# Prob. 22, Chap. 5 in Baby Rudin: Fixed Points of Real Functions

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?

• (d) This might help (different class of functions but the picture works): [Fixed Points - Graph Visualization][1] [1]: math.stackexchange.com/a/2289738/432081 May 21 '17 at 14:52

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.