Functions in reals such that inequality holds

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Discussion Overview

The discussion revolves around identifying functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfy the inequality \( |f(y) - f(x)| \leq (y-x)^2 \) for all \( x, y \in \mathbb{R} \). The scope includes mathematical reasoning and exploration of differentiability implications.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the specific inequality being discussed.
  • Another participant reformulates the inequality into a form involving the difference quotient and considers the implications for differentiability.
  • There is a suggestion that if the function is differentiable, applying limits could lead to familiar results regarding the type of function \( f \) must be.
  • Another participant proposes dividing the interval into subintervals to explore the implications if the function is not differentiable.
  • A claim is made that the inequality implies differentiability, leading to the conclusion that the derivative is zero everywhere, suggesting that \( f \) must be a constant function.
  • A counterexample is presented, questioning the assumption of differentiability by providing a function that does not satisfy the derivative condition while still adhering to the original inequality.
  • One participant asserts that the squeeze theorem can be applied, leading to the conclusion that \( f' \) must equal zero, reinforcing the idea that \( f \) is constant.

Areas of Agreement / Disagreement

Participants express differing views on whether the inequality implies differentiability and whether the conclusion that \( f \) must be constant follows necessarily. The discussion remains unresolved regarding the implications of the inequality and the validity of the counterexample presented.

Contextual Notes

There are unresolved assumptions regarding the differentiability of \( f \) and the conditions under which the inequality holds. The discussion also highlights the dependence on the definitions of differentiability and continuity.

maximus101
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for which f: [tex]R[/tex] [tex]\rightarrow[/tex] [tex]R[/tex] such that
[tex]\forall[/tex] [tex]x,y[/tex][tex]\in[/tex] [tex]R[/tex]

does

| [tex]f(y)[/tex] - [tex]f(x)[/tex] | [tex]\mid[/tex] [tex]\leq[/tex] [tex](y-x)^2[/tex]

hold
 
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Which inequality?
 
sorry just fixed it
 
so I did the following:

[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]

was obtained by [tex](y-x)^2[/tex]=[tex]|y-x||y-x|[/tex] and dividing both sides by [tex]|y-x|[/tex].


then trying to use the fact that

[tex]\displaystyle\lim_{y\to x}\frac{f(y)-f(x)}{y-x}[/tex] is the derivative at x

but not sure what's next how to combine the two?
 
Is the function differentiable? If it isn't you can't use the limit thing. But if it is, apply the limit to both sides. You should then get something very familiar and be able to deduce what kind of function f must be.
 
Okay, if f isn't differentiable, here's a hint. Divide the interval from [x,y] into n equal sub intervals. See if you can get anything from that.
 
Wouldn't this statement
[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]
automatically imply the function is differentiable?
 
Yuqing said:
Wouldn't this statement
[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]
automatically imply the function is differentiable?


Yes. Using the squeeze theorem you can see that the derivative is actually 0 everywhere, implying the only functions satisfying the inequality are constant functions.
 
Am I missing something? I don't see any derivatives anywhere, just the difference quotient.

If you indented for it to be the difference quotient, then a counter example would be [tex]r \mapsto r^3[/tex]. For all |x-y| >= 1, [tex]\left| x^3 - y^3 \right| > \left( x - y \right)^2[/tex]
 
  • #10
Assume there exists a function f satisfying [tex]0 \le |f(y) - f(x)| \le (y-x)^2 = |y-x|^2[/tex] for any y, x in the domain of f. This implies that [tex]0 \ le \frac{|f(y) - f(x)|}{|y-x|} \le |y-x|[/tex].

Let y approach x. By the squeeze theorem, we see that the limit exists, is equal to 0, and is in fact the derivative of f evaluated at x. So f'(x) = 0 for all x in its domain, which means that f must be a constant function.
 

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