# Is a function that is continuous and doesn't blow up always uniformly continuous?

A function $$f: D \rightarrow \mathbb{R}$$ is called uniformly continuous if and only if $$\forall \varepsilon>0 \exists \delta>0 \forall x, x_{0} \in D:\left|x-x_{0}\right|<\delta \Rightarrow\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$$.

I am having difficulty understanding what uniform continuity really means.

I think a continuous linear function is always uniformly continuous, right? But how can a function that is not linear be uniformly continuous?

I also looked at the Wikipedia article on uniform continuity, where they have an animation of what uniform continuity means visually. I think this animation tells us that whenever a function blows up to infinity, then it is not uniformly continuous - no matter how small we choose $$\delta$$ for some $$x_0$$ there is always an $$x$$ s.t. $$|x_0-x|<\delta$$ and $$f(x)=\infty$$ and so $$|f(x)-f(x_0)|>\varepsilon$$, right? Is it also true that a function is uniformly continuous if it is continuous and doesn't blow up?

My question may not be very specific, but in a sense I am just asking what uniform continuity really means.

• Continuity means that you can wiggle the input around some fixes $x$ a bit and then the output also just wiggles a bit. Uniform continuity means that you can quantify how much you can wiggle in the domain uniformly for all $x$ in the domain and still wiggle just a bit in the output. So you can control the rate of change over all the domain. Feb 7, 2021 at 10:56
• Of possible intuitive interest are the following pointwise notions: Continuity at $x=a$ means that the actual distance between $f(x)$ and $f(a)$ (i.e. distance using $y$-axis scale of units) approaches zero as $x$ approaches $a.$ Uniform continuity at $x=a$ means that for each "blow-up function" $h(t)$ (a strictly increasing function from some closed right neighborhood of $0$ to $[0,\infty)$ such that $h(0)=0),$ the $h$-blow-up distance distance between $f(x)$ and $f(a)$ (continued) Feb 7, 2021 at 11:19
• (i.e. take the function $h$ of the actual distance between $f(x)$ and $f(a))$ approaches zero as $x$ approaches $a.$ For example, $f(x) = \sqrt{x}$ is uniformly continuous at $x=0$ --- use $h(t) = \sqrt{t}.$ Note that the square root function "blows up" small positive numbers, and it does so by increasing relative amounts as the numbers get smaller and smaller, which can be seen by looking at the size of $\sqrt{0.1}$ as compared to $0.1,$ and the size of $\sqrt{0.01}$ as compared to $0.01,$ and so on. (continued) Feb 7, 2021 at 11:26
• It can be shown that if a function is not uniformly continuous on an interval (usual sense that everyone else here is using), then the interval contains at least one point at which the function is not uniformly continuous. Lipschitz continuous at $x=a$ means that we only need to blow-up by a linear function in the sense that the function can be shown to be uniformly continuous at $x=a$ by using a function of the form $h(t) = Kt$ for some positive real number $K.$ Equivalently, there exists $K$ such that $|f(x) - f(a)|$ isn't larger than $K|x-a|$ for all values of $x$ (continued) Feb 7, 2021 at 11:34
• sufficiently close to $a.$ It also means that the distances between $f(x)$ and $f(a)$ are at most a constant times the distances between $x$ and $a$ as $x$ approaches $a.$ Differentiable at $x=a$ means that the (signed) distances between $f(x)$ and $f(a)$ actually approach a constant times the distances between $x$ and $a$ as $x$ approaches $a.$ Thus, Lipschitz continuous at $x=a$ means the signed distances stay within the interval $[-K|x-a|,\;K|x-a|]$ as $x$ approaches $a,$ as opposed to the signed distances, when scaled up by $h(t) = Kt,$ actually approach one of $K$ or $-K.$ Feb 7, 2021 at 11:42

I'll try to describe it without the confusing math notation. For both continuity and uniform continuity, the game starts by choosing some usually small $$\varepsilon>0$$.

Continuity of $$f$$ means that at any given point $$x$$, if we get within a small enough $$\delta$$-neighborhood of $$x$$ (an open set containing all points whose distance to $$x$$ is less than $$\delta$$), then $$f$$ only takes values within an $$\varepsilon$$-neighborhood of $$f(x)$$. Essentially, we can force the function arbitrarily close to a particular value $$f(x)$$ by going close enough to $$x$$. What "close enough" means may depend on $$x$$.

Uniform continuity means that if we choose an arbitrary amount of values for $$x$$ under the condition that none of them are farther away from each other than a suitably small $$\delta$$, then their corresponding outputs under $$f$$ are no further away from each other than $$\varepsilon$$. Essentially, all intervals of length $$\delta$$ are mapped to within intervals of length $$\varepsilon$$, no matter where those $$\delta$$-intervals are. Or put another way, if $$x$$ varies by less than a suitably small $$\delta$$, then $$f$$ varies by less than $$\varepsilon$$.

Now to counterexamples to your hypothesis: bounded continuous functions are uniformly continuous. This is not true. Consider for instance $$f:(0,\infty)\to\mathbb R,~f(x)=\cos\left(\frac1x\right)$$. This is clearly bounded and continuous, but looking at a graph (use desmos or wolframalpha or whatever else you prefer), we can see that as we're getting closer to $$0$$, the distances between peaks and valleys of the cosine get arbitrarily small. But this also means that no matter how small we choose $$\delta$$, there are a peak and a valley which are closer to each other than $$\delta$$, so within the interval containing both, $$f$$ will never vary by less than $$2$$ (height of the peak to depth of the valley).

The other way around is also not true: uniformly continuous functions need not be bounded. Consider $$f:\mathbb R\to\mathbb R,~f(x)=x$$ as a simple counterexample.

However, what is true, and is also a pretty damn important thing to keep in mind, is that if $$f$$ is defined on a compact set and is continuous, then it is also uniformly continuous. This result is used in proving the fundamental theorem of calculus, for instance. But also more advanced stuff like Cauchy's generalized integral formula in complex analysis make use of this result.

Not sure if this is what you are looking for, but the essence in the difference of continuity and uniform continuity is the change in the order of the quantifiers:

• Continuity: for each $$\varepsilon>0$$ and for each $$x\in D$$ there exists $$\delta>0$$ so that... Note that $$\delta$$ depends on $$\varepsilon$$ and $$x$$.

• Uniform continuity: for each $$\varepsilon>0$$, there exists $$\delta>0$$ so that for each $$x\in D$$... Note that $$\delta$$ depends only on $$\varepsilon$$ here.

What this tells us is that, when a function is continuous, when we want to go $$\varepsilon$$-near the value $$f(x_0)$$ then we can find a quantity $$\delta>0$$ so that when we are $$\delta$$-near $$x$$, after applying $$f$$ we get values that are $$\varepsilon$$-near $$f(x_0)$$.

Imagine that you want to be at the same time $$\varepsilon$$-near $$f(x_n)$$ for an infinite set of points $$\{x_n\}\subset D$$. Can you find a distance $$\delta$$ so that, for all $$n\geq1$$, if $$x$$ is $$\delta$$-near some $$x_n$$ then $$f(x)$$ will be $$\varepsilon$$-near $$f(x_n)$$? Well, since $$f$$ is continuous, we find a $$\delta_n$$ depending on $$x_n$$ so that when $$x$$ is $$\delta_n$$-near $$x_n$$ then sure, $$f(x)$$ is $$\varepsilon$$-near $$f(x_n)$$. So a reasonable person would consider taking $$\delta:=\inf_{n}\delta_n$$. The problem is that this could be zero! In other words, what if the distance that you have to lie in so that $$f$$ maps you $$\varepsilon$$-near $$f(x_n)$$ gets arbitrarily small as $$n$$ becomes large?

On the other hand, uniform continuity does not get in such trouble: you can find a $$\delta>0$$ so that when you are $$\delta$$-near the set $$\{x_n\}$$ then your values through $$f$$ are $$\varepsilon$$-near the set $$\{f(x_n)\}$$.

Also, it depends on what you mean "blow up". For example, if a function is $$C^1$$ and its derivative is bounded (so it does not really "blow up": the rate under which its values change cannot get very large), then it is uniformly continuous. Can you see why? (hint: use the mean value theorem). On the other hand, there do exist bounded, continuous but not uniformly continuous functions: see for example Prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$.

I hope this helps.