0
$\begingroup$

Here is Prob. 22, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is a real function on $(-\infty, \infty)$. Call $x$ a fixed point of $f$ if $f(x)=x$.

(a) If $f$ is differentiable and $f^\prime(t) \neq 1$ for every real $t$, prove that $f$ has at most one fixed point.

(b) Show that the function $f$ defined by $$ f(t) = t + (1 + \mathrm{e}^t)^{-1} $$ has no fixed point, although $0 < f^\prime(t) < 1$ for all real $t$.

(c) However, if there is a constant $A < 1$ such that $\left| f^\prime(t) \right| \leq A$ for all real $t$, prove that a fixed point $x$ of $f$ exists, and that $x = \lim x_n$, where $x_1$ is an arbitrary real number and $$ x_{n+1} = f \left( x_n \right) $$ for $n = 1, 2, 3, \ldots$.

(d) Show that the process described in (c) can be visualized by the zig-zag path $$ \left( x_1, x_2 \right) \rightarrow \left( x_2, x_2 \right) \rightarrow \left( x_2, x_3 \right) \rightarrow \left( x_3, x_3 \right) \rightarrow \left( x_3, x_4 \right) \rightarrow \cdots.$$

My effort:

Part (a):

Suppose $f$ has two distinct fixed points $a$ and $b$ such that $a < b$. Then as $f$ is continuous on $[a, b]$ and differentiable in $(a, b)$, we can find a point $p$ in $(a, b)$ such that $$ b-a = f(b) - f(a) = (b-a) f^\prime(p),$$ and since $b-a > 0$, we have $$ f^\prime(p) =1, $$ a contradiction. Hence If $f$ is differentiable and $f^\prime(t) \neq 1$ for every real $t$, then $f$ has at most one fixed point.

Part (b):

Assuming that the dirivatve of $\mathrm{e}^t$ is $\mathrm{e}^t$ at every real $t$, and also that $\mathrm{e}^t > 0$ for every real $t$, we note that $1 < 1 + \mathrm{e}^t$, and so $$ 0 < \frac{1}{1 + \mathrm{e}^t } < 1, \tag{1} $$ which implies that $$ t < t + \frac{1}{1 + \mathrm{e}^t },$$ that is, $t < f(t)$ for every real $t$. But $$ f^\prime(t) = 1 - \frac{\mathrm{e}^t }{ \left( 1 + \mathrm{e}^t \right)^2 } = \frac{\mathrm{e}^{2t} + \mathrm{e}^t + 1}{ \left( 1 + \mathrm{e}^t \right)^2 } < 1, $$ so that $0 < f^\prime(t) < 1$ for every real $t$, by virtue of (1) above.

Part (c):

Under the hypotheses of this part, we note that, given any two real numbers $x$ and $y$, by the mean value theorem, there exists a point $t \in (x, y)$ such that $$ \left| f(x) - f(y) \right| = \left| f^\prime(t) \right| \cdot | x-y | \leq A | x-y|. \tag{2} $$ Thus $f$ satisfies the hypothesis of the well-known Banach fixed point theorem since $\mathbb{R}^1$ is a complete metric space. So $f$ has a (unique) fixed point $x$ given by $$ x = \lim_{n \to \infty} x_n, $$ as described in Part (c) of the problem itself.

Part (d):

Given $x_1$, we find $x_2 = f\left(x_1 \right)$, thus obtaining the point $\left( x_1, x_2 \right)$ in the plane. Next, we check if $x_1$ is a fixed point of $f$, which is the same as checking if the point $\left( x_1, x_2 \right)$ coincides with the point $\left(x_2, x_2 \right)$.

If not, then we find $x_3 = f\left( x_2 \right)$, and check if $x_2$ is a fixed point of $f$, which is the same as checking if the point $\left( x_2, x_3 \right)$ coincides with the point $\left(x_3, x_3 \right)$.

If not, then we find $x_4 = f\left( x_3 \right)$, and check if $x_3$ is a fixed point of $f$, which is the same as checking if the point $\left( x_3, x_4 \right)$ coincides with the point $\left( x_4, x_4 \right)$.

And, the process continues until we find a fixed point, (which we eventually do).

Is my reasoning correct in all four parts of this problem? If there are problems, then exactly where?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

All but (d) look correct to me. You don't ever actually find a fixed point even with the simplest of contractions $f(x) = \frac{1}{2}x$ and starting x value $x_1 = 1$: you only converge to a fixed point.

$\endgroup$
1
  • $\begingroup$ yes, you're right? We only converge to a fixed point. $\endgroup$ May 21, 2017 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .