Uniform continuity and Bounded Derivative

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Discussion Overview

The discussion centers around the relationship between uniform continuity and bounded derivatives for differentiable functions. Participants explore whether the condition of having a bounded derivative is necessary for a function to be uniformly continuous, particularly in the context of differentiability over the entire real line.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that if a function \( f: \mathbb{R} \to \mathbb{R} \) is differentiable and satisfies \( |f'(x)| < M < \infty \), then it is uniformly continuous, referencing the Mean Value Theorem.
  • Another participant counters this by providing the example of the function \( f(x) = \sqrt{x} \) on the interval \( (0,1) \), which is uniformly continuous but has an unbounded derivative as \( x \) approaches 0.
  • A third participant agrees with the previous point but notes that the original statement requires differentiability on all of \( \mathbb{R} \), suggesting that the example may not be applicable.
  • Another example is introduced, \( f(x) = x^{3/2} \sin(1/x) \) on \( [0,1] \), which is uniformly continuous and differentiable on \( \mathbb{R} \) but has an unbounded derivative.
  • One participant raises a question about the implications of extending functions outside of compact intervals and whether the bounded derivative condition must hold in those cases.

Areas of Agreement / Disagreement

Participants generally disagree on whether the condition of having a bounded derivative is necessary for uniform continuity. Multiple competing views remain, with examples provided to illustrate differing perspectives.

Contextual Notes

Some limitations include the dependence on the definitions of uniform continuity and differentiability, as well as the specific intervals considered for the functions discussed. The examples provided highlight the complexity of the relationship between these concepts.

Bacle
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Hi, All:

Let f R-->R be differentiable. If |f'(x)|<M< oo, then f is uniformly continuous, e.g.,

by the MVTheorem. Is this conditions necessary too, i.e., if f:R-->R is differentiable

and uniformly continuous, does it follow that |f'(x)|<M<oo ?

Thanks.
 
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Hi Bacle! :smile:

This is not true. Consider the function

f:]0,1[\rightarrow\mathbb{R}: x\rightarrow \sqrt{x}

then this is uniform continuous (continuous on compact interval). But the derivative f^\prime(x) grows very large if x gets closer to 0.

In fact, the condition you mention is equivalent to the Lipschitz-condition (for differentiable functions). That is, if f:[a,b]\rightarrow \mathbb{R} is continuous and differentiable on ]a,b[, then the following are equivalent
  • |f~{}^\prime(x)|\leq k~ for all x in ]a,b[
  • f is k-Lipschitz, that is |f(x)-f(y)|\leq k|x-y|~~ for all x,y in [a,b]

The proof uses the mid-value theorem.

It is easy to see that being Lipschitz is stronger than uniform continuity.
 
micromass said:
Hi Bacle! :smile:

This is not true. Consider the function

f:]0,1[\rightarrow\mathbb{R}: x\rightarrow \sqrt{x}

then this is uniform continuous (continuous on compact interval). But the derivative f^\prime(x) grows very large if x gets closer to 0.

Unless I'm reading something wrong, the statement in the original post requires differentiability on all R. This example isn't applicable.
 
Fine, consider

f:[0,1]\rightarrow \mathbb{R}:x\rightarrow x^{3/2}\sin(1/x)

this extends to an uniform continuous, differentiable function on \mathbb{R}, but it's derivative is not bounded.
 
Thanks, micromass; I agree, because continuous+compact implies uniformly-continuous;

I was thinking more of what if the extension outside of

[a,b] is non-trivial (e.g., a constant function, or without exhausting , by covering

the remainder by compact sets and pasting/smoothing at the endpoints), if the condition on

the derivative must be satisfied in (-00,a)\/(b,oo), i.e., if the function is not defined

piecewise in countably-many compact intervals.
 

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