# Baby Rudin, Ex. 24 Chapt. 5, possible imprecision?

In Rudin's "Principle of Mathematical Analysis", the following exercise is given in the chapter abut differentiation:

The process described in part (c) of Exercise 22 can of course also be applied to functions that map $$(0, \infty)$$ to $$(0, \infty)$$.

Fix some $$\alpha > 1$$, and put $$f(x) = \frac{1}{2} \left( x + \frac{\alpha}{x} \right), \qquad g(x) = \frac{\alpha+x}{1+x}.$$ Both $$f$$ and $$g$$ have $$\sqrt{\alpha}$$ as their only fixed point in $$(0, \infty)$$. Try to explain, on the basis of properties of $$f$$ and $$g$$, why the convergence in Exercise 16, Chap. 3, is so much more rapid than it is in Exercise 17. (Compare $$f^\prime$$ and $$g^\prime$$, draw the zig-zags suggested in Exercise 22.)

Do the same when $$0 < \alpha < 1$$.

where it is referring to the following exercise:

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

(c) 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 $$\left( x_1, x_2 \right) \rightarrow \left( x_2, x_2 \right) \rightarrow \left( x_2, x_3 \right) \rightarrow \left( x_3, x_3 \right) \rightarrow \left( x_3, x_4 \right) \rightarrow \cdots.$$

(Credit to @Saaqib Mahmood for having written these exercises before).

Now, to my actual problem with how the exercise is stated: in the first line Rudin suggests that the result of Ex. 22 Point (c) can be extended to functions $$f:]0, +\infty[ \rightarrow ]0, +\infty[$$ which, I think, isn't always true.

For instance consider the following counter example: $$f(x) = \frac{1-e^{-x}}{2}, x > 0$$. It is easy to see that it satisfies the hypothesis of Ex. 22 Point (c), however, any sequence defined by recursion through it, starting with a positive real, approaches $$0$$ (which is in fact its only fixed point on $$\Bbb R$$) as $$n \rightarrow +\infty$$.

I would like to know other opinions or thoughts on why Rudin has put it like that.

• You're quite right, as far as I can tell; the domain and codomain should probably have been the closed set [0, inf). Mar 19, 2022 at 23:11
• @Ovinus Real Yeah, I also think so. Mar 19, 2022 at 23:31

$$1-e^{-x}$$ does not satisfy the hypothesis of (c): Suppose $$A<1$$ and $$|f'(t)|\leq A$$ for all $$t \in (0,\infty)$$. Then $$e^{-t} \leq A$$ for all $$t >0$$. But this inequality is false for $$t$$ close to $$0$$. It is valid only when $$t \geq \ln (\frac 1 A)$$. [$$\ln (\frac 1 A)$$ is a positive number].
However, $$f(x)=\frac x 2$$ satisfies the hypothesis of $$(c)$$ and it has no fixed point in $$(0,\infty)$$. (c) holds for $$[0,\infty)$$ but not for $$(0,\infty)$$.