Difference between continuity and uniform continuity

Question

I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions. For example, my book defines continuity as:

Definition 4.3.1. A function $f:A \to \mathbb R$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-c| < \delta$ (and $x \in A$) it follows that $|f(x)-f(c)| < \epsilon$.

Uniform continuity is defined as:

Definition 4.4.5. A function $f:A \to \mathbb R$ is uniformly continuous on $A$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition? From what appears to me, it just seems like the only difference between Definition 4.3.1 and Definition 4.4.5 is that the letter $c$ was changed to a $y$.

My guess is that the first definition treats $c$ as a fixed point and it is only $x$ that varies, so in this case, $\delta$ can depend on $c$ since $c$ doesn't change. Whereas for the second definition, neither $x$ or $y$ are fixed, rather they can take on values across the whole domain, $A$. In this case, if we set a $\delta$ such that it depended on $y$, then when we pick a different $y$, the same $\delta$ may not work anymore. Is this somewhat a correct interpretation?

Anymore clarifications, examples, would be appreciated.

Answer 1

First of all, continuity is defined at a point $c$, whereas uniform continuity is defined on a set $A$. That makes a big difference. But your interpretation is rather correct: the point $c$ is part of the data, and is kept fixed as, for instance, $f$ itself. Roughly speaking, uniform continuity requires the existence of a single $\delta>0$ that works for the whole set $A$, and not near the single point $c$.

Answer 2

The difference is in the ordering of the quantifiers.

• Continuity:

For all $x$, for all $\varepsilon$, there exist such a $\delta$ that something something.

• Uniform continuity:

For all $\varepsilon$, there exists such a $\delta$ that for all $x$ something something.

For something to be continuous, you can check "one $x$ at a time", so for each $x$, you pick a $\varepsilon$ and then find some $\delta$ that depends on both $x$ and $\varepsilon$ so that $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta$. As you can see if you try it on $f(x)=1/x$ on $(0,1)$, you can find such a $\delta$ for every $x$ and $\varepsilon$. However, if you fix $\varepsilon$, the values for $\delta$ that you need become arbitrarily small as $x$ approaches $0$.

If you want uniform continuity, you need to pick a $\varepsilon$, then find a $\delta$ which is good for ALL the $x$ values you might have. As you see, for $f(x)=1/x$, such a $\delta$ does not exist.

Answer 3

Let me focus on this part of the question:

"I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition?"

This is apparent from the order of the quantifiers. When we write out these two statements into "Prenex normal form", we have that:

$$ \forall c \in A,\forall \epsilon >0, \exists \delta, \forall x \in A \:( |x-c|<\delta \implies |f(x)-f(c)|<\epsilon) $$

$$\forall \epsilon >0, \exists \delta, \forall x,c\in A \: (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon) $$

In the first statement, note that the universal ($\forall$) quantifier $\forall c$ precedes the existential ($\exists$) quantifier $\exists \delta$, and the universal quantifier $\forall x$ follows the existential quantifier $\exists \delta$.

Note that in the second definition, the universal quantifier $\forall c$ now also follows the existential quantifier $\exists \delta$.

To see the significance of the quantifier order, consider the following, where C is the set of cars, P is the set of people, and R is a relation such that cRp means c is owned by p.

$$ \forall c\in C, \exists p \in P: cRp $$ $$ \exists p\in P, \forall c \in C: cRp $$

Observe that in the first statement of the example, the universal quantifier precedes the existential quantifier. This statement means each car $c$ has an owner $p$. Observe that the person p depends on the car. In the second statement, the universal quantifier follows the existential quantifier. This statement means there is some person $p$ who owns EVERY car. Thus this person doesn't depend on the car (since he has all of them, or in other words; given every car, he has it).

To conclude, for any variables $x,y$, $y$ can depend on $x$ if and only if the universal quantifier for $\forall x$ precedes the existential quantifier for $\exists y$.

Applying this theorem to your definitions, we see that in the definition of continuity, the universal quantifiers $\forall x\in\mathbb{R}$ and $\forall\epsilon$ precede $\exists\delta$ . Thus here $\delta$ may depend on both $x,\epsilon$ . However, in the definition of uniform continuity, the only universal quantifier that precedes $\delta$ is $\forall\epsilon$ . Thus delta may only depend on $\epsilon$ and not $x$ .

In the definition of uniform continuity, $\exists \delta $ precedes neither $x$ nor $c$, therefore it can depend on neither of them, but only on $\epsilon$.

Answer 4

The subtle difference between these two definitions became more clear to me when I read their equivalent sequence definitions. First take the definition of a continuous function.

Definition A function $f: D\to\mathbb{R}$ is said to be continuous at the point $x_0$ in $D$ provided that for every sequence $\{x_n\}$ in $D$ that converges to $x_0$, the image sequence $\{f(x_n)\}$ converges to $f(x_0)$.

Now compare this to a uniformly continuous function.

Definition A function $f: D\to\mathbb{R}$ is said to be uniformly continuous provided that for any two sequences $\{y_n\}$ and $\{x_n\}$ in $D$ have the property $$\lim_{n\to\infty}(y_n-x_n)=0,$$ then $$\lim_{n\to\infty}(f(y_n)-f(x_n))=0$$

Notice how the second definition mentions no convergence to a point, but that two sequences are tending toward the same value and at the same rate. These sequences can both be divergent sequences when alone, but their terms can become arbitrarily close to each other.

The classic example is $f:\mathbb{R}\to\mathbb{R}, f(x)=x^2$ is continuous but not uniformly continuous. Take the two sequences $\{y_n\}=\{\sqrt{n^2+1}\}$ and $\{x_n\}=\{n\}$. (Note, both sequences diverge). Take the $\lim_{n\to\infty}{y_n-x_n}$, and solve by multiplying numerator and denominator by its conjugate. $$\lim_{n\to\infty}(\sqrt{n^2+1}-n)=\lim_{n\to\infty}\frac{n^2 +1-n^2 }{\sqrt{n^2+1}+n}=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}+n}=0.$$ Now, looking at $\lim_{n\to\infty}{f(y_n)-f(x_n)}$ we get the following $$\lim_{n\to\infty}{(\sqrt{n^2+1})^2-n^2}=\lim_{n\to\infty}{n^2+1-n^2}=1$$ So this goes against the definition of uniform continuity. We need the difference of function values to also go to $0$, as well, in order for it to be uniformly continuous.