I'm struggling somewhat with the difference between 'uniformly continuous' and 'continuous'. I understand the intuition, but I need to see it in an example.

Say $f=2x$, or some other easy function on the compact set $[1,2]$. Then is it correct to say that the function is uniformly continuous on that set? My question is what the difference is between the proofs for uniformly continuous, and just plain continuous, using the delta epsilon definition. I can't wrap my head around the proof at all.

  • $\begingroup$ roughly continuity is a pointwise property where as uniform continuity is a nbhd property . A continuous function which oscillates "highly" is not uniformly continuous . $\endgroup$ – Theorem May 15 '13 at 10:34

The difference is in the fact that the $\delta$ you find does not depend on $x$, but only on $\varepsilon$.

As you've said, $x \mapsto ax$ (for some $a \ne 0$) is uniformly continuous on $[1,2]$ but also on the whole real line. Indeed, pick a point $x\in\mathbb R$ and fix $\varepsilon > 0$: then $$ \vert f(x)-f(y) \vert = a \vert x -y \vert < \varepsilon $$ whenever $\vert x-y \vert < \frac{\varepsilon}{a}=: \delta$. Such a $\delta$ is independent on the point $x$, as you easily can see.

On the contrary, take the function $x \mapsto \frac{1}{x}$ with $x\in (0, 1]$. I show that this function is not uniformly continuous, i.e. I want to show that there exists a $\varepsilon>0$ s.t for every $\delta>0$ there are $x_1, x_2$ such that $\vert x_1-x_2\vert <\delta$ but $\vert f(x_1)-f(x_2)\vert \ge \varepsilon$. Pick a $\delta \in (0,1)$: if we choose $$ x_1 = \frac{\delta}{5}, \qquad x_2=\frac{\delta}{10} $$ we have $$ \vert x_1-x_2 \vert = \frac{\delta}{10} < \delta $$ but $$ \vert f(x_1)-f(x_2) \vert =\frac{5}{\delta}>5. $$ So, this is enough to take $\varepsilon \in (0, 5]$ to get our claim.

Some final notes

1. Every continuous function on a compact set is uniformly continuous (Heine-Cantor theorem).

2. If a continuous function $f \colon [0,+\infty) \to \mathbb R$ is s.t. $\lim_{x\to +\infty} f(x)$ exists and it is finite then $f$ is uniformly continuous.

  • 1
    $\begingroup$ I. Cannot thank you enough. Thank you so much for such a comprehensive answer. $\endgroup$ – Roger Tynes May 15 '13 at 10:57
  • 1
    $\begingroup$ My pleasure. You are welcome. $\endgroup$ – Romeo May 15 '13 at 11:56

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