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Q: Convex functions and global minimums ( Answered 5 out of 5 stars,   0 Comments )
Subject: Convex functions and global minimums
Category: Science > Math
Asked by: dime365-ga
List Price: $10.00
Posted: 05 Nov 2003 19:49 PST
Expires: 05 Dec 2003 19:49 PST
Question ID: 273064
Prove that a strictly convex function f: R --> R cannot possess more
than one global minimum.
Subject: Re: Convex functions and global minimums
Answered By: gwagner-ga on 05 Nov 2003 20:35 PST
Rated:5 out of 5 stars
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.

Best wishes,

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

Sorry i didnt state this, but im not allowed to assume
differentiability, and i have to prove that there is only one global

i think these ones show?
im not quite sure though

Clarification of Answer by gwagner-ga on 11 Nov 2003 05:13 PST
Hey Dime,

You are right. The first link
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.


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??

Request for Answer Clarification by dime365-ga on 11 Nov 2003 08:19 PST
would this be a good way to prove this aswell ?  (page 4)

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
[]. 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.

dime365-ga rated this answer:5 out of 5 stars and gave an additional tip of: $7.00
gwagner did an excellent job, answered all of my clarifications!  Thanks!

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