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I'm trying to have an intuition behind the uniformly continuous functions. Something to show to my students. For example, before giving the formal definition and some examples of continuous functions, we can say to beginners that roughly speaking, continuous functions are functions without "holes", that's why the name "continuous".

What about uniformly continuous functions? is there some ideas to give to the students to give them an intuition behind these functions, before to show to them the formal definition?

Thanks

user42912
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4 Answers4

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Here I will provide a real life heuristic in the lines of the explanation given by Michael Hardy.

Usually when one of us move his mouse a little bit (infinitesimal displacement) it happens that the arrow on the screen moves a little bit too (infinitesimal displacement too); it is what we expects from the mouse, this is "uniformly continuous". However, sometimes one try to move a little bit the mouse (infinitesimal displacement) but the arrow remains at the same place until it suddenly jumps to some "far" place (non-infinitesimal displacement); it is what we do not expects from the mouse, this is NOT uniformly continuous).

Idris Addou
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  • If $f$ is continuous and $x$ is real and $dx$ is infinitesimal, then $dy=f(x+dx)-f(x)$ is infinitesimal.

  • If $f$ is uniformly continuous and $dx$ is infinitesimal, then $dy=f(x+dx)-f(x)$ is infinitesimal even if $x$ is not real, i.e. if $x$ differs from a real number by an infinitely large or infinitely small amount.

And only if. In both cases above.

For example:

  • Suppose $f(x) = \sin\dfrac 1 x$. If $x\ne0$ is infinitely close to $0$ then $f(x)$ can change from $1$ to $-1$ while $x$ changes by an infinitely small amount, so the change in $f(x)$ is not infinitesimal. Thus this function is not uniformly continuous on $\mathbb R\setminus\{0\}$.
  • Suppose $f(x) = e^x$. If $x$ is infinitely large, then $f(x)$ can increase by $1$ while $x$ increases by an infinitely small amount. So this function is not uniformly continuous on $\mathbb R$.

This can function as a mere heuristic but it can also be proved rigorously in the context of Robinson's nonstandard analysis.

Michael Hardy
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  • How does this heuristic work for $\sin(1/(x-1))$? – Dirk May 17 '15 at 09:41
  • @Dirk : The same way as for $\sin(1/x)$. The variable $x$ can change from one number differing from $1$ by an infinitesimal to another number differing from $1$ by an infinitesimal, with the effect that $\sin(1/(x-1))$ changes from $1$ to $-1$. So $x\mapsto\sin(1/(x-1))$ is not uniformly continuous on $\mathbb R\setminus{1}$. ${}\qquad{}$ – Michael Hardy May 17 '15 at 16:55
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Uniformly continuons functions transforms Cauchy sequences into Cauchy sequences but continuons functions not necessarily (see $x_n=1/n$ and $f(x)=1/x$ on $(0,1]$)

Michael Hardy
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Idris Addou
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For uniform continuity, the value of $\delta$ depends only on $\varepsilon$ and not on the point $x$. (So the same $\delta$ works "uniformly" across all points in the interval under discussion.)

In contrast, for continuity, the value of $\delta$ can depend on $\varepsilon$ and the point $x$ under discussion.

JohnD
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