Uniform continuity and Bounded Derivative

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SUMMARY

The discussion centers on the relationship between uniform continuity and bounded derivatives for differentiable functions. It is established that while a bounded derivative implies uniform continuity, the converse is not true. Counterexamples, such as the function f(x) = √x and f(x) = x^(3/2)sin(1/x), demonstrate that a function can be uniformly continuous without having a bounded derivative. The equivalence of the Lipschitz condition to the bounded derivative condition is also highlighted, emphasizing that Lipschitz continuity is a stronger condition than uniform continuity.

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  • Understanding of differentiable functions
  • Familiarity with the Mean Value Theorem
  • Knowledge of Lipschitz continuity
  • Basic concepts of uniform continuity
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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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