# Are there any geometric interpretations to uniform continuity?

There are specifically the two forms, continuity and uniform continuity, I'm referring to. So a function is continuous if the graph "doesn't break," but this also applies to a uniform continuous function. What else is required for it to be uniformly continuous? I know the technical definitions (continuity: $\forall \varepsilon \, \forall x \, \exists \delta \, \forall y \, ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon \, )$, uniform continuity: $\forall \varepsilon \, \exists \delta \, \forall x \, \forall y \, ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon \,)$) but in geometric terms, what does it mean?

• Be careful with your geometric interpretation of continuity; while it's good for a globally continuous function, the Thomae function is continuous at almost all points. en.wikipedia.org/wiki/Thomae's_function Mar 24, 2013 at 0:52
• @JulienClancy Although we always have this pathologic (very interesting) cases, I think that intuitive interpretations for this things are important too. Mar 24, 2013 at 0:57
• @JulienClancy Yes, I actually think that is the best thing of mathematics (when outcomes of formality that looked like the intuitive thing at first, are so weird) Mar 24, 2013 at 1:05
• @AndreasHagen Thomae's function is discontinuous at all rational values. It is because of the density of the irrationals over the rationals that there is continuity at each irrational point. Basically, irrational numbers are always "closer together" than rationals. Mar 24, 2013 at 1:15
• There's a good definition of it in nonstandard analysis (i.e. infinitesimals but rigorously). A function is uniformly continuous if, for every (possibly infinite/infinitesimal) $x$ and $y$, if $x-y$ is infinitesimal, then $f(x)-f(y)$ is infinitesimal. For example, $f(x)=\frac1x$ isn't uniformly continuous, since, for infinitely large $N$, $\frac1N$ and $\frac1{N+1}$ are infinitely close but $f(\frac1N)$ and $f(\frac1{N+1})$ aren't. Similarly, $f(x)=x^2$ isn't uniformly continuous since $N$ and $N+\frac1N$ are infinitely close but $f(N)$ and $f(N+\frac1N)$ aren't. Aug 19, 2015 at 12:42

There's a great theorem about uniform continuity that helped me understand it quite a bit.

If $f \colon \mathbb{R} \to \mathbb{R}$ is a function and if $\lim_{x \to \infty} f(x)/x = \pm\infty$, then $f$ is not uniformly continuous.

Basically, what this means is that the "last" uniformly continuous functions are linear. If your function grows faster than every linear function, it cannot be uniformly continuous. You can apply this intuition to similar functions as well - $f(x) = \sin(x^2)$ is not uniformly continuous because eventually its local growth outpaces an arbitrary linear function.

Edit: Here's the proof. Assume that $f$ is uniformly continuous. Then there is $\delta > 0$ such that $|x - y| < \delta \Rightarrow |f(x) - f(y)| < 1$. If $f(x)/x$ has some limit as $x \to \infty$ then $(f(x) - f(0))/x$ should have the same limit, because $f(0)/x \to 0$. Now,

$\left|\frac{f(x) - f(0)}{x}\right| = \left|\frac{f(x) - f(x - \delta) + f(x - \delta) - ... + f(\delta_0) - f(0)}{x}\right| \leq \left|\frac{f(x) - f(x - \delta)}{x} \right| + \left| \frac{f(x - \delta) - f(x - 2\delta)}{x} \right| + ... + \left| \frac{f(\delta_0) - f(0)}{x}\right| < \frac{x}{\delta}\cdot\frac{1}{x} = \frac{1}{\delta}$

so the limit cannot be infinite. I used $\delta_0$ to be some positive number less than $\delta$; it could be that $x$ is not an exact multiple of $\delta$ and this takes the remainder. In the first equality I used a telescoping sum; in the second I used the triangle inequality; in the third I just counted the number of terms.

• What is the name of this theorem? I would like to refer this in my homework. Mar 24, 2013 at 1:01
• I'm not sure of the name but I'll try to recall the proof and post it. Mar 24, 2013 at 1:04
• However, if we bound a nonlinear function to some closed interval, it may still turn out uniformly continuous. For example $f(x)=x^2$. Mar 24, 2013 at 1:10
• Yes, that point is crucial. We're looking at asymptotic behavior (and in a sense my $\sin(x^2)$ example is asymptotic). It happens that every continuous function confined to a closed interval is uniformly continuous. Mar 24, 2013 at 1:16
• Would you say then that a full characterization of uniform continuity would be given by the theorem stated in your answer (asymptotic condition), and bounding continuous functions to closed intervals (local condition)? Mar 24, 2013 at 1:19

Continuity of $f$ at $a$ means that I can make $f(x)$ as close as I want to $f(a)$ by making $x$ close to $a$. Let us formalize: No matter how small an error tolerance (for all $\epsilon > 0$), one can make $f(x)$ close to $f(a)$ within this error tolerance ($|f(x)-f(a)|< \epsilon$) by finding a number $\delta$ such that if $x$ is less than $\delta$ off from $a$ then $f(x)$ is less than $\epsilon$ off from $f(a)$.

For uniform continuity, first take an example. Let our error tolerance ($\epsilon$) be $1/2$. If $f(x)$ = x, then if I want to make $f(x)$ at most $1/2$ off from $f(a)$, all I need to do is make $x$ $1/2$ off from $a$ (let $\delta = \epsilon = 1/2$). Usually, however, $\delta$ depends on $x$: if $f(x) = x^2$, then near zero all of the $f(x)$'s are fairly close to each other ($f(x)$ isn't changing very fast), so $\delta$ can be pretty big - if x is kinda close to zero, then $f(x)$ within 1/2 of zero. However, if x is near 100, $f(x)$ is changing very quickly, and thus for $f(x)$ to be close to $f(100) = 10000$, $x$ has to be really close to 100 for $f(x)$ to at most 1/2 off from $f(100)$, or $\delta$ must be teeny. Uniform continuity just says that a minimum $\delta$ works: if $x, y$ are in your interval, no matter how fast $f$ is changing, you are guaranteed that if $x$ and $y$ are within $\delta$ of each other, $f(x)$ and $f(y)$ are within $\epsilon$ of each other.

I hope that this helps: by an inscribed triangle I mean a right triangle s.t. the vertices of its longest side are on the graph of function $$f$$ and the adjacent sides to the right angle (legs) are parallel to $$x$$-axis and $$y$$-axis(unfortunately I don't know how to draw a picture here), we call the side parallel to $$x$$-axis its base and ,the side parallel to $$y$$-axis its altitude (just here not generally!)

$$f$$ is uniformly continuous if when we look at the set of all inscribed triangles, the altitude becomes small enough provided that the base becomes small enough.

for example $$f(x)=\sin{1/x}$$ is not uniformly continuous because near $$0$$ you can find inscribed triangles with arbitrary small base s.t. their altitude is not small enough. similarly, $$f(x)=x^2$$ is not uniformly continuous (I assume the domain is the whole real line) because you can find inscribed triangles with similar property, however, in this case the triangles go far away from $$0$$.

One can find a similar interpretation for continuity in terms of inscribed triangles which have one of their vertices (other than the vertex with the right angle) at point $$(x,f(x))$$.

Suppose $$\forall x,y\in E, \frac{|f(x)-f(y)|}{|x-y|}. In other words, the slopes of the secant lines between any two points on the curve are bounded. It also follows that the derivative is bounded if the function is differentiable. If we select $$\delta=\epsilon/M$$, and $$|x-y|<\delta$$ then $$|f(x)-f(y)| and so $$f(x)$$ is uniformly continuous. So bounded slopes of secant lines imply uniform continuity.

Suppose the $$|f'(x)| on $$[a,b]$$.Then $$|f(b)-f(a)|. Then by the above the function is uniformly continuous.

This is very useful for sines and cosines. Up to a sign, they are each others' derivatives. The maximum absolute value is $$1$$, so no line with a slope greater than $$1$$ is ever parallel to a tangent line of sine or cosine, their powers, or their products of powers.

So some geometric properties imply uniform continuity.