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Statement of L'Hospital's Rule

Let $F$ be an ordered field.

L'Hospital's Rule. Let $f$ and $g$ be $F$-valued functions defined on an open interval $I$ in $F$. Let $c$ be an endpoint of $I$. Note $c$ may be a finite number or one of the symbols $-\infty$,$+\infty$. Suppose $f'(x)$ and $g'(x)$ are defined everywhere on $I$. Suppose $g(x)$ and $g'(x)$ are never zero and never change sign on $I.$ Suppose one of the following two hypothesis is satisfied:

  1. $\displaystyle \lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} g(x) = 0$
  2. $\displaystyle \lim_{x \rightarrow c} g(x) = \pm \infty$

Suppose $$\displaystyle \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)} = L$$ where $L$ is a finite number or one of the symbols $-\infty$, $+\infty$.

Then $$\displaystyle \lim_{x\rightarrow c} \frac{f(x)}{g(x)} = L.$$

Motivation

The standard proof of l'Hospital's rule (when $F=\mathbb{R}$) uses Cauchy's mean value theorem. See:

For an ordered field, the mean value theorem is equivalent to the least upper bound property. See:

Another proof of l'Hospital's rule can be based on the convergence of bounded monotone sequences and the statement that $f' > 0$ on an interval $I$ implies $f$ strictly increasing on $I$. See Taylor, A. E. (1952), "L'Hospital's rule", Amer. Math. Monthly, 59: 20–24.

The convergence of bounded monotone sequences is equivalent to the least upper bound property. The statement that $f' > 0$ on $I$ implies $f$ strictly increasing on $I$ is also equivalent to the least upper bound property.

L'Hospital's rule is false for $F=\mathbb{Q}$. The following proof outline is adapted from Exercise 7.8 of Korner's book "A Companion to Analysis." Choose a sequence $a_n \in \mathbb{R} \setminus \mathbb{Q}$ with $4^{-n-1} < a_n < 4^{-n}$ for $n=1,2,\ldots$. Define $I_0 = \{x \in \mathbb{Q} : a_0 < x \}$ and $I_n = \{x \in \mathbb{Q} : a_n < x < a_{n-1}\}$. Notice $4^{-n-1} < x< 4^{-n}$ whenever $x \in I_n$. Define $f:\mathbb{Q} \rightarrow \mathbb{Q}$ by $f(0)=0$ and $f(x) = 8^{-n}$ if $|x| \in I_n$. Remember we work in the ordered field $F=\mathbb{Q}$. We have $f'(x)=0$ for all $x \in \mathbb{Q}$, and $$ \lim_{x \to 0}\frac{f(x)}{x^2} = \infty \quad \text{and} \quad \lim_{x \to 0}\frac{f'(x)}{2x} = 0. $$

Question

What is the logical relationship between l'Hospital's rule and the least upper bound property?

Does l'Hospital's rule imply the least upper bound property? Or is there an ordered field with l'Hospital's rule but not the least upper bound property?

Related

Here is a related question. Limit of the derivative and LUB

It asks (in my notation) whether the following differentiability criteria implies the least upper bound property.

Differentiability Criteria. Let $f$ be an $F$-valued function defined on open interval $I$. Suppose $f$ is continuous on $I$. Suppose $f$ is differentiable on $I$ except at one point $c$ in $I$. If $\lim_{x \rightarrow c} f'(x)$ exists, then $f'(c)$ exists and equals this limit.

L'Hospital's rule implies this differentiability criteria. See for example https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Corollary

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    $\begingroup$ I guess one would have to have some kind of definition of derivtives and limits on the field $F$, even to formulate a relation of the kind you seek for an ordered field $F.$ Do you have such definitions in mind? $\endgroup$
    – coffeemath
    Sep 6, 2016 at 13:36
  • $\begingroup$ The definitions of limit and derivative in $\mathbb{R}$ work in any ordered field. Just replace statements like $\epsilon \in \mathbb{R}, \epsilon > 0$ by $\epsilon \in F, \epsilon > 0$. $\endgroup$ Sep 6, 2016 at 17:12
  • $\begingroup$ So would you be considering functions $f:\mathbb{F}\to\mathbb{F}$, then? $\endgroup$ Sep 6, 2016 at 17:33
  • $\begingroup$ @CameronWilliams You could. Or more generally $f:I \rightarrow F$, where is $I$ is an interval in $F$. $\endgroup$ Sep 6, 2016 at 17:49

1 Answer 1

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Here let's assume $F$ is non-archimedean.

For each infinitesimal* $x \in F^{\times}$, let $[x]$ denote the archimedean class $\bigcup \limits_{n \in \mathbb{N}^*} \left]-n |x|;-\frac{1}{n} |x|\right[ \cup \left]\frac{1}{n}\cdot|x|;n\cdot|x|\right[$ of $x$ in $F^{\times}$. Let $C$ denote the set of acrhimedean classes in $F^{\times}$.

($C$ is at least countable since for each infinitesimal $x$, $[x];[x^2];\ldots;[x^n];\ldots$ are distinct.)

Now let $\varphi$ be a choice function on $C$, and let $f$ be the map defined by $f(x) = \varphi([x])$ if $x$ is infinitesimal, and $f(y) = 0$ if $y$ is not an infinitesimal.

$f$ is differentiable on $F^{\times}$ with $f' = 0$ since for $x \neq 0$, $f$ is constant in the neighborhood $[x]$ of $x$. Moreover, $\lim \limits_{x \to 0} f(x) = 0$, since for $\varepsilon > 0$ infinitesimal, $\forall x \in \left]-\varepsilon^2;\varepsilon^2\right[$, $|f(x)| \leq \varepsilon$.

However, $\frac{f}{\operatorname{id}_{F^{\times}}}$ does not converge at $0$: given $r > 0$ and infinitesimal, and $n \in \mathbb{N}^*$, $[\frac{\varphi(r)}{n}] = [\varphi(r)]$, so $\dfrac{f(\frac{\varphi([r])}{n})}{\frac{\varphi([r])}{n}} = \dfrac{\varphi([r])}{\frac{\varphi([r])}{n}} = n$.

I believe this counterexample (inspired by that which you proposed for $\mathbb{Q}$), along with an adaptation of your argument for $\mathbb{Q}$ to any archimedean proper subfield of $\mathbb{R}$, proves that l'Hospital's rule is equivalent to the LUB property. (Incidently, the counterexample also fits my question since $f$ is not differentiable at $0$.)


*$x$ such that $\forall n \in \mathbb{N}, n |x| < 1$.

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  • $\begingroup$ Very interesting. So the argument is: F with L'Hopital's Rule (or Differentiability Criteria) implies $F$ archimedean (proved by contrapositive). And no proper archimedean subfield of $\mathbb{R}$ can have L'Hopital's Rule (or Differentiability Criteria), so $F$ must be (isomorphic to) $\mathbb{R}$, the only (up to ismorphishm) ordered field with LUB. $\endgroup$ Oct 2, 2016 at 17:53
  • $\begingroup$ Clarification: $\mathbb{N}^{\ast}$ is the smallest inductive set in $F$? In other words, it's the copy of $\mathbb{N}$ in $F$? Why the $\ast$? $\endgroup$ Oct 2, 2016 at 17:55
  • $\begingroup$ Is every archimedean ordered field a proper subfield of $\mathbb{R}$? Up to isomorphism, of course. $\endgroup$ Oct 2, 2016 at 17:58
  • $\begingroup$ Indeed, every archimedean ordered field is (isomorphic to) a subfield of $\mathbb{R}$. planetmath.org/archimedeanorderedfieldsarereal $\endgroup$ Oct 2, 2016 at 18:05
  • $\begingroup$ $\mathbb{N}$ would be the smallest inductive set in $F$ containing $0$, and $\mathbb{N}^*$ would be $\mathbb{N} - \{0\}$. It might be a french notation, I always forget. And yes, archimedean fields embed uniquely in $\mathbb{R}$, so if an archimedean field isn't isomorphic to $\mathbb{R}$, you can copy your argument for $\mathbb{Q}$ since it is not difficult to produce sequences of the sort using one element from $\mathbb{R} - F$ and elements form $F$. $\endgroup$
    – nombre
    Oct 2, 2016 at 20:23

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