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 + (1-t) y) < t * f(x) + (1-t) * 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 (y-x)) - f(x)]/t = [f(x + t (y-x)) - f(x)] *
(y-x) / (t * (y-x))
Take the limit as t goes to 0 and you have the expression in terms of derivatives:
f(y) - f(x) > f'(x) * (y-x)
So now we have the statement that f is strictly convex iff
f(y) - f(x) > f'(x) * (y-x), 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,
gwagner-ga
Source: Simon, Carl P. and Lawrence Blume. Mathematics for Economists.
Norton. 1994. [an excellent book, by the way.] |
Request for Answer Clarification by
dime365-ga
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
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Clarification of Answer by
gwagner-ga
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 less-than-or-equal signs to less-than signs and
you should be all set.
Best,
gwagner-ga
|
Request for Answer Clarification by
dime365-ga
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 + (1-t)x)
Since f is convex, we get:
1) f(x) < f(ty + (1-t)x)
2) f(x) < tf(y) + (1-t)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
dime365-ga
on
11 Nov 2003 08:19 PST
would this be a good way to prove this aswell ?
http://web.mit.edu/6.291/www-old/Lecture_5.pdf (page 4)
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Clarification of Answer by
gwagner-ga
on
11 Nov 2003 11:54 PST
To answer your previous question, you get from (1) to (2) by the fact
that f(ty + (1-t)x) < tf(y) + (1-t)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/www-old/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 + (1-t)x) < tf(y) + (1-t)f(x).
Now we have Hence, f(x) < tf(y) + (1-t)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,
gwagner-ga
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