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.

  • 1
    $\begingroup$ You're quite right, as far as I can tell; the domain and codomain should probably have been the closed set [0, inf). $\endgroup$ Mar 19, 2022 at 23:11
  • $\begingroup$ @Ovinus Real Yeah, I also think so. $\endgroup$ Mar 19, 2022 at 23:31

1 Answer 1


$1-e^{-x}$ 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 $(0,\infty)$.

  • $\begingroup$ Yeah, you are entirely right! Also, I could have just divided my function by 2 thus obtaining the wanted result. However I would like to know what do you think about Rudin’s “imprecision”. $\endgroup$ Mar 19, 2022 at 23:29

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