Home page for accesible maths 4 Continuity vs. discontinuity 4 Continuity vs. discontinuity 4.2 Uniform continuity

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

4.1 Continuous functions

Definition 4.1.1 (Plain English Definition of Continuity)

Let $f:\mathbb{R}\to\mathbb{R}$ be a function and $x\in\mathbb{R}$ . We say that $f$ is continuous at $x$ if $f(x)$ is close to $f(y)$ provided that $x$ is (really) close to $y$ .

Definition 4.1.2 (The so-called Sequential Definition of Continuity)

Let $f:\mathbb{R}\to\mathbb{R}$ be a function and $x\in\mathbb{R}$ . We say that $f$ is continuous at $x$ if for any sequence $\{x_{n}\}^{\infty}_{n=1}$ that tends to $x$ , $\lim_{n\to\infty}f(x_{n})=f(x)\,.$ In this case we say that $x$ is a point of continuity of the function $f$ . If $f$ is not continuous at a certain point $x$ we say that $x$ is a point of discontinuity of the function $f$ .

Remark 4.1.1

The definition above has some advantages and disadvantages as well. Theoretically, if want to prove that $x$ is a point of continuity of $f$ then we must prove something for ALL the sequences converging to $x$ . This can be harsh. On the other hand, if we want to prove that $x$ is a point of discontinuity then it is enough to find one single test sequence $\{x_{n}\}^{\infty}_{n=1}$ tending to $x$ to prove the claim.

Definition 4.1.3

Let $f:\mathbb{R}\to\mathbb{R}$ be a function that is continuous at any real number. Then we call the function $f$ continuous. If $f:[a,b]\to R$ is a function that is continuous for any $a\leq x\leq b$ , we call $f$ a continuous function on the interval $[a,b]$ .

Example 4.1.1

The function $f(x)=x$ (a.k.a the identity function) and the function $g(x)=C$ (a.k.a the constant function, where the constant is any real number $C$ ) are continuous at any real number $x$ . This immediately follows from the definition above.

Remark 4.1.2 (Our favourite functions)

The $\sin$ , $\cos$ and $\exp$ are continuous at any real number. In Calculus, you even took the derivatives of this function, you know that they are continuous. The problem is that $\sin$ is defined in a geometrical way, you do not have a rigorous analytical definition of, say, $\sin{1/2}$ or $e^{\pi}$ . However, by understanding convergence and continuity you will actually understand a completely rigorous definition of $\sin{1/2}$ that does not use geometry, circles, angles, whatever (but it is still the $\sin{1/2}$ you can compute with a calculator). Just for fun: $\sin{1/2}$ will be defined as the limit of the following convergent sequence:

x_{1}=1/2,x_{2}=1/2-\frac{(1/2)^{3}}{6},x_{3}=1/2-\frac{(1/2)^{3}}{6}+\frac{(1% /2)^{5}}{120},

x_{4}=1/2-\frac{(1/2)^{3}}{6}+\frac{(1/2)^{5}}{120}-\frac{(1/2)^{7}}{5040}% \dots....

Proposition 4.1.1

1.

Let $f, g$ be functions that are continuous at $x$ . Then both $f+g$ and $f g$ are continuous at $x$ . [Sum and Product Rules for Continuous Functions]
2.

Let $h$ be a positive continuous function that is continuous at $x$ . Then $\frac{1}{h}$ is also continuous at $x$ . [Quotient Rules for Continuous Function]
3.

Let $f:\mathbb{R}\to\mathbb{R}$ be continuous at $y$ and $g:\mathbb{R}\to\mathbb{R}$ be continuous at $x$ , such that $g(x)=y$ . Then the composition function $f\circ g$ is continuous at $x$ [Composition Rule]

Proof: The first statement follows from the Product Rule/Sum Rule immediately. If $x_{n}\to x$ and $f(x_{n})\to f(x),g(x_{n})\to g(x)$ , then $(f+g)(x_{n})\to f(x)+g(x)$ and $(fg)(x_{n})\to f(x)g(x)$ . Also, if $y_{n}\to y$ and $h(y_{n})\to h(y)$ for some positive function, then by the Quotient Rule, $\frac{1}{h(y_{n})}\to\frac{1}{h(y)}$ . Now, let us prove the third statement. Suppose that $x_{n}\to x$ . We need to prove that $(f\circ g)(x_{n})\to(f\circ g)(x)$ . First by the continuity of $g$ , $g(x_{n})\to g(x)=y$ . Now we use the continuity of $f$ at $g(x)=y$ . Since $\{g(x_{n})\}^{\infty}_{n=1}$ tends to $g(x)$ , $(f\circ g)(x_{n})=f(g(x_{n}))\to f(g(x))=(f\circ g)(x)\,.$ $\square$ .

Question 4.1.1

Show that a polynomial $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}$ is continuous at any real number.

Solution: By the Product Rule for Continuous Functions, $x^{n}$ is continuous at any real number. Hence $a_{n}x^{n}$ is continuous at any real number, so by the Sum Rule for Continuous Functions our polynomial is also continuous at any given real number. Also, by the Composition Rule the function $exp(x^{2})$ and $\frac{sin(cos(x))}{exp(x)}$ is continuous at any given number.

We have another definition for continuity.

Definition 4.1.4 (The infamous $(\varepsilon,\delta)$ -definition)

The function $f:\mathbb{R}\to\mathbb{R}$ is continuous at $x$ if for any $\varepsilon>0$ , there exists some $\delta>0$ such that if $|y-x|<\delta$ , then $|f(y)-f(x)|<\varepsilon$ .

Note that this is again a mathematized version of the fact that if we are getting closer and closer to $x$ , then $f(y)$ is getting closer and closer to $f(x)$ . If $y$ is $\delta$ -close to $x$ , then $f(y)$ is $\varepsilon$ -close to $f(x)$ . So, it is not surprising (but awfully important) that:

Theorem 4.1.1

The Sequential Definition and the $(\varepsilon,\delta)$ -definition are equivalent.

Proof: How can we prove something like that? We need to prove that if $f$ is continuous at $x$ according to the Sequential Definition, then $f$ is continuous at $x$ according to the $(\varepsilon,\delta)$ -definition and vice versa. We use our favourite:

$A$ follows from $B$ if and only if not- $B$ follows from not- $A$

trick. Negate the statement that “ $f$ is continuous at $x$ according to the $(\varepsilon,\delta)$ -definition”. The negated statement is that there exists some $\varepsilon>0$ for which there is no $\delta$ such that if $|y-x|<\delta$ , then $|f(y)-f(x)|<\varepsilon$ . So, in particular there exists $y_{n}$ such that $|y_{n}-x|<\frac{1}{n}$ but $|f(y_{n})-f(x)|\geq\varepsilon$ . This means that

•

$\{y_{n}\}^{\infty}_{n=1}$ converges to $x$ .
•

$\{f(y_{n})\}^{\infty}_{n=1}$ does not converge to $f(x)$ .

So, $f$ is not convergent at $x$ according to the Sequential Definition. What did we prove??? We proved that if $f$ is continuous at $x$ according to the Sequential Definition, then $f$ is continuous at $x$ according to the $(\varepsilon,\delta)$ -definition. Here comes the “vice versa” part. Negate the statement that “ for all convergent sequences $x_{n}\to x$ , $f(x_{n})\to f(x)$ ”. It is not very hard: there exists a sequence $x_{n}\to x$ such that $\{f(x_{n})\}^{\infty}_{n=1}$ does not converge to $f(x)$ . What does it mean that the sequence $\{f(x_{n})\}^{\infty}_{n=1}$ is not converging to $f(x)$ . Recall Question 2.2.1. It means that there exists a subsequence $\{f(x_{n_{k}})\}^{\infty}_{k=1}$ so that for some $\varepsilon>0$ , $|f(x_{n_{k}})-f(x)|\geq\varepsilon$ . This will be our “bad” $\varepsilon$ . No matter how small $\delta$ is, there will be some element $x_{n_{k}}$ in the subsequence such that $|x_{n_{k}}-x|$ . For this element $x_{n_{k}}$ , $|f(x_{n_{k}})-f(x)|\geq\varepsilon$ . Hence $f$ is not continuous at $x$ according to the $(\varepsilon,\delta)$ -definition. $\square$

Question 4.1.2

Let $f$ be a continuous function on the real numbers. Suppose that $f(q)=0$ if $q$ is rational. Show that $f(x)=0$ for all $x\in\mathbb{R}$ .

Solution: Let $x$ be a real number and $\{x_{n}\}^{\infty}_{n=1}$ be a sequence of rational numbers tending to $x$ . Then by the continuity of $f$ , $0=\lim_{n\to\infty},f(x_{n})=f(x)\,.$ $\square$