5
$\begingroup$

Background:

Let $ f : \mathbb R ^ 2 \to \mathbb R $ be a continuous function such that $ f ( x , x ) = 0 $ for all real numbers $ x $, and $ f ( x , y ) > 0 $ for all real numbers $ x , y $ with $ x > y $.

Hypothesis: For any such $ f $, there exists a strictly increasing real function $ g $ such that $ g ' ( x ) - g ' ( y ) \ge f ( x , y ) \big( g ( x ) - g ( y ) \big) $ for all $ x , y $ that $ x > y > 0 $.

Is it possible to prove this hypothesis?

Motivation: Since $ f ( x , y ) $ can be arbitrarily large, this question translates to: Can the derivative difference be arbitrarily larger than the function difference? From English the answer seems to be obviously no, but looking at the mathematical formula, the existence seems also obvious.

Example 1: Let my try a a small example when $ f ( x , y ) $ has an upper bound, say $ k $.

Then, suppose that $ g ( x ) = e ^ { m x } $ where $ m > k $.

Can we verify that $ g ' ( x ) - g ' ( y ) \ge f ( x , y ) \big( g ( x ) - g ( y ) \big) $?

Example 2: $ f ( x , y ) = ( x - y ) ^ 2 $.

(my try)

Using Taylor expansion, one sufficient condition for the hypothesis is: for any $ n $, for any $ y \in ( 0 , x ) $, $$ ( x - y ) ^ n g ^ { ( n + 1 ) } ( x ) \ge ( x - y ) ^ 2 ( x - y ) ^ n g ^ n ( x ) $$ $$ g ^ { ( n + 1 ) } ( x ) \ge ( x - y ) ^ 2 g ^ n ( x ) \text . $$

A little work will show that, for any $ n $, the solution set for this inequality does exist. However, I am not sure if the solution sets overlap.

Related: Is it possible for the derivative of a function to grow arbitrarily faster than the function itself?

First answer.

Note: I took Martin's suggestion to make everything positive.

$\endgroup$
12
  • 2
    $\begingroup$ At least two modifications are needed to give this statement a chance at being true: first, we should assume that $f$ is continuous (otherwise it could be unbounded in every neighborhood of every point); second, we should put absolute values in the desired inequality, otherwise it's unlikely that it will be true for both $(x,y)$ and $(y,x)$ simultaneously. $\endgroup$ Apr 3, 2021 at 4:33
  • 2
    $\begingroup$ For what kind of examples functions $f(x,y)$ have you found a mathcing $g(x)$? If $f(x,y)$ is constant $k$ then $g(x)=e^{kx}$ will do nicely. But otherwise? For example $g(x)=e^{x^3}$ does not work for the entire line because $g'(x)$ is not injective on the negative half. I seem to bang my head against that wall, but may be I'm approaching this question from a wrong direction :-) $\endgroup$ Apr 3, 2021 at 5:42
  • 2
    $\begingroup$ (increasing should be interpreted as strictly increasing, else $g\equiv c$ solves the problem) $\endgroup$ Apr 3, 2021 at 6:10
  • 2
    $\begingroup$ With $g$ strictly increasing we immediately get that $g'$ is also strictly increasing. What functions other than $g(x)=Ax+B+e^{kx}, k,A>0$ have this property? $\endgroup$ Apr 3, 2021 at 6:15
  • 2
    $\begingroup$ @JyrkiLahtonen $e^{e^x}$? :) $\endgroup$ Apr 3, 2021 at 6:26

1 Answer 1

4
$\begingroup$

$ \def \R {\mathbb R} \def \Rp {\mathbb R _ +} \def \Rz {\mathbb R _ { 0 + }} $ Let $ A = \left\{ ( x , y ) \in \Rp ^ 2 \big| x > y \right\} $. You can show that for any locally bounded $ f : A \to \Rz $ there is a strictly increasing $ g : \Rp \to \Rp $ as smooth as you want (in fact analytic) satisfying $$ g ' ( x ) - g ' ( y ) > f ( x , y ) \big( g ( x ) - g ( y ) \big) $$ for all $ x , y \in \Rp $ with $ x > y $. It's sufficient to prove the existence of a strictly increasing analytic function $ h : \Rp \to \Rp $ such that $ f ( x , y ) < h ( x ) $ for all $ x , y \in \Rp $ with $ x > y $. That's because we can then define $ g : \Rp \to \Rp $ with $$ g ( x ) = \exp \int _ 0 ^ x h ( t ) \ \mathrm d t \text , $$ which gives $ g ' ( x ) = h ( x ) g ( x ) $, and thus $$ \frac { g ' ( x ) - g ' ( y ) } { g ( x ) - g ( y ) } = \frac { h ( x ) \big( g ( x ) - g ( y ) \big) + g ( y ) \big( h ( x ) - h ( y ) \big) } { g ( x ) - g ( y ) } \\ = h ( x ) + \frac { \overbrace { h ( x ) - h ( y ) } ^ { > 0 } } { \exp \underbrace { \int _ y ^ x h ( t ) \ \mathrm d t } _ { > 0 } - 1 } > h ( x ) \text . $$ To prove that such $ h $ exists, first define $ h _ 0 : \Rp \to \Rp $ with $$ h _ 0 ( z ) = 1 + \sup _ { x , y \le z } f ( x , y ) $$ (which can be done since $ f $ is locally bounded), then find a continuous function $ h _ 1 : \Rp \to \Rp $ dominating $ h _ 0 $ as is done here, and use $ h _ 1 $ to get $ h $ as is done here. Note that $ h _ 0 $ is increasing by definition, $ h _ 1 $ will become increasing by the construction in the first link, and $ h $ will become strictly increasing since it is defined by a power series with positive coefficients, and every monomial with positive coefficient is strictly increasing on the positive half-line. The fact that $ f ( x , y ) < h ( x ) $ comes from the definition of $ h _ 0 $ and the fact that $ h _ 0 $ is dominated by $ h _ 1 $ and $ h _ 1 $ is dominated by $ h $. Note that in the given links, the meaning of being dominated is taken to be something about the limit at infinity. But in the case of the current problem, the constructions given there in fact result in dominating functions in the stronger sense that we need; i.e. ones which are greater than the dominated function at every point of the domain $ \Rp $.


It can be shown that local boundedness of $ f : A \to \Rz $ is necessary for existence of a strictly increasing differentiable function $ g : \Rp \to \R $ satisfying $$ g ' ( x ) - g ' ( y ) \ge f ( x , y ) \big( g ( x ) - g ( y ) \big) $$ for all $ x , y \in \Rp $ with $ x > y $. To see how, first note that the right-hand side of the inequality is nonnegative by the assumptions, and thus $ g ' $ is increasing (possibly not strictly). On the other hand, by Darboux's theorem $ g ' $ has the intermediate value property, which together with being increasing, proves that $ g ' $ is continuous. Therefore, we can define the function $ q : A \to \Rz $ with $$ q ( x , y ) = \frac { g ' ( x ) - g ' ( y ) } { g ( x ) - g ( y ) } \text , $$ and see that $ q $ is a continuous function satisfying $ q ( x , y ) \ge f ( x , y ) $ for all $ x , y \in \Rp $ with $ x > y $. Since $ q $ is continuous, it is locally bounded, and thus $ f $ is locally bounded above. As $ f $ is bounded below by assumption, the result follows.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .