# analysis question

• November 17th 2007, 11:18 AM
agentZERO
analysis question
Hi. I'm taking an Analysis I course and we just started differentiation. I'm not 100% comfortable with it yet and would appreciate some help on a problem in my textbook. Suppose that f:[0,1] to R is continuous on f(0)= 0, differentiable for x in [0,1], and 0 <= f '(x) <= af(x) for a > 0. Prove that f = 0. (Hint: consider the derivative of (e^(-ax))f(x).) Any help you offer would be greatly appreciated.
• November 17th 2007, 12:41 PM
Plato
Is it exactly as it appears in your text book? (What is the text?)
It just does not seem to be correct.
• November 17th 2007, 12:50 PM
agentZERO
It has that f --- 0 where --- is three horizontal lines on top of each other (basically an equal sign with an extra line). I figured this meant equals, but I guess it does not. What exactly does it mean and how would you solve this? Thank you for trying to help.
• November 17th 2007, 01:36 PM
Plato
Quote:

Originally Posted by agentZERO
It has that f --- 0 where --- is three horizontal lines on top of each other (basically an equal sign with an extra line). I figured this meant equals

That is correct, but that is not what I asked.
What does "f:[0,1] to R is continuous on f(0)= 0" mean?
I suppect that it is not written that way in the text.
What is the textbook you are using?
• November 17th 2007, 01:44 PM
agentZERO
Yikes. I did make a serious mistake. It should be: Suppose that f:[0,1] to R is continuous with f(0)= 0, differentiable for x in [0,1], and 0 <= f '(x) <= af(x) for a > 0. Prove that f = 0. (Hint: consider the derivative of (e^(-ax))f(x).)
I'm using a textbook outside of class just to study. It's called Elementary Real Analysis by Brian S. Thomson.
• November 17th 2007, 02:25 PM
Plato
Use the hint:
$- ae^{ - ax} f(x) + e^{ - ax} f'\left( x \right) = e^{ - ax} \left( { - af(x) + f'\left( x \right)} \right) \le e^{ - ax} \left( { - af(x) + af\left( x \right)} \right)
$
• November 17th 2007, 02:45 PM
agentZERO
Ah, that makes sense. How do you know f is zero though?
• November 17th 2007, 03:14 PM
Plato
Does that mean that the derivative of $e^{ - ax} f(x)$ is zero everywhere?
What does that mean, recalling $f(0)=0$?
• November 17th 2007, 07:17 PM
agentZERO
You don't have to be a grouch. You're the one helping and I thought you wanted to help. Plus, it's not always easy for people who don't know the code to type out their work. It doesn't mean they're not thinking and trying to work it out themselves.
• November 18th 2007, 08:51 AM
topsquark
Quote:

Originally Posted by agentZERO
You don't have to be a grouch. You're the one helping and I thought you wanted to help. Plus, it's not always easy for people who don't know the code to type out their work. It doesn't mean they're not thinking and trying to work it out themselves.

He isn't being a grouch. He simply wants to see evidence that you are doing something to try to solve the problem. Hence even if you have no work to show, you could mention what you have tried to do to solve the problem.

And, as most members here don't know how to use LaTeX, you can post your work in the following kind of format:
-ae^{ - ax} f(x) + e^{ - ax} f'(x) = e^{ - ax} (- af(x) + f'(x)} ) <= e^{ - ax} ( - af(x) + af(x))

It's harder to read, but still quite understandable.

-Dan