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Let us consider the function $$f(x)=αx^{θ}+β$$ where $α>0,β<0,θ>0$.

My question is: Find sufficient and necessary conditions in which this function has a unique fixed point in the interval $[0,1]$.

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  • $\begingroup$ Note that a fixed point amounts to a solution of $f(x) = x$. Since $f(x)-x$ is continuous on $[0,1]$, you can use a variety of standard propositions to explore when the function has a unique root in that interval. $\endgroup$ – hardmath Dec 25 '17 at 15:15
  • $\begingroup$ @hardmath: Can you give an example on that methods. $\endgroup$ – China Dec 25 '17 at 15:17
  • $\begingroup$ Try the Intermediate Value Theorem and check values for $f(x)-x$ at the endpoints. $\endgroup$ – hardmath Dec 25 '17 at 15:18
  • $\begingroup$ @hardmath: This result show the existence if $α+β-1>0$ but not the unicity. $\endgroup$ – China Dec 25 '17 at 15:21
  • $\begingroup$ Proving the root is unique amounts to showing the function remains positive after its first root. So one sufficient condition is $f(x)-x$ increasing on $[0,1]$. You can think about ways to tighten that. $\endgroup$ – hardmath Dec 25 '17 at 15:28
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(Quoting @hardmath as well since he commented first about a similar way of solving it)

Recall that a fixed point means :

$$f(x_0) = x_0 \Rightarrow ax_0^\theta + \beta = x_0 \Leftrightarrow ax_0^\theta+\beta - x_0 = 0 \Leftrightarrow x_0(ax_0^{\theta - 1}-1) + \beta=0$$

Thus all you need to do, is determine when the function :

$$g(x) = x(ax^{\theta - 1}-1) + \beta $$

has a unique solution in the interval $[0,1]$.

The values of $g(x)$ at the end points of the given interval, are :

$$g(0) = \beta$$

$$g(1) = a-1+\beta$$

For the function $g(x)$ to have at least one root in the interval $[0,1]$, one can apply Bolzano's Theorem since the function is continuous in it. Then, if :

$$g(0) \cdot g(1) < 0 \Rightarrow (a-1+\beta)\beta < 0$$

the function $g(x)$ will have at least one root in $(0,1)$. Since $\beta <0$ then you'll need :

$$a-1+\beta > 0 $$

For this root to be unique, it would be enough if you could show that $g(x)$ is strictly monotonic. Studying the derivative of $g(x)$, we'll have :

$$g'(x) =a\theta x^{\theta-1}-1$$

To be strictly monotonic, you want one of the following, since strictly monotonic means increasing or decreasing in the given interval :

$$g'(x) > 0 \Rightarrow a\theta x^{\theta -1} -1>0 \quad \forall x\in (0,1)$$

$$g'(x) < 0 \Rightarrow a\theta x^{\theta -1} -1<0 \quad \forall x\in (0,1)$$

Joining these $2$ with the first case in $2$ different cases, you'll have sufficient conditions for strictly one root of $g(x)$ in $[0,1]$ or in other words a unique fixed point of $f(x)$, if you can give solutions to the following systems of inequalities for all $x \in (0,1)$ :

$$\begin{cases} a-1+\beta < 0 \\ a\theta x^{\theta -1} -1>0 \end{cases}$$

$$\text{or}$$

$$\begin{cases} a-1+\beta < 0 \\ a\theta x^{\theta -1} -1<0 \end{cases}$$

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  • $\begingroup$ Your last conditions gives $x>((1/(θα)))^{(1/(θ-1))}$ for all $x$ in $[0,1]$ and this is impossible since the term is positive. $\endgroup$ – China Dec 25 '17 at 15:38
  • $\begingroup$ @ZeraouliaElhadj Hi, why is this impossible ? $\theta, a$ are positive reals which means the root of $\theta-1$ is defined and is positive, which also is logical since $x\in [0,1]$. $\endgroup$ – Rebellos Dec 25 '17 at 15:44
  • $\begingroup$ The problem is with the statement: for all $x$ in $[0,1]$. Notice that not all these numbers are $>((1/(θα)))^{(1/(θ-1))}$ $\endgroup$ – China Dec 25 '17 at 15:46
  • $\begingroup$ @ZeraouliaElhadj How is this ? $\theta > 0, a>0$ gives you an infinite combinations of both and you also have another inequality that must be true. Also, it's easy to understand that such conditions are necessary and sufficient because both theorems are strict and are cases which prove what we need in an analysis way but involving geometry too in their initial definitions. Also, an approach with the Intermediate Value Theorem also provides necessary and sufficient conditions and is similar to this one that I've elaborated over. $\endgroup$ – Rebellos Dec 25 '17 at 15:53
  • $\begingroup$ Just note that there is one more case to be examined; that of Bolzano's Theorem not being satisfied, however, there is a fixed point of $f$. So, the cases described are not all the cases. $\endgroup$ – Βασίλης Μάρκος Dec 25 '17 at 15:58

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