# Twice-differentiable, mean value theorem, fixed point

1. Sep 25, 2009

### fmam3

1. The problem statement, all variables and given/known data
Let $$g:[0,1] \to \mathbb{R}$$ be twice-differentiable (i.e. both $$g$$ and $$g'$$ are differentiable functions) with $$g''(x) > 0$$ for all $$x \in [0,1]$$. If $$g(0) > 0$$ and $$g(1) = 1$$, show that $$g(d) = d$$ for some $$d \in (0,1)$$ if and only if $$g'(1) > 1$$.

2. Relevant equations

3. The attempt at a solution
I can prove one direction but not the other!

Suppose $$g(d) = d$$ for some $$d \in (0,1)$$ and let's show $$g'(1) > 1$$. Since g is continuous on $$[0,1]$$ and differentiable on $$(0,1)$$, by the Mean Value Theorem applied to $$[d,1]$$, then $$\exists c_1 \in (d,1)$$, then we have
\begin{align*} g'(c_1) &= \frac{g(1) - g(d)}{1 - d} \\ &= \frac{1 - d}{1 - d} \\ &= 1 \end{align*}
Note that since $$g$$ is differentiable on $$[0,1]$$, $$g'(1)$$ exists and is finite. Then since $$g'$$ is a differentiable and hence continuous function on $$[0,1]$$, we can apply the Mean Value Theorem on $$[c_1,1]$$ to have that, $$\exists c_2 \in (c_1, 1)$$ such that
\begin{align*} g''(c_2) &= \frac{g'(1) - g'(c_1)}{1 - c_1} \\ &= \frac{g'(1) - 1}{1 - c_1} \\ &> 0 \end{align*}
where the last strict inequality follows from the hypothesis that $$g''(x) > 0, \forall x \in [0,1]$$ and $$1 - c_1 > 0$$. This implies $$g'(1) > 1$$ as required.

Now for the converse, suppose we have $$g'(1) > 1$$ and we want to show $$g(d) = d$$ for $$\exists d \in (0,1)$$. Define the function $$h(x) = g(x) - x$$. Then clearly, $$h$$ is also twice differentiable on $$[0,1]$$. Now, see that $$h'(1) = g'(1) - 1 > 0$$..... (this is where I'm stuck!)

I'm having trouble showing the converse! I've tried to apply the Mean Value Theorem, Intermediate Value Theorem for continuous functions on a compact set, and also Intermediate Value Theorem for derivatives --- but still, I feel that I'm so close but yet not quite there. Any small hint would be greatly appreciated!

Thanks :)

Last edited: Sep 25, 2009
2. Sep 25, 2009

### Billy Bob

Very nice job so far.

Small nudge: Apply defn of derivative to h'(1).

3. Sep 25, 2009

### fmam3

I hope you wouldn't mind giving me a bigger nudge! Here's my attempt on applying your tip.

Define $$h(x) = g(x) - x$$ on $$[0,1]$$ and clearly, it is continuous and differentiable on $$[0,1]$$. Now, note that since $$g'(x) > 1$$, it follows that $$h'(1) = g'(1) - 1 > 0$$. Then by the definition of the derivative, we have that
\begin{align*} h'(1) = \lim_{x \to 1} \frac{h(x) - h(1)}{x - 1} \end{align*}
which means, $$\forall \varepsilon >0, \exists \delta > 0$$ such that $$\forall x \in [0,1]$$ and $$0 < |x - 1| < \delta$$, we have
\begin{align*} \left| \frac{h(x) - h(1)}{x - 1} - h'(1) \right| < \varepsilon \end{align*}
In particular, we can set $$\varepsilon = h'(1) > 0$$, and thus we have that
\begin{align*} -h'(1) < \frac{h(x) - h(1)}{x - 1} - h'(1) < h'(1) \\ 0 < \frac{h(x) - h(1)}{x - 1} < 2h'(1) \end{align*}
or simply, $$h(x) > h(1)$$, implying $$g(x) - x > g(1) - 1 = 1 - 1 = 0$$, for $$x \in [0,1]$$ and $$0 < |x - 1| < \delta$$.

I have up to that but I still can't see how we can have $$h(d) = 0$$, which is sufficient to show $$g(d) = d$$. Any more nudges would be greatly appreciated! :)

4. Sep 25, 2009

### Office_Shredder

Staff Emeritus
Here's my take on it (informally).

If g'(1)>1 we know that g' is increasing, so g'(0)<1 (if it was greater than or equal to 1, g(x) would always increase faster than the line y=x so g(1) would not be 1). That means somewhere, g'(a)=1. Is g(a)<a or g(a)>a, and why?

5. Sep 25, 2009

### fmam3

Thanks for the input!

But may I ask how did you conclude that if $$g'(1) > 1$$, then $$g'$$ is increasing?

Perhaps my knowledge is still not solid enough but I have so far only learned that if a function $$f$$ is defined on a set $$A \subseteq \mathbb{R}$$ and if $$f'(x) > 0$$ on $$A$$, then $$f$$ is strictly increasing on A; we have increasing if we replace the $$>$$ with an $$\geq$$.

Thanks again :)

Last edited: Sep 25, 2009
6. Sep 25, 2009

### fmam3

Thanks Billy Bob & Office_Shredder for the tips above!

I think I have figured out a solution. Here it is :)

For the converse, suppose we have $$g'(1) > 1$$ and we want to show $$g(d) = d$$ for some $$d \in (0,1)$$. For contradiction, suppose we have $$g(d) \ne d$$ for $$\forall d \in (0,1)$$.

Suppose for a fixed $$d \in (0,1)$$, we have $$g(d) > d$$. Applying the Mean Value theorem on $$[d,1]$$, we have that $$\exists c_1 \in (d,1)$$ such that
\begin{align*} g'(c_1) &= \frac{g(1) - g(d)}{1 - d} \\ &= \frac{1 - g(d)}{1 - d} \end{align*}
and since $$g'$$ is differentiable, applying the Mean Value Theorem again on $$[c_1, 1$$, we have that $$\exists c_2 \in (c_1, 1)$$ such that
\begin{align*} g''(c_2) &= \frac{g'(1) - g'(c_1))}{1 - c_1} \\ &> 0 \end{align*}
implying that $$g'(1) > g'(c_1)$$. But note that since we have assumed $$g(d) > d$$, it implies that we have $$g'(c_1) = \frac{1 - g(d)}{1 - d} < \frac{1 - d}{1 - d} = 1$$ --- contradiction to the assumption that $$g'(1) > 1$$.

Next, we assume that for a fixed $$d \in (0,1)$$, we have $$g(d) < d$$. Again, applying the Mean Value Theorem to $$[0,d]$$, we have that $$\exists c_3 \in (0,d)$$ such that
\begin{align*} g'(c_3) &= \frac{g(d) - g(0))}{d - 0} \\ &= \frac{g(d) - g(0)}{d} \end{align*}
and since $$g'$$ is differentiable and applying the Mean Value Theorem on $$[c_3, 1$$, we have that $$\exists c_4 \in (c_3, 1)$$ such that
\begin{align*} g''(c_4) &= \frac{g'(1) - g'(c_3)}{1 - c_3} \\ &> 0 \end{align*}
implying $$g'(1) > g'(c_3)$$. But note that $$g'(c_3) = \frac{g(d) - g(0)}{d} = \frac{(g(d) - d) + d - g(0)} {d} = \frac{g(d) - d}{d} + 1 - \frac{g(0)}{d}$$. But since we have that $$g(0) > 0$$ and assumed that $$g(d) < d$$, it follows that $$\frac{g(d) - d}{d} < 0$$ and $$\frac{g(0)}{d} > 0$$. Thus, it follows that $$g'(c_3) = \frac{g(d) - d}{d} + 1 - \frac{g(0)}{d} < 1$$ --- contradiction, since $$g'(1) > g'(c_3)$$ but we have that $$g'(1) > 1$$.

Thus, neither $$g(d) < d$$ nor $$g(d) > d$$ holds for $$\forall d \in (0,1)$$. Hence, we must have that $$\exists d \in (0,1)$$ such that $$g(d) = d$$. This completes the proof of the converse :)

7. Sep 25, 2009

### Office_Shredder

Staff Emeritus
You start off by assuming for a fixed d that g(d)>d leads to a contradiction. Then you do the same for g(d)<d. That doesn't prove for some d g(d)=d, that proves for ALL d g(d)=d!

g'(1)>g'(c1 and g'(c1)=1. No problem, since that gives g'(1)>1 which is true

I concluded g' is increasing by the fact that g''(x)>0

8. Sep 25, 2009

### fmam3

Yes... you're right. I screwed up on the negation. The negation of the statement "$$\exists d \in (0,1)$$ s.t. $$g(d) = d$$" should be $$\forall d \in (0,1), g(d) = d$$.

Actually, I showed that $$g'(c_1) < 1$$ in the above. But this doesn't matter, since my negated statement is wrong anyways.

I'll have to go back and think a little more on this... thanks for the reply!

9. Sep 25, 2009

### fmam3

I'm actually wondering --- can my proof still be repaired? So if we start of by assuming that $$\forall d \in (0,1), g(d) = d$$, then we must have the two cases:
(1) $$\forall d \in (0,1), g(d) \geq d$$; and
(2) $$\forall d \in (0,1), g(d) \leq d$$

Now, if I were two replace in my above proof $$g(d) > d$$ with $$g(d) \geq d$$ and $$g(d) < d$$ with $$g(d) \leq d$$, use the same method to construct the contradictions, the result would be that we cannot have $$\forall d \in (0,1), g(d) = d$$ and hence this gives the desired result. Is this correct or am I losing my mind over this problem?

10. Sep 25, 2009

### Office_Shredder

Staff Emeritus
If g(d) >= d for all d, that's not a problem. Since it could be g(d)=d.

I think you need to start with a more intuitive approach.. Stop trying to bang out mean value theorems and start drawing a couple example graphs. The graph of g(x) starts out above the graph of the function x. And at the end they meet at the same point. BUT! g(x) is also increasing faster than x at that point. So draw the line y=x. Put a point on the y-axis above 0 to represent g(0). Now draw a curve from the point (1,1) with a slope greater than 1 back towards 0 to represent g(x).

Do you notice how your graph of g(x) is immediately below the graph of x? What the problem here is asking you to show is that since g(0)>0, g(x) has to cross over the line y=x in order to look like this

11. Sep 25, 2009

### fmam3

Thanks for the reply again. I mean, geometrically, I see exactly what you mean. But since it is necessary to prove this rigorously, I'm sure there must be a way of applying the Mean Value Theorem (and I'm dead sure it must be used in the converse case) to solve it. *sigh*....

But thanks for your patience with me on this problem :)

12. Sep 25, 2009

### Billy Bob

Very good, but with one crucial error. Since x<1 (after all, 0<x<1) you should obtain h(x)<h(1)=0 for some x close to but less than. But h(0)>0.

13. Sep 25, 2009

### fmam3

Ah... yes perfect! Yes, I should have caught that error myself (i.e. $$x \in (0,1)$$ part). Then we have that $$h(x) < h(1) = 0$$ for $$\exists x \in (0,1)$$. And since $$h(0) = g(0) - 0 > 0$$, it follows that we have $$h(x) < 0 < h(0)$$ for $$\exists x \in (0,1)$$. Hence, since h is clearly continuous on $$[0,1]$$, then by the Intermediate Value Theorem, $$\exists d \in (0,x) \subseteq [0,1]$$ such that $$h(d) = g(d) - d = 0$$. This completes the proof.

Wow! Thanks! This problem has been killing me for the last 36 hrs or so! :)

14. Sep 25, 2009

### fmam3

Thanks again to both Billy Bob and Office_Shredder! You guys were great help :)