[Paper Review] "Non-strict" l'Hospital-Type Rules for Monotonicity: Intervals of Constancy

Iosif Pinelis|arXiv (Cornell University)|Nov 21, 2006

Mathematical Inequalities and Applications30 references17 citations

TL;DR

This paper establishes that the original ratio $ r = f/g $ can have at most one maximal interval of constancy (m.i.c.), and such an interval must coincide exactly with a level set of the derivative ratio $ \tilde{\rho} = \frac{f'}{g'} - \frac{f}{g} \cdot \frac{g'}{g} $, which is derived from the monotonicity behavior of $ \rho = f'/g' $. The key contribution is a complete characterization of when and where $ r $ remains constant, showing that constancy intervals of $ r $ are uniquely determined by the zero set of $ \tilde{\rho} $, and that $ r $ cannot have multiple disjoint m.i.c.s.

ABSTRACT

Assuming that a "derivative" ratio rho:=f'/g' of the ratio r:=f/g of differentiable functions f and g is strictly monotonic (that is, rho is increasing or decreasing), it was shown in previous papers that then r can switch at most once, from decrease to increase or vice versa. In the present paper, it is shown that, if rho is non-strictly monotonic (that is, non-increasing or non-decreasing), then r can have at most one maximal interval of constancy (m.i.c.); on the other hand, any one m.i.c. of a given derivative ratio rho is the m.i.c. of an appropriately constructed original ratio r.

Motivation & Objective

• To characterize the structure of maximal intervals of constancy (m.i.c.) for the ratio $ r = f/g $ in terms of the derivative ratio $ \rho = f'/g' $.
• To determine under what conditions $ r $ can be constant on a non-degenerate interval, and how such constancy relates to the monotonicity and zero set of $ \tilde{\rho} = \rho - r \cdot \frac{g'}{g} $.
• To show that $ r $ can have at most one m.i.c., and that any such interval must be an m.i.c. of both $ \rho $ and $ \tilde{\rho} $.
• To construct, for any given m.i.c. of $ \rho $, a corresponding $ f $ such that $ r = f/g $ has exactly that interval as its unique m.i.c.

Proposed method

• Define $ \tilde{\rho} = \frac{f'}{g'} - \frac{f}{g} \cdot \frac{g'}{g} $, a transformed derivative ratio that captures the rate of change of $ r $.
• Use the condition $ gg' \neq 0 $ on $ (a,b) $ to ensure $ g $ is strictly monotonic and of bounded variation, enabling Riemann-Stieltjes integration.
• Construct $ f $ via the formula $ f(x) = Kg(z) + \int_z^x \rho(u) \, dg(u) $, ensuring $ f'/g' = \rho $ and $ f/g = K $ on the desired interval.
• Analyze the level set $ \ell_0(\tilde{\rho}) = \{ u \in (a,b) : \tilde{\rho}(u) = 0 \} $, showing it equals the m.i.c. of $ r $.
• Leverage the continuity and monotonicity of $ \tilde{\rho} $ to prove that its zero set is a closed interval, which becomes the unique m.i.c. of $ r $.
• Prove that any m.i.c. of $ \rho $ can be realized as the m.i.c. of some $ r = f/g $, via explicit construction of $ f $.

Experimental results

Research questions

• RQ1 Under what conditions can the ratio $ r = f/g $ be constant on a non-degenerate interval?
• RQ2 How does the structure of the derivative ratio $ \rho = f'/g' $ constrain the possible intervals of constancy of $ r $?
• RQ3 Can $ r $ have more than one maximal interval of constancy, and if not, why?
• RQ4 What is the precise relationship between the zero set of $ \tilde{\rho} $ and the m.i.c. of $ r $?
• RQ5 Given any m.i.c. of $ \rho $, can one always construct $ f $ such that $ r = f/g $ has that interval as its unique m.i.c.?

Key findings

• The original ratio $ r = f/g $ can have at most one maximal interval of constancy (m.i.c.).
• Any m.i.c. of $ r $ must also be an m.i.c. of the derivative ratio $ \rho = f'/g' $, and hence of the derived function $ \tilde{\rho} = \rho - r \cdot \frac{g'}{g} $.
• The unique m.i.c. of $ r $, if it exists, is exactly the level-0 set $ \ell_0(\tilde{\rho}) = \{ x \in (a,b) : \tilde{\rho}(x) = 0 \} $.
• The set of all m.i.c.’s of $ \tilde{\rho} $ is identical to that of $ \rho $, and both are equal to the m.i.c. of $ r $ when it exists.
• For any given m.i.c. $ I $ of $ \rho $, there exists a function $ f $ such that $ r = f/g $ has $ I $ as its unique m.i.c.,
• The construction of such $ f $ is achieved via Riemann-Stieltjes integration: $ f(x) = Kg(z) + \int_z^x \rho(u) \, dg(u) $, ensuring $ f'/g' = \rho $ and $ f/g = K $ on $ I $.