Prove: Derivative Proof: f(x)=0 in (0,1)

[JG89]

Homework Statement

Suppose f is a differentiable function. Prove that if f(0) = 0 and |f'(x)| <= |f(x)| then f(x) = 0 for x in (0,1).

Homework Equations

The Attempt at a Solution

The actual question asks me to prove f(x) = 0 for all real x, but if I can prove f(x) = 0 for x in (0,1) then the rest follows easily.

So:

Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that \frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f&#039;(c*). We know that f&#039;(c*) \le f(c*). So \frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) &lt; f(c*) since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?

 

[JG89]

I can't view the latex in my post for some reason now. It was fine when I first posted it. Now I just see a banner saying physicsforums.com wherever the latex was before.

Here is a repost of my proof:


The actual question asks me to prove f(x) = 0 for all real x, but if I can prove f(x) = 0 for x in (0,1) then the rest follows easily.

So:

Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that \frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f&#039;(c*). We know that f&#039;(c*) \le f(c*). So \frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) &lt; f(c*) since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?

 

[snipez90]

Although you have the right idea, I don't think the proof works as stands. For one, |a| \leq |b| does not imply a \leq b, and I think this makes many of your inequalities very tenuous.

 

[JG89]

We've assumed f(c) > 0, and c is also positive since c is in (0,1). So f(c)/c = f'(c*), making f'(c*) also positive and so |f&#039;(c*)| \le |f(c*)| \Leftrightarrow f&#039;(c*) \le f(c*) so in the inequality follows.

If we assume f(c) < 0, then the inequality signs work out to find the same contradiction.

 

[snipez90]

ok, good point. but how do you know that f(c*) won't be negative and greater in magnitude than f'(c*). i may as well have overlooked something.

 

[JG89]

f(c*) can't be negative, because f(c)/c is positive, and f(c)/c = f'(c*) <= f(c*)

Remember we've assumed that |f'(x)| <= |f(x)|, but all the values we're considering right now are positive, so we can drop the absolute value bars.

 

[snipez90]

Err, so you're essentially precluding the possibility of there being an x in (0,c) for which f(x) < 0? Maybe I'm missing something, but I still don't see how this follows immediately from the assumptions. Yes I'm aware that f(c)/c = f'(c) <= |f(c*)|, but how do you know that the c* in (0,c) isn't so that f(c*) < 0 (note even if f(c*) < 0, it can still be true that 0 < f'(c) =|f'(c)| <= |f(c*)|).

 

[JG89]

:( Now I see your point.

I'm going to catch some sleep then work on this more tomorrow. I'm sure I'm on the right path to finishing up the proof.

 

[VietDao29]

Homework Statement

Suppose f is a differentiable function. Prove that if f(0) = 0 and |f'(x)| <= |f(x)| then f(x) = 0 for x in (0,1).

Homework Equations

The Attempt at a Solution

The actual question asks me to prove f(x) = 0 for all real x, but if I can prove f(x) = 0 for x in (0,1) then the rest follows easily.

So:

Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that \frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f&#039;(c*). We know that f&#039;(c*) \le f(c*). So \frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) &lt; f(c*) since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?

Well, how about carrying on using the MVT for some more times, like this:

Proof by Contradiction
Assume that there exists some x_1 \in (0, 1) : f(x_1) \neq 0.
By the MVT, we have:
\exists x_2 \in (0, x_1) : \frac{f(x_1) - f(0)}{x_1 - 0} = f&#039;(x_2) \Rightarrow |f(x_1)| = |f&#039;(x_2)| x_1 \leq |f(x_2)| x_1

Since f(x_1) \neq 0, it must follow that: f(x_2) \neq 0, we then once again, apply the MTV:

\exists x_3 \in (0, x_2) : \frac{f(x_2) - f(0)}{x_2 - 0} = f&#039;(x_3) \Rightarrow |f(x_2)| = |f&#039;(x_3)| x_2 \leq |f(x_3)| x_2 \Rightarrow |f(x_1)| \leq |f(x_2)| x_1 \leq |f(x_3)| x_1 x_2

And so on, you'll be able to construct a sequence (x_n), such that:
\left\{ \begin{array}{c} f(x_n) \neq 0, \forall n \\ | f(x_{1}) | \leq |f(x_{k})| \prod\limits_{i = 1}^{k - 1} x_i &lt; |f(x_{k})| x_1 ^ {k - 1}, \forall k \geq 2 \end{array} \right.

What can you say about the sequence (x_n)? Is it convergent, or divergent?

From there, can you see what contradiction it leads to?

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Hopefully you can go from here, right? :)

 

[snipez90]

The second approach you mentioned is closer to the approach I had in mind that uses JG89's idea. In fact, I think it is possible to argue directly using this idea as well.

The problem with the first approach seems to be that we cannot assume that f ' is actually integrable.

There is even a nicer approach via proof by contradiction, I think, that makes full use of the continuity of f. If you consider the right kind of closed interval with a specific delta guaranteed by uniform continuity and use the fact that a continuous function on a closed interval attains maximum and minimum values, then you easily obtain a contradiction via the mean value theorem.

 

[VietDao29]

The second approach you mentioned is closer to the approach I had in mind that uses JG89's idea. In fact, I think it is possible to argue directly using this idea as well.

The problem with the first approach seems to be that we cannot assume that f ' is actually integrable.

Yes, my bad. I don't really know what gets into my mind. :(

Ok, stand corrected. Thanks.. :)

 

[JG89]

Thanks for the help guys.

Someone showed me a simple solution to this that uses VietDao29's method, but proves it directly.