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I'm sorry in advance for how specific this problem is. I've been trying to make it as generic as possible but this is as far as I could get. I want to prove that a fixed point is attractive.

I didn't find much on Google about proving the attractiveness of a fixed point, but if there is a general methodology, that would be very helpful already.

The function is: $$ f(x) = (ax + b)^{1/\alpha} $$ with $a>0$, $\alpha\in(0,1)$ and $b\in(0,b_0)$ where $\displaystyle{b_0 = (1-\alpha)\left(\frac{\alpha}{a}\right)^{\alpha/(1-\alpha)}}$. The domain of this function is $[0,x_0)$, with $\displaystyle{x_0 = \left(\frac{\alpha}{a}\right)^{1/(1-\alpha)}}$.

I proved that this function has one unique fixed point in its domain, but I don't know how to prove that it is attractive. Specifically, the derivative is not absolutely lesser than 1. What I know already about the derivative is that it is strictly positive and increasing, and takes values in $[0,x_0)$.


I made a mistake in my calculations; the derivative takes values in $[0,1)$ which is sufficient to prove the fixed-point is attractive. Thanks to Julian for his answer.

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It is only the derivative at the fixed point itself that matters. – Will Jagy Jan 30 '13 at 22:30
Isn't the domain of the function $(-b/a,+\infty)$? – 1015 Jan 30 '13 at 22:30
@WillJagy The fixed point has no analytical expression, this is why it is dificult. – Sh3ljohn Jan 30 '13 at 22:32
@julien It is important to keep the domain of the functino as it is stated. It has not much to do with where the quantity exists, more than where we're looking for a fixed point. – Sh3ljohn Jan 30 '13 at 22:33
Actually, it is automatic if you have two distinct fixed points, as $f'$ is always positive, and $f(0) > 0$ as well. So the first fixed point occurs with the graph having positive derivative but smaller than 1. The second fixed point, the derivative is larger than 1, as the graph is passing upwards through the $y=x$ line. You really should pick some values for your numbers and draw some careful graphs. – Will Jagy Jan 30 '13 at 22:59
up vote 2 down vote accepted

We will work on $[0,+\infty[$.

First observe that for $x\geq 0$: $$ f(x)=x\quad\Leftrightarrow\quad ax+b=x^\alpha. $$ Now draw the graphs of $ax+b$ and $x^\alpha$ to convince yourself that there could be $0$, $1$ or $2$ positive fixed points in general.

It turns out that the condition $b<b_0$ guarantees that there are $2$ positive fixed points.

Indeed, consider the point $$ x_2:=\frac{\alpha b}{(1-\alpha)a}. $$ At this point, we have, after simplifications: $$ ax_2+b<x_2^\alpha\quad\Leftrightarrow\quad b< (1-\alpha)\left( \frac{a}{\alpha}\right)^\alpha=b_0. $$ The latter is one hypothesis given by the OP, so now we know why it is here. Since the graph of $x^\alpha$ is above $ax+b$ at $x_2$ whereas it is below at $0$ and at $+\infty$, we know that there are two fixed points $x_1,x_3$ such that $0<x_1<x_2<x_3$.

Next compute the derivative at $x_1$ and use the fixed point condition: $$ f'(x_1)=\frac{a}{\alpha}(ax_1+b)^{1/\alpha-1}=\frac{ax_1}{\alpha(ax_1+b)}. $$ Finally, we see after simplification that: $$ f'(x_1)<1\quad\Leftrightarrow\quad x_1<x_2. $$ So the smallest fixed point is attractive.

Note that the last paragraph is actually the consequence of the more general observation made by Will Jagy.

Finally, the condition $x_1<x_0$ is also fulfilled. Indeed, we have $ax_0+b<x_0^\alpha$ so $x_1<x_0$, since the former turns out to be equivalent to $b<b_0$ again.

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well, there you go then. – Will Jagy Jan 31 '13 at 4:15
Silly me, I wrote $f^\prime(x) = \frac{ax}{\alpha} (ax+b)^{\frac{1}{\alpha} -1}$ in my calculations.. Now I understand why my upper bound was $x_0$! Thanks a lot :) – Sh3ljohn Jan 31 '13 at 4:50
By the way, this is exactly why $x_0$ and $b_0$ were there. I was precisely trying to solve the problem $ax+b = x^\alpha$ when there are two solutions! For those who are interested, $x_0$ is the point where the derivative of $x^\alpha$ is equal to $a$. – Sh3ljohn Jan 31 '13 at 4:59
@Sh3ljohn, you have the responsibility to write down correct things when asking strangers to do you the favor of explaining your calculations. – Will Jagy Jan 31 '13 at 6:01
@WillJagy I made a mistake when writing that the derivative took values in $[0,x_0)$ (now edited). I think you misunderstood my comment; I was not asking to solve $ax+b = x^\alpha$. My question was a subpart of a subcase of this equation, but I find it nice that Julien took the trouble of getting higgher up to the root of this problem. – Sh3ljohn Feb 3 '13 at 16:05

The second derivative $f''$ is positive as is $f'$ and $f$ itself. In particular, $f(0) > 0.$ So, one possibility is no fixed points. Another is exactly one fixed point, as in $g(x) = e^{x-1}$ at $x=1,$ in which case the derivative is exactly one.

If, as you say, there are two fixed points, call them $0 < u < v.$ So, $f(u) = u$ and $ 0 < f'(u) < 1, $ which is what you wanted to know. Then $f(x) < x$ for $u < x < v.$ Finally $f(v)=v,$ with $f'(v) > 1,$ and for $x > v$ we get $f(x) > x.$

EEDDIITT: One way to see all this is simply to notice that $\frac{1}{\alpha} > 1,$ so $f(x) > x$ for large $x.$ So, define $$ h(x) = f(x) - x. $$ We know $h(0) > 0,$ also $h(x) > 0$ for large $x.$ And we know $h'' > 0.$ So, if $h(u) = h(v) =0,$ we know from the Mean Value Theorem that there is some $u < w < v$ with $h'(w) = 0.$ Furthermore, $h'(u) < 0, \; h'(v) > 0,$ and $h(w) < 0.$ The part you want is $h'(u) < 0.$

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I don't know why you say the derivative at the first fixed point is lesser than 1, but I can prove that the upper bound (and I mean upper bound, not majorant) of the derivative on $(0,x_0)$ is $x_0$, which could be greater than 1. The second fixed point is in $(x_0,+\infty)$, and iterating this function does not yield convergence to it unless the starting point is after the second fixed point. – Sh3ljohn Jan 30 '13 at 23:34
@Sh3ljohn mostly because, if the derivative were too large, there would be no fixed point. Draw some pictures. – Will Jagy Jan 30 '13 at 23:41
Thank you for your answer :) I understand your point, and I would really like it to be true (actually, it is true for the function that converges to the other fixed point, and that's how I proved its convergence, but it is out-of-scope..), but nothing indicates that the derivative of this function is lesser than 1 on the interval $(0,x_0)$. – Sh3ljohn Jan 30 '13 at 23:46
Nice observation, +1. – 1015 Jan 31 '13 at 2:16
Thanks, Will. I think I finally understood the condition $b<b_0$. It is meant to assure that there are indeed two positive fixed points. The fact that the smallest one is attractive is automatic by your observation. – 1015 Jan 31 '13 at 3:32

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