Continuous Functions

Continuity marks a new classification of functions, especially prominent when the theorems explained later on in this page will be put to use. However, if one is reading this wikibook linearly, then it will be good to note that the wikibook will describe functions with even more properties than continuity. As an example, the functions in elementary mathematics, such as polynomials, trigonometric functions, and the exponential and logarithmic functions, contain many levels more properties than that of a continuous function. We will also see several examples of discontinuous functions as well, to provide some remarks of common functions that do not fit the bill.

Continuity at a Point

Definition

The function $f(x)={\tfrac {1}{x}}$ is continuous on the domain $\mathbb {R} \setminus \{0\}$, but is not continuous over the domain $\mathbb {R}$ because it is undefined at $x=0$

A real function, that is a function from real numbers to real numbers, can be represented by a graph in the Cartesian plane; such a function is continuous if, roughly speaking, the graph is a single unbroken curve whose domain is the entire real line. A more mathematically rigorous definition is given below.

A rigorous definition of continuity of real functions is usually given in a first course in calculus in terms of the idea of a limit. First, a function f with variable x is said to be continuous at the point c on the real line, if the limit of $f(x),$ as x approaches that point c, is equal to the value $f(c)$; and second, the function (as a whole) is said to be continuous, if it is continuous at every point. A function is said to be discontinuous (or to have a discontinuity) at some point when it is not continuous there. These points themselves are also addressed as discontinuities.

There are several different definitions of continuity of a function. Sometimes a function is said to be continuous if it is continuous at every point in its domain. In this case, the function $f(x)=\tan(x),$ with the domain of all real $x\neq \pi {\frac {2n+1}{2}},$ $n$ any integer, is continuous. Sometimes an exception is made for boundaries of the domain. For example, the graph of the function $f(x)={\sqrt {x}},$ with the domain of all non-negative reals, has a left-hand endpoint. In this case only the limit from the right is required to equal the value of the function. Under this definition f is continuous at the boundary $x=0$ and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers. In this case, the previous two examples are not continuous, but every polynomial function is continuous, as are the sine, cosine, and exponential functions. Care should be exercised in using the word continuous, so that it is clear from the context which meaning of the word is intended.

Using mathematical notation, there are several ways to define continuous functions in each of the three senses mentioned above.

Let

$f:D\to \mathbb {R} \quad$ be a function defined on a subset $D$ of the set $\mathbb {R}$ of real numbers.

This subset $D$ is the domain of f. Some possible choices include

$D=\mathbb {R} \quad$ ($D$ is the whole set of real numbers), or, for a and b real numbers,

$D=[a,b]=\{x\in \mathbb {R} \mid a\leq x\leq b\}\quad$ ($D$ is a closed interval), or

$D=(a,b)=\{x\in \mathbb {R} \mid a<x<b\}\quad$ ($D$ is an open interval).

In case of the domain $D$ being defined as an open interval, $a$ and $b$ do not belong to $D$, and the values of $f(a)$ and $f(b)$ do not matter for continuity on $D$.

Definition in terms of limits of functions

The function f is continuous at some point c of its domain if the limit of $f(x),$ as x approaches c through the domain of f, exists and is equal to $f(c).$ In mathematical notation, this is written as

$$\lim _{x\to c}{f(x)}=f(c).$$

In detail this means three conditions: first, f has to be defined at c (guaranteed by the requirement that c is in the domain of f). Second, the limit on the left hand side of that equation has to exist. Third, the value of this limit must equal $f(c).$

(Here, we have assumed that the domain of f does not have any isolated points.)

Definition in terms of neighborhoods

A neighborhood of a point c is a set that contains, at least, all points within some fixed distance of c. Intuitively, a function is continuous at a point c if the range of f over the neighborhood of c shrinks to a single point $f(c)$ as the width of the neighborhood around c shrinks to zero. More precisely, a function f is continuous at a point c of its domain if, for any neighborhood $N_{1}(f(c))$ there is a neighborhood $N_{2}(c)$ in its domain such that $f(x)\in N_{1}(f(c))$ whenever $x\in N_{2}(c).$

This definition only requires that the domain and the codomain are topological spaces and is thus the most general definition. It follows from this definition that a function f is automatically continuous at every isolated point of its domain. As a specific example, every real valued function on the set of integers is continuous.

Definition in terms of limits of sequences

The sequence $\exp(1/n)$ converges to exp(0)

One can instead require that for any sequence $(x_{n})_{n\in \mathbb {N} }$ of points in the domain which converges to c, the corresponding sequence $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ converges to $f(c).$ In mathematical notation, $\forall (x_{n})_{n\in \mathbb {N} }\subset D:\lim _{n\to \infty }x_{n}=c\Rightarrow \lim _{n\to \infty }f(x_{n})=f(c)\,.$

Weierstrass and Jordan definitions (epsilon–delta) of continuous functions

Illustration of the $\varepsilon -\delta$-definition: for $\varepsilon =0.5,c=2,$ the value $\delta =0.5$ satisfies the condition of the definition.

Explicitly including the definition of the limit of a function, we obtain a self-contained definition: Given a function $f:D\to R$ as above and an element $x_{0}$ of the domain D, f is said to be continuous at the point $x_{0}$ when the following holds: For any number $\varepsilon >0,$ however small, there exists some number $\delta >0$ such that for all x in the domain of f with $x_{0}-\delta <x<x_{0}+\delta ,$ the value of $f(x)$ satisfies

$$f\left(x_{0}\right)-\varepsilon <f(x)<f(x_{0})+\varepsilon .$$

Alternatively written, continuity of $f:D\to R$ at $x_{0}\in D$ means that for every $\varepsilon >0,$ there exists a $\delta >0$ such that for all $x\in D$:

$$\left|x-x_{0}\right|<\delta {\text{ implies }}|f(x)-f(x_{0})|<\varepsilon .$$

More intuitively, we can say that if we want to get all the $f(x)$ values to stay in some small neighborhood around $f\left(x_{0}\right),$ we simply need to choose a small enough neighborhood for the x values around $x_{0}.$ If we can do that no matter how small the $f(x)$ neighborhood is, then f is continuous at $x_{0}.$

In modern terms, this is generalized by the definition of continuity of a function with respect to a basis for the topology, here the metric topology.

Weierstrass had required that the interval $x_{0}-\delta <x<x_{0}+\delta$ be entirely within the domain D, but Jordan removed that restriction.

Continuity on an Interval

Definition of a Continuous Function on I

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Given an interval $I$ and a function $f:I\to \mathbb {R}$, continuous on I is defined as upholding the following property:

$\forall \varepsilon >0:\exists \delta >0:\forall x\in I:\exists c\in \mathbb {R} :|x-c|<\delta \implies |f(x)-f(c)|<\varepsilon$

and it is notated as $\lim _{x\to c}f(x)=f(c)$

Readers may note the similarity between this definition to the definition of a limit in that unlike the limit, where the function $f$ can converge to any value, continuity restricts the returning value to be only the expected value when the function $f$ is evaluated. This added restriction provides many new theorems, as some of the more important ones will be shown in the following headings.

Operations/Combinations

Since limits are preserved under algebraic operations, let's check whether this is also the case with continuity.

Algebraic

We see that if $f(x)$ and $g(x)$ are both continuous at c, continuity still works out fine for the following situations:

List of Continuous Functions that are Preserved Under Algebraic Operations

Addition	$\lim _{x\rightarrow c}{(f+g)(x)}=f(c)+g(c)$
Subtraction	$\lim _{x\rightarrow c}{(f-g)(x)}=f(c)-g(c)$
Product	$\lim _{x\rightarrow c}{(f\cdot g)(x)}=f(c)\cdot g(c)$
Multiple of a Function	$\lim _{x\rightarrow c}{\lambda \cdot g(x)}=\lambda \cdot g(c)$
Reciprocal	$\lim _{x\rightarrow c}{\left({\dfrac {1}{g}}\right)(x)}={\dfrac {1}{g(c)}}$
Division	$\lim _{x\rightarrow c}{\left({\dfrac {f}{g}}\right)(x)}={\dfrac {f(c)}{g(c)}}$

Note that of course, for any division, g(c) must be a valid number i.e. not 0.

This is actually a corollary when you look at the proofs for the preservation of algebraic operation for limits. Simply replace the limit values L and M with ƒ(c) and g(c) respectively.

We can use sequential limits to prove that functions are discontinuous as follows:

• $f(x)$ is discontinuous at $c$ if and only if there are two sequences $(x_{n})\rightarrow c$ and $(y_{n})\rightarrow c$ such that $\lim _{n\rightarrow \infty }(f(x_{n}))\not =\lim _{n\rightarrow \infty }(f(y_{n}))$.

Composition

Composition is a lot trickier though, as always, but it still works as intuition would suggest; composition of two continuous functions is still a continuous function.

Theorem

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If $f:B\rightarrow \mathbb {R}$ is continuous on the range of $g(x)$ and $g:A\rightarrow B$ is continuous on any interval $A$, then the composition $(f\circ g)(x)=f(g(x))$ is continuous on A.

The proof simply works by fulfilling the definition of continuity for the composition function of $f$ and $g$ using variable substitutions based off fulfilling all requirements for those variables. As such, there is no algebra and no theorems used other than purely definitions.

Proof that Composition works for Continuous Functions at c

We know what is needed for continuity first. As such, we will define epsilon using the most basic definition that will fit the requirement for continuity.	Let $\epsilon >0$
Since f must be continuous, we will also write down what we know is true—its checklist of properties fulfilling continuity. For now, we will make a slight modification to the delta variable for reasons that will be justified later.	$\exists \delta _{1}>0:|x-c|<\delta _{1}\implies |f(x)-f(c)|<\epsilon$.
However, $f$ has more properties than that. The key ones are what the value of $c$ and $x$ refer to. Since the function $f$ is continuous on the range of $g(x)$, that means that the input value of $f$ is actual the output value of $g$. Thus, we can validly substitute the value $c$ and $x$ for the function $f$ with $g(x)$'s output value.	$\exists \delta _{1}>0:|g(x)-g(c)|<\delta _{1}\implies |f\circ g(x)-f\circ g(c)|<\epsilon$.
Since g must be continuous, we will also write down what we know is true, which is the definition of continuity.	$\exists \delta _{2}>0:|x-c|<\delta _{2}\implies |g(x)-g(c)|<\epsilon$.
The expression $|g(x)-g(c)|<\epsilon$ and $|g(x)-g(c)|<\delta _{1}$ is awfully similar. We can use this to our advantage by seeing if we can abuse any properties we know of. Given that the only requirement for $\epsilon$ is that it must be positive and its inequality relationship is valid for any number, and that $\delta _{1}$ is positive and a number, then we can put two and two together and define $\epsilon$ as $\delta _{1}$.	Let $\epsilon =\delta _{1}$
Thus, we abstractly strung together a definition of continuity built off what we know is true; the continuity of $f$ and $g$. Reading the valid results of what this new implication statement suggests; that continuity for the composition of $f$ and $g$, we are assured in our claim that the composition of $f$ and $g$ is also continuous. QED.	Thus $|x-c|<\delta _{2}\implies |g(x)-g(c)|<\delta _{1}\implies |f(g(x))-f(g(c))|<\epsilon$ so $f\circ g(x)$ is continuous on A.
$\square$

The Three Continuity Theorems

Think about what an intuitive notion of continuity is. If you can’t the image of a polynomial function always works. The smooth curve as it travels through the domain of the function is a graphical representation of continuity. However, how do we mathematically know that it’s continuous? Well, we’ll start with the Three Continuity Theorems that will verify this notion.

The Intermediate Value Theorem

This is the big theorem on continuity. Essentially it says that continuous functions have no sudden jumps or breaks.

Theorem

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Let f(x) be a continuous function. If $a<b$ and $f(a)<m<f(b)$, then $\exists c\in (a,b):f(c)=m$.

The intermediate value theorem: Given a continuous function on [a,b] and three variables a < c < b it must be the case that ƒ(a) < ƒ(c) < ƒ(b)

Proof

Let $S=\{x\in (a,b):f(x)<m\}$, and let $c=\sup S$.

Let $\epsilon =|f(c)-m|$. By continuity, $\exists \delta :|x-c|<\delta \implies |f(x)-f(c)|<\epsilon$.

If f(c) < m, then $|f(c+{\frac {\delta }{2}})-f(c)|<\epsilon$, so $f(c+{\frac {\delta }{2}})<f(c)+\epsilon =m$. But then $c+{\frac {\delta }{2}}\in S$, which implies that c is not an upper bound for S, a contradiction.

If f(c) > m, then since $c=\sup S$, $\exists x:x\in S,c>x>c-\delta$. But since $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$, so $f(x)>f(c)-\epsilon$ = m, which implies that $x\notin S$, a contradiction. $\Box$

We will now prove the Minimum-Maximum theorem, which is another significant result that is related to continuity. Essentially, it states that any continuous image of a closed interval is bounded, and also that it attains these bounds.

Minimum-Maximum Theorem

This theorem functions as a first part in another bigger theorem. However, on its own, it helps bridge the gap between supremums and infimums in regards to functions.

Theorem

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Given a continuous function ƒ on [a,b] i.e. $f:[a,b]\to \mathbb {R}$, if $a<b$ and $f(a)<m<f(b)$, then $f([a,b])$ is bounded.

Proof

Assume if possible that $f$ is unbounded.

Let $x_{1}={\tfrac {a+b}{2}}$. Then, $f$ is unbounded on at least one of the closed intervals $[a,x_{1}]$ and $[x_{1},b]$ (for otherwise, $f$ would be bounded on $[a,b]$ contradicting the assumption). Call this interval $I_{1}$.

Similarly, partition $I_{1}$ into two closed intervals and let $I_{2}$ be the one on which $f$ is unbounded.

Thus we have a sequence of nested closed intervals $[a,b]\supseteq I_{1}\supseteq I_{2}\supseteq \ldots$ such that $f$ is unbounded on each of them.

We know that the intersection of a sequence of nested closed intervals is nonempty. Hence, let $x_{0}\in I_{1}\cap I_{2}\cap \ldots$

As $f(x)$ is continuous at $x=x_{0}$, there exists $\delta >0$ such that $x\in V_{\delta }(x_{0})\implies f(x)\in (f(x_{0})-1,f(x_{0})+1)$ But by definition, there always exists $k\in \mathbb {N}$ such that $I_{k}\subseteq V_{\delta }(x_{0})$, contradicting the assumption that $f$ is unbounded over $I_{k}$. Thus, $f$ is bounded over $[a,b]$

Extreme Value Theorem

This is the second part of the theorem. It is the more assertive version of the previous theorem, stating that not only is there a supremum and a infimum, it also is reachable by the function ƒ and will be in between the interval you specified.

Theorem

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Given a continuous function ƒ on [a,b] i.e. $f:[a,b]\to \mathbb {R}$, if $M,m$ are respectively the upper and lower bounds of $f([a,b])$, then there exist $c,d\in [a,b]$ such that $f(c)=M,f(d)=m$.

A depiction of the Extreme Value Theorem: Given a continuous function on [a,b], there must exist a maximum c and d such that ƒ(c) is the greatest value in the interval and ƒ(d) is the littlest.

Proof

Assume if possible, $M=\sup(f([a,b]))$ but $M\notin f([a,b])$.

Consider the function $g(x)={\frac {1}{M-f(x)}}$. By algebraic properties of continuity, $g:[a,b]\to \mathbb {R}$ is continuous. However, $M$ being a cluster point of $f([a,b])$, $g(x)$ is unbounded over $[a,b]$, contradicting (i). Hence, $M\in f([a,b])$. Similarly, we can show that $m\in f([a,b])$.

Appendix

Continuity will come again in other branches in mathematics. You will come across not only different variations of continuity, but you will also come across different definitions of continuity too.

Uniform Continuity

Let $A\subseteq \mathbb {R}$
Let $f:A\to \mathbb {R}$

We say that $f$ is Uniformly Continuous on $A$ if and only if for every $\varepsilon >0$ there exists $\delta >0$ such that if $x,y\in A$ and $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$

Lipschitz continuity

Let $A\subseteq \mathbb {R}$
Let $f:A\to \mathbb {R}$

We say that $f$ is Lipschitz continuous on $A$ if and only if there exists a positive real constant $K$ such that, for all $x,y\in A$, $|f(x)-f(y)|\leq K|x-y|$.

The smallest such $K$ is called the Lipschitz constant of the function $f$.

Topological Continuity

As mentioned, the idea of continuous functions is used in several areas of mathematics, most notably in Topology. A different characterization of continuity is useful in such scenarios.

Theorem

Let $A\subseteq \mathbb {R}$
Let $f:A\to \mathbb {R}$

$f(x)$ is continuous at $x=c$ if and only if for every open neighbourhood $V$ of $f(x)$, there exists an open neighbourhood $U$ of $x$ such that $U\subseteq f^{-1}(V)$

It must be mentioned here that the term "Open Set" can be defined in much more general settings than the set of reals or even metric spaces, and hence the utility of this characterization.

Licensing

Content obtained and/or adapted from:

• Continuity, Wikibooks: Real Analysis under a CC BY-SA license