looking at the line joining f(y) and f(x). The value 0.5f(y) + 0.5f(x)
takes at the halfway point is even higher than f(0.5y + 0.5x).
Hence, f(x) < f(ty + (1t)x) < tf(y) + (1t)f(x).
Now we have Hence, f(x) < tf(y) + (1t)f(x). Subtract f(x) from both
sides and add tf(x) at both, and you will get tf(x) < tf(y). Divide by
t (which we can do since t>0) and you have f(x) < f(y). Exactly what
you wanted to show.
If you want a complete proof, I would go with this version. The
version on the MIT lecture slides skips even more steps.
Best,
gwagnerga
