Supremum of Continuous Functions is Continuous Rudin

evianpring Asks: Spivak, Ch. 22, "Infinite Sequences", Step in Problem 26c: Given $e^{-1}\leq b\leq e$ and $a=b^{1/b}$, how to obtain $e^{-e}\leq a \leq e^{1/e}$?
Given $$e^{-1}\leq b\leq e\tag{1}$$

and

$$a=b^{1/b}\tag{2}$$

How do we derive the result that

$$e^{-e}\leq a \leq e^{1/e}\tag{3}$$

For context about how this question arose

The following is a problem from Ch. 22 of Spivak's Calculus

1. This problem investigates for which $a>0$ the symbol

$$a^{a^{a^{a^{...}}}}$$

makes sense. In other words, if we define a sequence $\{b_n\}$ by

$$b_1=a$$

$$b_n+1=a^{b_n}$$

when does

$$\lim\limits_{n\to\infty} b_n=b$$

exist?

(a) If $b$ exists, then $a$ can be written in the form $y^{1/y}$ for some $y$. Describe the graph of $g=y^{1/y}$ and conclude that $0<a\leq e^{1/e}$.

(b) Suppose that $1\leq a \leq e^{1/e}$. Show that $\{b_n\}$ is increasing and also that $b_n\leq e$. This proves that $b$ exists (and also that $b\leq e$)

The analysis for $a<1$ is more difficult.

(c) Using Problem 25, show that if $b$ exists, then $e^{-1}\leq b\leq e$. Then show that $e^{-e}\leq a \leq e^{1/e}$.

My question is about a single step in item $(c)$ that seems to involve essentially an understanding of inequalities.

Let me go through the items to reach the point where the question arises. If you want to see the question directly, just skip to the last part.

Note that $\{b_n\}$ is the sequence

$$a, f(a), f(f(a)), f(f(f(a))), ...$$

The general task is to figure out for which $a$ this sequence converges.

If it does exist then (as proved in a previous problem) the limit is a fixed point of $f$, ie it is some value $b$ such that

$$f(b)=b=a^b$$

and thus we can write

$$a=b^{1/b}$$

If we plot the function $g(x)=x^{1/x}$ we arrive at

What this graph tells us is that $a$ can't be just any value. Since $x^{1/x}$ reaches a max value of $e^{1/e}$ at $x=e$, then $a$ can only be between $0$ and $e^{1/e}$.

Part b

Assume that $1\leq a\leq e^{1/e}$.

We can show that for such values of $a$, $\{b_n\}$ is increasing and bounded above.

$$b_1=a\geq 1$$

$$b_2=a^a\geq b_1\geq 1$$

If we assume $b_n\geq b_{n-1}$, then $b_{n+1}=a^{b_n}>a^{b_{n-1}}=b_n$.

By induction, we can infer that $\{b_n\}$ is increasing.

Using another induction argument we can show that $b_n<e$. Thus, since $\{b_n\}$ is bounded above and increasing, it converges.

Part c

Here is what what proved in Problem 25

If $\{b_n\}$ is a sequence defined by $b_1=a$ and $b_{n+1}=f(b_n)$, then if $b=\lim\limits_{n\to\infty} b_n$ exists and $f'$ is continuous at $b$, then $|f'(b)|\leq 1$ (provided we don't already have $b_n=b$ for some $n$).

We can apply this directly to the current problem. The only thing to check is the condition on $f'$.

$$f'(x)=a^x\log{(a)}$$

which is continuous.

Therefore, if we assume that $b=\lim\limits_{n\to\infty} b_n$ exists, then by the result above we infer that $|f'(b)|\leq 1$ which means

$$|f'(b)|=|a^b\log{(a)}|=|b\log{(a)}|=|\log{(a^b)}|=|log{(b)}|\leq 1$$

$$-1\leq\log{(b)}\leq 1$$

$$e^{-1}\leq b\leq e\tag{1}$$

Now, for the question

How do we derive the result that

$$e^{-e}\leq a \leq e^{1/e}\tag{3}$$

?

The solution manuals says

Since $$a=b^{1/b}\tag{2}$$ we have

$$(e^{-1})^{1/e^{-1}}\leq a \leq e^{1/e}\tag{4}$$

or

$$e^{-e}\leq a \leq e^{1/e}\tag{3}$$

How do we get from $(1)$ to $(3)$ using $(2)$?

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