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#1 2006-03-14 00:32:57

Jake
Guest

real analysis problem

a function f: R -> R has the property that for any four real numbers a, b, c, d such that a - b > c - d, we have f(a) - f(b) > f(c) - f(d). prove that f is a linear function, ie f(x)  = mx + n for all x belonging to R, where m, n belong to R and m > 0


does anyone know the name of this theorom or have any ideas how to prove it?!

#2 2006-03-14 03:26:23

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: real analysis problem

Are you supposed to prove that f can be a linear function, or f has to be a linear function?

If it's the first:

We know a - b > c - d.  We also know that f(x) = mx + n.  So let's apply m to both sides of our original equation.  m(a-b) > m(c-d).  Note that m is positive, and thus, this is legal.  So ma - mb > mc - md.  Now let's add and subtract n.  ma - mb + n - n > mc - md + n - n.  After a bit of rearrangement, we get ma + n - mb - n > mc + n - md - n.  Grouping that negative, we get ma + n - (mb + n) > mc + n - (md + n).  Now we can write this as f(a) - f(b) > f(c) - f(d).  So the linear function holds.

Edit:

To do this problem, you must work it backwards first.

Start out assuming f(a) - f(b) > f(c) - f(d).  Then you can replace that with ma + n - (mb + n) > mc + n - (md + n).  After this, you can get ride of the n's, divide the whole thing through by m, and you should end up with a - b > c - d.  Then when writing the proof, you start at the bottom (a - b > c - d) and work your way back up to end at f(a) - f(b) > f(c) - f(d).

Last edited by Ricky (2006-03-14 03:28:26)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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