Expert: Paul Klarreich - 12/12/2006

Question
Hello, I have a problem.It;s hard for me to find the right proof for the following problem.We have the function f(x) which is defined for all x.And we must prove that the case when f(x)<o and f''(x)>o (for all x)is impossible.If I think that since the function is negative, it means that we can present it as some g(x) multiplied by the constant: -1.And it doesn't matter how many times we derivate it, the constant will stay the same.I mean the initial and the double prime functions will have the same signs.Is it enough, how do you think?Does it sound a little bit persuasive?:)
Thanx.
Please reply if you can ASAP

Answer
Questioner:  Ann
Category:  Calculus
 
Subject:  Derivatives.HELP NEEDED!!!
Question:  Hello, I have a problem.It;s hard for me to find the right proof for the following problem.  We have the function f(x) which is defined for all x.And we must prove that the case when f(x)<o and f''(x)>o (for all x)is impossible.  If I think that since the function is negative, it means that we can present it as some g(x) multiplied by the constant: -1.  And it doesn't matter how many times we derivate it, the constant will stay the same.  I mean the initial and the double prime functions will have the same signs.  Is it enough, how do you think?Does it sound a little bit persuasive?:)
Thanx.
Please reply if you can ASAP
........................................
Hi, Ann,

As I understand it, you want to prove that if f''(x) > 0 for all x, then there must exist  x such that  f(x) > 0

Geometrically, if a graph is always concave upward, it must get up, in fact, get way up.

BIG CLUE:  Things like this usually are applications of the Mean Value Theorem in some way.

Suppose that  f''(x) > 0  for all x.  Then f'(x) is monotonic increasing.  (Monotonic is a fancy word for 'always'.)

What about f'(x)?  There are two possibilities:

A. f'(x) is positive somewhere.
B. f'(x) is always negative.  (We do this later.)

Now we do A.  Basically we are going to say that if f'(x) is positive, f(x) is rising and we can make it go really really high, so even if it started really really low it will have to cross the x-axis.

Assume f'(x0) > 0  for some x0.

If so, then f'(x) > 0  for all larger x, and f(x) is increasing.  

Now let x0 be some value of x where f' is positive.

Let  x1 > x0.
Let  Dx = x1 - x0

By the Mean Value Theorem, (I told you) there exists a value x = xm1 between x0 and x1 such that:

         f(x1) - f(x0)
f'(xm1) = -------------  and we can solve for f(x1):
            x1 - x0

f(x1) - f(x0) = f'(xm1)(x1 - x0)
f(x1) = f(x0) + f'(xm1)(x1 - x0)
f(x1) = f(x0) + f'(xm1) Dx

Now take  x2 = x1 + Dx  and try the same process:

f(x2) = f(x1) + f'(xm2)Dx

But f' is increasing, so f'(xm2) > f'(xm1)  and:

f(x2) > f(x1) + f'(xm1)Dx

And substituting:

       <----- f(x1)---->
f(x2) > f(x0) + f'(xm1)Dx + f'(xm1)Dx

f(x2) > f(x0) + 2*f'(xm1)Dx

And extending this,

f(xn) > f(x0) + n*f'(xm1)Dx

Now it becomes easy to prove that we will find a positive function value.  Since f'(xm1) and Dx are both positive, so is their product, and thus

n*f'(xm1)Dx  can be made as large as we want.  Formally, we say that

Given that  f(x0) is a (huge?) negative number, can we make f(xn) be positive?

We want   f(x0) + n f'(xm1)Dx > 0,  where f(x0) = - K. (I'd like to make a bigger K, but all I have is a 12-point font.)

- K + n f'(xm1)Dx > 0

 n f'(xm1)Dx > K

      K
n > ---------
   f'(xm1)Dx

That's it.  As long as we take  n > that number, we will get f(xn) > 0.
......................................................
B. That was the easy case.  Now we assume  f'(x) is always negative.  Here we will do the same argument, but moving to the left.  We say something like: Even if f(x0) is really really low, it had to come so far down that it must have been really really up at some time. (some point, we mean)

NOTE: The reasoning in this part is very similar, but the negative numbers play havoc with the inequalities and require a lot of care.

Take any  x0.  Of course, f'(x0) < 0, but remember that f' is an increasing function of x.  THAT MEANS IF WE TAKE A SMALLER X WE GET A SMALLER ANSWER.

Let  x1 < x0 and apply the MVT.  The arrangement looks different, but remember that x0 is the larger number.

         f(x0) - f(x1)
f'(xm1) = -------------,  where  xm1 is between x1 and x0
           x0 - x1

Also: We make  Dx = x0 - x1, so it will be positive.

Now we will solve this for  f(x1)  [Yes, for f(x1).]

f(x0) - f(x1) = f'(xm1) Dx

- f(x1) =  - f(x0) + f'(xm1) Dx

f(x1) =  f(x0) - f'(xm1) Dx

Now take  x2 = x1 - Dx  and try the same process:

f(x2) = f(x1) - f'(xm2)Dx

But remember that  f' is increasing, so f'(xm2) < f'(xm1)
Is that right?  xm2 < xm1, and smaller x's give smaller values.

f(x2) > f(x1) - f'(xm1)Dx

Is that right?  f'(xm1) is bigger, and it is subtracted, and subtracting a bigger thing gives you a smaller thing.  Yes, the inequality is in the right direction.

Substitute as before:

       <----- f(x1)---->
f(x2) > f(x0) - f'(xm1)Dx - f'(xm1)Dx

f(x2) > f(x0) - 2*f'(xm1)Dx

And extending as before:

f(xn) > f(x0) - n*f'(xm1)Dx

Now how does this make  f(xn) big enough?  Remember that  f'(x) is always negative in this part of the proof.  Suppose we call it  -e.

If  f'(xm1) = - e,  then

f(xn) > f(x0) + n*e*Dx

And we do the same stuff as before, to show that regardless of how huge a negative number f(x0) may be, when we add lots and lots of  e*Dx's to it, we finally get a positive number.

And that's it.