# Small Challenge

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• Jul 21st 2010, 06:27 AM
Dinkydoe
Small Challenge
This is a challenge question:

Let $\displaystyle f:[0,1]\to [0,1]$ with $\displaystyle f(0)=0, f(1)=1$ differentiable

Show there exists distinct points $\displaystyle x,y\in [0,1]$ such that $\displaystyle f'(x)f'(y)=1$.

Moderator edit: Approved Challenge question.
• Jul 21st 2010, 11:56 AM
Unbeatable0
From Lagrange theorem it follows that there exist $\displaystyle c\in[0,1]$ such that

$\displaystyle f'(c) = \frac{f(1)-f(0)}{1-0} = 1$ so we can assume $\displaystyle f'(x)>1$ or $\displaystyle f'(x)<1$ for all $\displaystyle x\in[0,1]\setminus\{c\}$

(otherwise the proof it trivial). Suppose we have $\displaystyle f'(a)<1$ and $\displaystyle f'(b)>1$ for some $\displaystyle a,b\in[0,1]$.

If $\displaystyle f'(a)f'(b)=1$ then we're done.

If $\displaystyle f'(a)\le0$ then $\displaystyle f'(a)<\frac{1}{f'(b)}<f'(c)$ and thus the existence of $\displaystyle d$ such

that $\displaystyle f'(d)=\frac{1}{f'(b)}$ is guaranteed by the intermediate value theorem for derivatives and we're done.

Otherwise $\displaystyle 0<f'(a)<1$ and $\displaystyle f'(b)>1$ so either $\displaystyle f'(a)<\frac{1}{f'(b)} < f'(c) (=1)$ or $\displaystyle f'(c)<\frac{1}{f'(a)}<f'(b)$. Either

way, using the intermediate value theorem for derivatives we're done again.

Now suppose the previous case is not true. That is, $\displaystyle f'(x)<1$ for all $\displaystyle x\in[0,1]\setminus\{c\}$ or $\displaystyle f'(x)>1$ for all $\displaystyle x\in[0,1]\setminus\{c\}$

Therefore the function $\displaystyle g(x) = x-f(x)$ is either monotone increasing or monotone decreasing (by looking at the derivative).

Either way, it means that $\displaystyle g(0)\ne g(1)$. But from the conditions on $\displaystyle f(x)$ it follows that $\displaystyle g(0)=g(1)=0$, a contradiction.
• Jul 21st 2010, 03:23 PM
Dinkydoe
:)

Well done.
• Jul 21st 2010, 03:28 PM
Unbeatable0
Quote:

Originally Posted by Dinkydoe
$\displaystyle 0< f'(a)< 1$ doesn't guarantee $\displaystyle \frac{1}{f'(a)}< f'(b)$

Not $\displaystyle 0<f'(a)<1$ alone.

$\displaystyle f'(a)<\frac{1}{f'(b)} < f'(c)$ corresponds to the option $\displaystyle f'(a)f'(b) < 1$.

$\displaystyle f'(c)<\frac{1}{f'(a)}<f'(b)$ corresponds to the option $\displaystyle f'(a)f'(b) > 1$

I'm interested - what's your solution?
• Jul 21st 2010, 03:34 PM
Dinkydoe
I allready corrected my statement ;p

Didnt expect such fast reply xp
• Jul 21st 2010, 03:41 PM
Dinkydoe
My solution was a little shorter:

We may assume $\displaystyle f(x)\neq x$. (otherwise there's nothing to prove).
By the fundamental theorem of calculus we have $\displaystyle \int_{0}^1f'(x)dx= f(1)-f(0)= 1$. Hence we can't have $\displaystyle f'(x)\leq 1$ or $\displaystyle f'(x)\geq 1$ for all x. (since we also assumed $\displaystyle f(x)\neq x$)

But then there exist points $\displaystyle a,b \in [0,1]$ such that f'(a)<1 and f'(b)>1 and the intermediate value theorem now gives that f'(x) reaches all points in (f'(a),f'(b)) and trivially we can find p,q in this interval such that pq= 1.
• Jul 21st 2010, 03:55 PM
Unbeatable0
Quote:

Originally Posted by Dinkydoe
My solution was a little shorter:

We may assume $\displaystyle f(x)\neq x$. (otherwise there's nothing to prove).
By the fundamental theorem of calculus we have $\displaystyle \int_{0}^1f'(x)dx= f(1)-f(0)= 1$. Hence we can't have $\displaystyle f'(x)\leq 1$ or $\displaystyle f'(x)\geq 1$ for all x. (since we also assumed $\displaystyle f(x)\neq x$)

But then there exist points $\displaystyle a,b \in [0,1]$ such that f'(a)<1 and f'(b)>1 and the intermediate value theorem now gives that f'(x) reaches all points in (f'(a),f'(b)) and trivially we can find p,q in this interval such that pq= 1.

The fundumental theorem of calculus has the assumption that $\displaystyle f'(x)$ is continuous,

which is not given, and there are example of derivatives which are not continuous, so I think your argument is

not perfectly justified.
• Jul 21st 2010, 04:00 PM
Dinkydoe
Quote:

The fundumental theorem of calculus has the assumption that is continuous,

which is not given, and there are example of derivatives which are not continuous, so I think your argument is

not perfectly justified.
Check Fundamental theorem of calculus - Wikipedia, the free encyclopedia

Second Fundamental Theorem of Calculus

We don't have to assume f'(x) is continuous.
• Jul 21st 2010, 04:12 PM
Unbeatable0
Quote:

Originally Posted by Dinkydoe
Check Fundamental theorem of calculus - Wikipedia, the free encyclopedia

Second Fundamental Theorem of Calculus

We don't have to assume f'(x) is continuous.

It still does require $\displaystyle f'(x)$ to be integrable, which is not always the case (you can google some examples).
• Jul 21st 2010, 04:29 PM
Dinkydoe
Quote:

It still does require to be integrable, which is not always the case (you can google some examples).
f'(x) is by definition integrable, since we start out with the 'antiderivative' f(x) as a continuous function. ;p
• Jul 22nd 2010, 01:05 AM
Unbeatable0
Quote:

Originally Posted by Dinkydoe
f'(x) is by definition integrable, since we start out with the 'antiderivative' f(x) as a continuous function. ;p

That's not the definition.

See here: A differentiable function whose derivative is not integrable
• Jul 22nd 2010, 02:53 AM
tonio
Quote:

Originally Posted by Unbeatable0

I think what he meant is that since $\displaystyle f(x)$ is defined and derivable in $\displaystyle [0,1]$ and $\displaystyle f(x)$ is clearly a primitive function of $\displaystyle f'(x)$ , the fundamental

theorem of integral calculus gives us that $\displaystyle f'(x)$ is (Riemann) integrable and $\displaystyle \int\limits^1_0f'(x)\,dx=f(1)-f(0)$

Tonio
• Jul 22nd 2010, 02:57 AM
Unbeatable0
Quote:

Originally Posted by tonio
I think what he meant is that since $\displaystyle f(x)$ is defined and derivable in $\displaystyle [0,1]$ and $\displaystyle f(x)$ is clearly a primitive function of $\displaystyle f'(x)$ , the fundamental

theorem of integral calculus gives us that $\displaystyle f'(x)$ is (Riemann) integrable and $\displaystyle \int\limits^1_0f'(x)\,dx=f(1)-f(0)$

Tonio

That's not true. A derivative need not in general be integrable over a closed interval.
• Jul 22nd 2010, 03:16 AM
Dinkydoe
The examples given in your link are of oscillating functions that are unbounded. (at least from what I read) From say, $\displaystyle f:[-1,1]\to \mathbb{R}$?

I hope you didn't forget our function is $\displaystyle f:[0,1]\to [0,1]$ and is by definition bounded.

And I remember that a function can still be integrable, even though it may have a countable set of problem-points.

The second last post gives a function, from where they agree it is integrable....(oscillating but still bounded)

But, maybe we need help of an expert here.
• Jul 22nd 2010, 03:23 AM
Unbeatable0
Quote:

Originally Posted by Dinkydoe
The examples given in your link are of oscillating functions that are unbounded. (at least from what I read) From say, $\displaystyle f:[-1,1]\to \mathbb{R}$?

I hope you didn't forget our function is $\displaystyle f:[0,1]\to [0,1]$ and is by definition bounded.

Well, I think we need help of an expert here.

It's $\displaystyle f'(x)$ that's not bounded in the example there.

$\displaystyle f(x) = x^2 \sin\frac{1}{x^3}$ is bounded on every closed interval. $\displaystyle [0,1]$, for example.

(of course while defining $\displaystyle f(0) = 0$)

The above function indeed satisfies $\displaystyle f:[0,1]\to [0,1]$.
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