2.6. Real Functions

In this section, we will deal with functions of type $f:\mathbb{R}\to \mathbb{R}$. Main references are [2, 76].

Our goal is a cursory review of the relevant theory. We will state the main definitions and results and provide a few examples wherever needed. Detailed proofs will be skipped.

Definition 2.62 (Real function)

A (partial) function of type $f:\mathbb{R}\to \mathbb{R}$ mapping real values to real values is called a real function.

The domain and range of a real function are both subsets of $\mathbb{R}$.

Definition 2.63 (Arithmetic operators)

Let $f,g$ be real functions with $\mathrm{dom}f\cap \mathrm{dom}g\ne \varnothing$.

We define $f+g$ as:

$$(f+g)(x)=f(x)+g(x)\mathrm{\forall }x\in \mathrm{dom}f\cap \mathrm{dom}g.$$

We define $f-g$ as:

$$(f-g)(x)=f(x)-g(x)\mathrm{\forall }x\in \mathrm{dom}f\cap \mathrm{dom}g.$$

We define $fg$ as:

$$(fg)(x)=f(x)g(x)\mathrm{\forall }x\in \mathrm{dom}f\cap \mathrm{dom}g.$$

We define the quotient $f/g$ as:

$$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\mathrm{\forall }x\in \mathrm{dom}f\cap \mathrm{dom}g\text{ such that }g(x)\ne 0.$$

For some $c\in \mathbb{R}$, we define $cf$ as:

$$(cf)(x)=cf(x)\mathrm{\forall }x\in \mathrm{dom}f.$$

Example 2.10 (Function sum, product, powers)

Let $f_1,f_2,\dots ,f_n$ be real functions.

Then, their sum is defined by:

$$(f_1+f_2+\cdots +f_n)(x)=f_1(x)+f_2(x)+\cdots +f_n(x)\mathrm{\forall }x\in \bigcap _{i=1}^n\mathrm{dom}{f}_i$$

and their product is defined by:

$$(f_1{f}_2\dots f_n)(x)=f_1(x)f_2(x)\dots f_n(x)\mathrm{\forall }x\in \bigcap _{i=1}^n\mathrm{dom}{f}_i$$

provided the domain $D=\bigcap _{i=1}^n\mathrm{dom}{f}_i$ is nonempty.

If $f=f_1=f_2=\cdots =f_n$, then the n-th power of $f$ is defined as:

$$(f^n)(x)=(f(x){)}^n\mathrm{\forall }x\in \mathrm{dom}f.$$

2.6.1. Limits

Definition 2.64 (Limit)

We say that $f$ approaches the limit $a$ as $x$ approaches $x_0$ and write:

$$\underset{x\to x_0}{lim}f(x)=a$$

if $f$ is defined on some deleted neighborhood of $x_0$ and for every $ϵ>0$, there is $\delta >0$ such that:

$$|f(x)-a|<ϵ\text{ whenever }0<|x-x_0|<\delta .$$

• $f$ may or may not be defined at $x=x_0$. But, it must be defined in the neighborhood around $x_0$.

• If $f$ is defined at $x=x_0$, then $f(x_0)$ doesn’t need to be equal to $a$.

Example 2.11 (Limit outside domain points)

Let

$$f(x)=x\sin \frac{1}{x},x\ne 0.$$

We have $\mathrm{dom}f=\mathbb{R}\setminus \{0\}=(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$. We can compute the limit of $f$ at $x=0$. We will show that:

$$\underset{x\to 0}{lim}f(x)=0.$$

Assume, $ϵ>0$ and let $\delta =ϵ$.

Then, for any $x$ satisfying $0<|x-0|<\delta$, we have

$$|f(x)-0|=|x\sin \frac{1}{x}|=|x||\sin \frac{1}{x}|\le |x|<\delta =ϵ.$$

Thus, for every $ϵ>0$, there exists $\delta >0$, given by $\delta =ϵ$, such that:

$$0<|x-0|<\delta ⟹|f(x)-0|<ϵ.$$

Thus,

$$\underset{x\to 0}{lim}f(x)=0.$$

Theorem 2.35 (Limit uniqueness)

If $\underset{x\to x_0}{lim}f(x)$ exists, then it is unique.

Proof. Assume that the limit exists and assume that:

$$\underset{x\to x_0}{lim}f(x)=a_1\text{ and }\underset{x\to x_0}{lim}f(x)=a_2$$

hold true. We will show that $a_1=a_2$ must be true.

Let $ϵ>0$. Since, the limit exists, hence there are positive numbers ${\delta }_1$ and ${\delta }_2$ such that:

$$|f(x)-a_i|<ϵ\text{ whenever }0<|x-x_0|<{\delta }_i,i=1,2.$$

Choose $\delta =min({\delta }_1,{\delta }_2)$.

Then:

$$|a_1-a_2|\le |f(x)-a_1|+|f(x)-a_2|\le 2ϵ\text{ whenever }0<|x-x_0|<\delta .$$

Thus, $|a_1-a_2|<2ϵ$. Since this is valid for every $ϵ>0$, hence $a_1=a_2$ must be true (Proposition 2.5).

Theorem 2.36 (Arithmetic of limits)

Let $\underset{x\to x_0}{lim}f(x)=a$ and $\underset{x\to x_0}{lim}g(x)=b$.

Then, we have the following rules.

Addition of limits:

$$\underset{x\to x_0}{lim}(f+g)(x)=a+b.$$

Subtraction of limits:

$$\underset{x\to x_0}{lim}(f-g)(x)=a-b.$$

Multiplication of limits:

$$\underset{x\to x_0}{lim}(fg)(x)=ab.$$

Division of limits. If $b\ne 0$, then:

$$\underset{x\to x_0}{lim}\left(\frac{f}{g}\right)(x)=\frac{a}{b}.$$

Definition 2.65 (One sided limits)

We that that $f$ approaches the left hand limit $p$ as $x$ approaches $a$ from the left and write:

$$\underset{x\to a^{-}}{lim}f(x)=p$$

if $f$ is defined on some open interval $(a-b,a)$ (with $b>0$) and for each $ϵ>0$, there is a $\delta >0$ such that

$$|f(x)-p|<ϵ\text{ whenever }a-\delta <x<a.$$

We that that $f$ approaches the right hand limit $p$ as $x$ approaches $a$ from the right and write:

$$\underset{x\to a^{+}}{lim}f(x)=p$$

if $f$ is defined on some open interval $(a,a+b)$ (with $b>0$) and for each $ϵ>0$, there is a $\delta >0$ such that

$$|f(x)-p|<ϵ\text{ whenever }a<x<a+\delta .$$

The left and right hand limits are called one sided limits.

We often write simplify the notation as:

$$\underset{x\to a^{-}}{lim}f(x)=f(a^{-})\text{ and }\underset{x\to a^{+}}{lim}f(x)=f(a^{+}).$$

Theorem 2.37

A function $f$ has a limit at $x=a$ if and only if it has left and right hand limits at $x=a$ and they are equal.

$$\underset{x\to a}{lim}f(x)=p⟺f(a^{-})=f(a^{+})=p.$$

Definition 2.66 (Limits at infinities)

We say that $f$ approaches the limit $p$ as $x$ approaches $\mathrm{\infty }$ and write:

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=p$$

if $f$ is defined on an interval $(a,\mathrm{\infty })$ and for each $ϵ>0$, there exists a number $b$ such that

$$|f(x)-p|<ϵ\text{ whenever }x>b.$$

We say that $f$ approaches the limit $p$ as $x$ approaches $-\mathrm{\infty }$ and write:

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=p$$

if $f$ is defined on an interval $(-\mathrm{\infty },a)$ and for each $ϵ>0$, there exists a number $b$ such that

$$|f(x)-p|<ϵ\text{ whenever }x<b.$$

We sometimes write:

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=f(\mathrm{\infty })\text{ and }\underset{x\to -\mathrm{\infty }}{lim}f(x)=f(-\mathrm{\infty }).$$

Definition 2.67 (Infinite limits)

We say that $f$ approaches $\mathrm{\infty }$ as $x$ approaches $a$ from the left, and write:

$$\underset{x\to a^{-}}{lim}f(x)=\mathrm{\infty }$$

if $f$ is defined on an interval $(a-b,a)$ (with $b>0$), and, for each real number $M$, there exists $\delta >0$ such that

$$f(x)>M\text{ whenever }a-\delta <x<a.$$

We say that $f$ approaches $\mathrm{\infty }$ as $x$ approaches $a$ from the right, and write:

$$\underset{x\to a^{+}}{lim}f(x)=\mathrm{\infty }$$

if $f$ is defined on an interval $(a,a+b)$ (with $b>0$), and, for each real number $M$, there exists $\delta >0$ such that

$$f(x)>M\text{ whenever }a<x<a+\delta .$$

We say that $f$ approaches $-\mathrm{\infty }$ as $x$ approaches $a$ from the left, and write:

$$\underset{x\to a^{-}}{lim}f(x)=-\mathrm{\infty }$$

if $f$ is defined on an interval $(a-b,a)$ (with $b>0$), and, for each real number $M$, there exists $\delta >0$ such that

$$f(x)<M\text{ whenever }a-\delta <x<a.$$

We say that $f$ approaches $-\mathrm{\infty }$ as $x$ approaches $a$ from the right, and write:

$$\underset{x\to a^{+}}{lim}f(x)=-\mathrm{\infty }$$

if $f$ is defined on an interval $(a,a+b)$ (with $b>0$), and, for each real number $M$, there exists $\delta >0$ such that

$$f(x)<M\text{ whenever }a<x<a+\delta .$$

If left hand and right hand limits are equal, we say that the limit at $x=a$ given by $\underset{x\to a}{lim}f(a)$ exists and is equal to $f(a^{+})$ or $f(a^{-})$.

Remark 2.15

Theorem 2.35 can be extended for the following too:

• If a left hand limit exists, it is unique.

• If a right hand limit exists, it is unique.

• If a limit at infinity exists, it is unique.

• If the limit value is infinite, it is unique.

Theorem 2.36 remains valid for the following too:

• Left hand limits

• Right hand limits

• Limits at infinity

Addition, subtraction and multiplication rules remain valid if either or both limits are infinite, provided that the R.H.S. is not indeterminate. E.g., if $limf=\mathrm{\infty }$ and $limg=\mathrm{\infty }$, then $lim(f-g)=limf-limg$ doesn’t make sense as $\mathrm{\infty }-\mathrm{\infty }$ is indeterminate.

Division rule for limits remains valid if $limf/limg$ is not indeterminate and $limg\ne 0$.

2.6.2. Monotonicity

Definition 2.68 (Monotonic function)

A function $f$ is increasing on an interval $I$ if

$$f(x_1)\le f(x_2)\text{ whenever }{x}_1<x_2\mathrm{\forall }{x}_1,x_2\in I.$$

A function $f$ is decreasing on an interval $I$ if

$$f(x_1)\ge f(x_2)\text{ whenever }{x}_1<x_2\mathrm{\forall }{x}_1,x_2\in I.$$

If $f$ is increasing or decreasing on $I$, then we say that $f$ is monotonic on $I$.

A function $f$ is strictly increasing on an interval $I$ if

$$f(x_1)<f(x_2)\text{ whenever }{x}_1<x_2\mathrm{\forall }{x}_1,x_2\in I.$$

A function $f$ is strictly decreasing on an interval $I$ if

$$f(x_1)<f(x_2)\text{ whenever }{x}_1<x_2\mathrm{\forall }{x}_1,x_2\in I.$$

If $f$ is strictly increasing or strictly decreasing on $I$, then we say that $f$ is strictly monotonic on $I$.

Theorem 2.38 (Monotonicity and one sided limits)

Let $f$ be monotonic of an interval $(a,b)$. Let

$$\alpha =\underset{a<x<b}{inf}f(x)\text{ and }\beta =\underset{a<x<b}{sup}f(x).$$

1. If $f$ is increasing, then $f(a^{+})=\alpha$ and $f(b^{-})=\beta$.

2. If $f$ is decreasing, then $f(a^{+})=\beta$ and $f(b^{-})=\alpha$.

3. If $a<c<b$, then $f(c^{+})$ and $f(c^{-})$ exist and are finite. Moreover, if $f$ is increasing, then:

   $$f(c^{-})\le f(c)\le f(c^{+})$$

   and, if $f$ is decreasing, then:

   $$f(c^{-})\ge f(c)\ge f(c^{+}).$$

2.6.3. Continuity

If the limit of a function at a point matches its value at that point, then the function is continuous at that point.

Definition 2.69 (Continuity)

1. We say that $f$ is continuous at $x=c$ if $f$ is defined on an interval $(a,b)$ containing $c$ (i.e. $a<c<b$) and $\underset{x\to c}{lim}f(x)=f(c)$.

2. We say that $f$ is continuous from the left at $x=c$ if $f$ is defined on an interval $(a,c]$ and $f(c^{-})=f(c)$.

3. We say that $f$ is continuous from the right at $x=c$ if $f$ is defined on an interval $[c,b)$ and $f(c^{+})=f(c)$.

Theorem 2.39 (Characterization of continuity)

A function $f$ is continuous at $x=c$ if and only if $f$ is defined on an interval $(a,b)$ containing $c$ and for each $ϵ>0$, there exists $\delta >0$ such that:

$$|f(x)-f(c)|<ϵ\text{ whenever }|x-c|<\delta .$$

A function $f$ is continuous from the left at $x=c$ if and only if $f$ is defined on an interval $(a,c]$ and for each $ϵ>0$, there exists $\delta >0$ such that:

$$|f(x)-f(c)|<ϵ\text{ whenever }c-\delta <x\le c.$$

A function $f$ is continuous from the right at $x=c$ if and only if $f$ is defined on an interval $[c,b)$ and for each $ϵ>0$, there exists $\delta >0$ such that:

$$|f(x)-f(c)|<ϵ\text{ whenever }c\le x<c+\delta .$$

$f$ is continuous at $x=c$ if and only if

$$f(c^{-})=f(c)=f(c^{+}).$$

This theorem is a restatement of continuity definition in the form of $ϵ-\delta$ rules.

Definition 2.70 (Continuity on an interval)

A function $f$ is continuous on an open interval $(a,b)$ if it is continuous at every point in $(a,b)$.

• If $f(b^{-})=f(b)$ holds too (i.e. $f$ is continuous from the left at $b$), then $f$ is continuous on $(a,b]$.

• If $f(a^{+})=f(a)$ holds too (i.e. $f$ is continuous from the right at $a$), then $f$ is continuous on $[a,b)$.

• If both $f(a^{+})=f(a)$ and $f(b^{-})=f(b)$ hold true, then $f$ is continuous on the closed interval $[a,b]$.

If $S$ is a subset of $\mathrm{dom}f$ consisting of finitely or infinitely many disjoint intervals, then $f$ is continuous on $S$ if $S$ is continuous on every interval in $S$.

When we say that $f$ is continuous on some set $S\subseteq \mathrm{dom}f$, we mean that $S$ is a union of finitely or infinitely many disjoint intervals.

Proposition 2.22 (Continuity on a closed interval $[a,b]$)

Let $f$ be continuous on $[a,b]$. Then, for every $t\in [a,b]$, and for every $ϵ>0$, there exists an open interval $I_t=(c,d)$ containing $t$ such that

$$|f(x)-f(t)|<ϵ\text{ whenever }x\in I_t\cap [a,b].$$

Proof. Assume $a<t<b$. Then, there exists $\delta >0$, such that:

$$|f(x)-f(t)|<ϵ\text{ whenever }t-\delta <x<t+\delta .$$

Choose $I_t=(t-\delta ,t+\delta )$.

Now, assume $t=a$. $f$ must be continuous from the right at $t=a$. There exists $\delta >0$ such that:

$$|f(x)-f(t)|<ϵ\text{ whenever }t\le x<t+\delta .$$

Choose $I_t=(t-\delta ,t+\delta )$. Then $I_t\cap (a,b)=[t,t+\delta )$.

Finally, assume $t=b$. $f$ must be continuous from the left at $t=b$. There exists $\delta >0$ such that:

$$|f(x)-f(t)|<ϵ\text{ whenever }t-\delta <x\le t.$$

Choose $I_t=(t-\delta ,t+\delta )$. Then $I_t\cap (a,b)=(t-\delta ,t]$.

The set $\mathcal{O}\triangleq \{I_t\mathrm{\forall }t\in [a,b]\}$ forms an open cover of $[a,b]$.

Proposition 2.23

Let $f$ be continuous at $x=a$

1. If $f(a)>\mu$, then $f(x)>\mu$ in some neighborhood of $a$.

2. If $f(a)<\mu$, then $f(x)<\mu$ in some neighborhood of $a$.

Proof. (1) Let $ϵ=f(a)-\mu$. $ϵ>0$ as $f(a)>\mu$. There exists $\delta >0$ such that

$$|f(x)-f(a)|<ϵ\text{ whenever }|x-a|<\delta .$$

Thus, in the neighborhood $x\in (a-\delta ,a+\delta )$:

$$f(a)-ϵ<f(x)<f(a)+ϵ⟹\mu <f(x)<2f(a)-\mu .$$

(2) Let $ϵ=\mu -f(a)$. $ϵ>0$ as $f(a)<\mu$. There exists $\delta >0$ such that

$$|f(x)-f(a)|<ϵ\text{ whenever }|x-a|<\delta .$$

Thus, in the neighborhood $x\in (a-\delta ,a+\delta )$:

$$f(a)-ϵ<f(x)<f(a)+ϵ⟹\mu -2f(a)<f(x)<\mu .$$

Theorem 2.40 (Continuity and arithmetic)

If $f$ and $g$ are continuous on a set $S$, then so are $f+g$, $f-g$, and $fg$. $\frac{f}{g}$ is continuous at each $x\in S$ such that $g(x)\ne 0$.

2.6.4. Discontinuities

A function $f$ is discontinuous at some $x=c$ in its domain $\mathrm{dom}f$ if either $\underset{x\to c}{lim}f(x)$ doesn’t exist or $\underset{x\to c}{lim}f(x)\ne f(c)$.

Definition 2.71 (Jump discontinuity)

Let $f:\mathbb{R}\to \mathbb{R}$ be a real function and let $[a,b]\subseteq \mathrm{dom}f$.

1. $f$ has a jump discontinuity at a point $c\in (a,b)$ if both the left hand limit $f(c^{-})$ and the right hand limit $f(c^{+})$ exist but $f(c^{-})\ne f(c^{+})$. In this case, $\underset{x\to c}{lim}f(x)$ doesn’t exist and $f$ is not continuous at $x=c$.

2. $f$ has a jump discontinuity at $x=a$ if the right hand limit $f(a^{+})$ exists but $f(a)\ne f(a^{+})$. In this case, $f$ is not continuous from the right at $x=a$.

3. $f$ has a jump discontinuity at $x=b$ if the left hand limit $f(b^{-})$ exists but $f(b)\ne f(b^{-})$. In this case, $f$ is not continuous from the left at $x=b$.

Definition 2.72 (Piecewise continuity)

A function $f$ is piecewise-continuous on $[a,b]$ if

1. $f(x^{+})$ exists for all $a\le x<b$;

2. $f(x^{-})$ exists for all $a<x\le b$;

3. $f(x^{+})=f(x^{-})=f(x)$ for all but finitely many points in $(a,b)$.

Jump discontinuities:

• If (3) fails to hold at some $x=c$ in $(a,b)$, $f$ has a jump discontinuity at $x=c$.

• $f$ has a jump discontinuity at $a$ if $f(a^{+})\ne f(a)$.

• $f$ has a jump discontinuity at $b$ if $f(b^{-})\ne f(b)$.

In other words:

• Left hand limits exist everywhere (except at $x=b$).

• Right hand limits exist everywhere (except at $x=a$).

• There are only a finite number of points where these two limits don’t match.

• $f$ is not continuous at those finite number of points.

• $f$ is continuous everywhere else in the open interval $(a,b)$.

• $f$ may not be continuous at the boundaries $x=a$ and $x=b$.

• At a jump discontinuity, $f$ may be continuous from the right or continuous from the left or neither.

Definition 2.73 (Removable discontinuity)

Let $f$ be discontinuous at some $x=a$. If $\underset{x\to a}{lim}f(x)$ exists, then we say that the discontinuity at $x=a$ is a removable discontinuity. Moreover, the function:

$$\begin{array}{r}g(x)\triangleq \{\begin{array}{ll}f(x)& x\in \mathrm{dom}f\setminus \{a\}\\ \underset{x\to a}{lim}f(x)& x=a\end{array}\end{array}$$

is continuous at $x=a$.

Definition 2.74 (Essential discontinuity)

Let $x$ be an interior point of $\mathrm{dom}f$. If either of the one sided limits $f(x^{+})$, or $f(x^{-})$ don’t exist, then $f$ has an essential discontinuity at $x$.

If $x\in \mathrm{dom}f$ is a boundary point, then only a one sided limit is possible (left or right hand limit). If such a limit doesn’t exist, then $f$ has an essential discontinuity at the boundary point $x$.

2.6.5. Continuity with Function Composition

Next, we look at continuity w.r.t. function composition.

Theorem 2.41

Suppose $f$ is continuous at $x=a$; $f(a)$ is an interior point of $\mathrm{dom}g$ and $g$ is continuous at $f(a)$. Then $g\circ f$ is continuous at $x=a$.

Proof. We proceed as follows:

1. Let $ϵ>0$ be arbitrary.

2. Since $g$ is continuous at $f(a)$, there is ${\delta }_1>0$ such that

   $$|g(t)-g(f(a))|<ϵ\text{ whenever }|t-f(a)|<{\delta }_1.$$

3. Since $f$ is continuous at $a$, hence, there exists $\delta >0$ such that

   $$|f(x)-f(a)|<{\delta }_1\text{ whenever }|x-a|<\delta .$$

4. Together, they imply that

   $$|g(f(x))-g(f(a))|<ϵ\text{ whenever }|x-a|<\delta .$$

5. Therefore, $g\circ f$ is continuous at $x=a$.

2.6.6. Boundedness

Definition 2.75 (Bounded function)

A real function $f$ is bounded below on a set $S\subseteq \mathrm{dom}f$ if there is a real number $m$ such that

$$f(x)\ge m\mathrm{\forall }x\in S.$$

Then, the set

$$V=\{f(x)|x\in S\}$$

has a infimum (due to Corollary 2.2), and we write:

$$\alpha =\underset{x\in S}{inf}f(x).$$

If there is a point $c\in S$ such that $\alpha =f(c)$, we say that $\alpha$ is the minimum of $f$ on $S$ and $f$ attains the minimum at $x=c$.

A real function $f$ is bounded above on a set $S\subseteq \mathrm{dom}f$ if there is a real number $M$ such that

$$f(x)\le M\mathrm{\forall }x\in S.$$

Then, the set $V$ has a supremum (due to Axiom 2.1), and we write:

$$\beta =\underset{x\in S}{sup}f(x).$$

If there is a point $d\in S$ such that $\beta =f(d)$, we say that $\beta$ is the maximum of $f$ on $S$ and $f$ attains the maximum at $x=d$.

If $f$ is bounded above and below on a set $S$, we say that $f$ is bounded on $S$.

Theorem 2.42

If $f$ is continuous on a finite closed interval $[a,b]$, then $f$ is bounded on $[a,b]$.

Proof. Let $f$ be continuous on $[a,b]$. Let $t\in [a,b]$. Choose $ϵ=1$. Then, due to Proposition 2.22, there exists an interval $I_t$ such that

$$|f(x)-f(t)|<1\text{ whenever }x\in I_t\cap [a,b].$$

The set $\mathcal{O}\triangleq \{I_t\mathrm{\forall }t\in [a,b]\}$ forms an open cover of $[a,b]$.

Due to Heine-Borel theorem, $[a,b]$ has a finite subcover contained in $\mathcal{O}$.

Let $t_1<t_2<\cdots <t_n$ index the finite open subcover. We have:

$$[a,b]\subset \bigcup _{i=1}^n(I_{t_i}\cap [a,b]).$$

Then, for each $t_i$:

$$|f(x)-f(t_i)|<1\text{ whenever }x\in I_{t_i}\cap [a,b].$$

Therefore,

$$|f(x)|=|f(x)-f(t_i)+f(t_i)|\le |f(x)-f(t_i)|+|f(t_i)|\le 1+|f(t_i)|\text{ whenever }x\in I_{t_i}\cap [a,b].$$

Thus, for every $x\in [a,b]$, there exists a $t_i\in [a,b]$ such that,

$$|f(x)|\le 1+|f(t_i)|.$$

Taking the maximum on the R.H.S. over all the inequalities, we get:

$$|f(x)\le 1+\underset{1\le i\le n}{max}|f(t_i)|.$$

Thus, $f$ is bounded with a bound:

$$M=1+\underset{1\le i\le n}{max}|f(t_i)|.$$

Corollary 2.8

If $f$ is continuous on a finite closed interval $[a,b]$, then $f$ has an infimum and a supremum.

Proof. By Theorem 2.42, $f$ is bounded. Let

$$V=\{f(x)|x\in [a,b]\}.$$

Since $f$ is bounded, hence $V$ is bounded, hence $V$ has an infimum as well as a supremum.

Theorem 2.43

Let $f$ be continuous on a finite closed interval $[a,b]$. Let

$$\alpha =\underset{a\le x\le b}{inf}f(x)\text{ and }\beta =\underset{a\le x\le b}{sup}f(x).$$

Then, $\alpha$ and $\beta$ are respectively, the minimum and maximum values of $f$ on $[a,b]$ and $f$ attains these values at some points in $[a,b]$.

I.e., there exists $x_1,x_2\in [a,b]$ such that:

$$f(x_1)=\alpha \text{ and }f(x_2)=\beta .$$

Proof. Assume that $\alpha$ is the infimum of $f$ over $[a,b]$ and there is no $x_1\in [a,b]$ such that $f(x_1)=\alpha$. Then $f(x)>\alpha$ for all $x\in [a,b]$.

Let $t\in [a,b]$. Then, $f(t)>\alpha$. Thus,

$$f(t)>\frac{f(t)+\alpha }{2}>\alpha .$$

By Proposition 2.22 and Proposition 2.23, there is a open interval $I_t$ at $t$ such that

$$f(x)>\frac{f(t)+\alpha }{2}\text{ whenever }x\in I_t\cap [a,b].$$

The set $\mathcal{O}=\{I_t|a\le t\le b\}$ is an open cover of $[a,b]$.

Due to Heine-Borel theorem, $[a,b]$ has a finite subcover contained in $\mathcal{O}$. Thus, there are finitely many points $t_1,t_2,\dots ,t_n$ such that:

$$[a,b]\subset \bigcup _{i=1}^n(I_{t_i}\cap [a,b]).$$

Define:

$${\alpha }_1=\underset{1\le i\le n}{min}\frac{f(t_i)+\alpha }{2}.$$

Due to the finite cover; for every $x\in [a,b]$, there exists $t_i$ such that $x\in I_{t_i}\cap [a,b]$ and thus, $f(x)>\frac{f(t_i)+\alpha }{2}\ge {\alpha }_1$. Thus,

$$f(x)>{\alpha }_1\mathrm{\forall }x\in [a,b].$$

But ${\alpha }_1>\alpha$. Thus, $\alpha$ cannot be the infimum of $f$ over $[a,b]$. We arrive at the contradiction.

Thus, there exists some $x=x_1$ such that $f(x_1)=\alpha$.

A similar argument shows that $f$ attains $\beta$ at some $x_2\in [a,b]$.

Theorem 2.44 (Intermediate value theorem)

Let $f$ be continuous on a finite closed interval $[a,b]$. Assume that $f(a)\ne f(b)$. Let $\mu$ be in between $f(a)$ and $f(b)$. Then $f$ attains the value of $\mu$ at some $x=c\in (a,b)$.

In other words, there exists $c\in (a,b)$ such that $f(c)=\mu$.

Proof. Let us assume that $f(a)<\mu <f(b)$.

Define the set:

$$A=\{x\in [a,b]|f(x)\le \mu \}.$$

1. The set is bounded since $a\le x\le b$.

2. The set is nonempty since $f(a)<\mu$. And since $f$ is continuous from the right at $a$, hence there exists an interval $[a,a+\delta )$ such that $f(x)<\mu$ in this interval.

3. Thus, $A$ has an infimum(obviously $a$) and a supremum.

4. Let $c=supA$.

5. We will claim that $f(c)=\mu$.

If $f(c)>\mu$, then:

1. $c>a$.

2. $f$ is continuous at $x=c$.

3. Thus, there exists an $ϵ>0$ such that $f(x)>\mu$ whenever $c-ϵ<x\le c$.

4. Therefore, the interval $(c-ϵ,c]$ is not included in $A$.

5. Therefore, $c-ϵ$ is an upper bound for $A$.

6. This contradicts the assumption that $c=supA$.

If $f(c)<\mu$, then:

1. $c<b$.

2. $f$ is continuous at $x=c$.

3. Thus, there exists an $ϵ>0$ such that $f(x)<\mu$ whenever $c\le x<c+ϵ$.

4. Therefore, the interval $[c,c+ϵ)$ is included in $A$.

5. Therefore, $c$ is not an upper bound for $A$.

6. This also contradicts the assumption that $c=supA$.

Therefore $f(c)=\mu$ must be true.

A similar argument can be pursued when $f(b)<\mu <f(a)$.

Note that the proof picks up just one possible value of $x$ such that $f(x)=\mu$. It is quite possible that $f$ attains $\mu$ more than one times in the interval $[a,b]$. The proof doesn’t claim to find all such values of $x$.

In this sense, this theorem is a weak result as it claims the existence of just one point at which $f(x)=\mu$. It doesn’t claim to characterize the set of points at which $f(x)=\mu$.

2.6.7. Uniform Continuity

Definition 2.76 (Continuity over a set)

Let $A\subseteq \mathrm{dom}f$. We say that $f$ is continuous over the set $A$ if for each $ϵ>0$ and each $x_0\in A$, there exists $\delta >0$ (depending on $x_0$ and $ϵ$) such that

$$|f(x)-f(x_0)|<ϵ\text{ whenever }|x-x_0|<\delta \text{ and }x\in \mathrm{dom}f.$$

The clause $x\in \mathrm{dom}f$ in the definition is important.

1. If $x_0$ is an interior point of $\mathrm{dom}f$, we can pick an interval $(x_0-\delta ,x_0+\delta )$.

2. If $x_0$ is a non-interior point of $\mathrm{dom}f$, we can pick up a half-open interval $(x_0-\delta ,x_0]$ or $[x_0,x_0+\delta )$ whichever is applicable.

3. If $x_0$ is an isolated point, we pick the degenerate interval $[x_0,x_0]$ with suitable choice of $\delta$.

4. Thus, on the non-interior points, $f$ is either continuous from the left or right while on the interior points, $f$ is continuous.

The key issue here is that the size of the interval (decided by $\delta$) may depend on both $ϵ$ as well as the point $x_0$. This is not always desirable. Uniform continuity addresses this concern.

Definition 2.77 (Uniform continuity)

Let $f:\mathbb{R}\to \mathbb{R}$ be a real function. Let $A\subseteq \mathrm{dom}f$. We say that $f$ is uniformly continuous over the set $A$ if for every $ϵ>0$, there exists $\delta >0$ (depending on $ϵ$) such that

$$|f(x)-f(y)|<ϵ\text{ whenever }|x-y|<\delta \text{ and }x,y\in A.$$

Few observations on this definition:

1. $\delta$ depends on $ϵ$.

2. $\delta$ is independent of the choice of $x$ and $y$.

3. $\delta$ might depend on the set $A$. E.g., if $A$ is a bounded set, it may depend on its size.

4. The definition is restricted to points in $A$. It doesn’t consider points in $\mathrm{dom}f\setminus A$.

Remark 2.16

If $f$ is not uniformly continuous on a set $A$, then there is an ${ϵ}_0>0$ such that for any $\delta >0$, there are points $x,y$ in $A$ such that:

$$|x-y|<\delta \text{ but }|f(x)-f(y)|\ge {ϵ}_0.$$

While a continuous function may not be uniformly continuous in general, it is so on a compact subset.

Theorem 2.45

If $f$ is continuous on a closed and bounded (compact) interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.

Proof. Let $ϵ>0$ be arbitrary. Since $f$ is continuous on $[a,b]$, for every $t\in [a,b]$, there exists ${\delta }_t>0$ such that

$$|f(x)-f(t)|<\frac{ϵ}{2}\text{ whenever }|x-t|<2{\delta }_t\text{ and }x\in [a,b].$$

Choose an open interval $I_t=(t-{\delta }_t,t+{\delta }_t)$ for every $t\in [a,b]$.

The collection $\mathcal{O}=\{I_t|t\in [a,b]\}$ is an open cover for $[a,b]$. Since $[a,b]$ is closed and bounded (compact), hence due to Heine-Borel theorem, there are finitely many points $t_1,t_2,\dots ,t_n$ in $[a,b]$ such that $\mathcal{P}=\{I_{t_1},I_{t_2},\dots ,I_{t_n}\}$ form a finite open cover of $[a,b]$.

Define:

$$\delta =min\{{\delta }_{t_1},{\delta }_{t_2},\dots ,{\delta }_{t_n}\}.$$

Assume that $|x-y|<\delta$ and $x,y\in [a,b]$. Assume that $x\in I_{t_r}$ for some $r$. This is true since $\mathcal{P}$ is a cover for $[a,b]$.

Now, from triangle inequality:

$$|f(x)-f(y)|=|f(x)-f(t_r)+f(t_r)-f(y)|\le |f(x)-f(t_r)|+|f(t_r)-f(y)|.$$

Since $x\in I_{t_r}$, hence $|x-t_r|<{\delta }_{t_r}$, hence:

$$|f(x)-f(t_r)|<\frac{ϵ}{2}.$$

On the other hand:

$$|y-t_r|\le |y-x|+|x-t_r|<\delta +{\delta }_{t_r}\le 2{\delta }_{t_r}.$$

Thus,

$$|f(t_r)-f(y)|<\frac{ϵ}{2}.$$

Together, we have:

$$|f(x)-f(y)|<ϵ.$$

We have shown that for any $ϵ>0$, a $\delta >0$ can be chosen such that, $|x-y|<\delta$ with $x,y\in [a,b]$ implies $|f(x)-f(y)|<ϵ$. Thus, $f$ is uniformly continuous over $[a,b]$.

Corollary 2.9

If $f$ is continuous on a set $A$, then $f$ is uniformly continuous on any finite closed interval contained in $A$.

2.6.8. Continuity and Monotonic Functions

Proposition 2.24

If $f$ is monotonic on an interval $I$, then $f$ is either continuous or has a jump discontinuity at each $x\in I$.

Proof. Let $I$ be an open interval $(a,b)$. Then, due to Theorem 2.38, both left hand and right hand limits exist at each point in $(a,b)$.

Consider some $c\in (a,b)$.

1. If $f(c^{-})=f(c^{+})$, then by monotonicity, $f(c^{-})=f(c)=f(c^{+})$, thus $f$ is continuous at $c$.

2. If $f(c^{-})\ne f(c^{+})$, then we have a jump discontinuity.

Next, the boundary points. Consider $x=a$.

1. The right hand limit $f(a^{+})$ exists due to monotonicity.

2. If $f(a)=f(a^{+})$, then $f$ is continuous from the right at $x=a$.

3. Otherwise, we have a jump discontinuity at $x=a$.

Similarly, for $x=b$:

1. The left hand limit $f(b^{-})$ exists due to monotonicity.

2. If $f(b)=f(b^{-})$, then $f$ is continuous from the left at $x=b$.

3. Otherwise, we have a jump discontinuity at $x=b$.

Theorem 2.46

If $f$ is monotonic on $[a,b]$, then $f$ is continuous on $[a,b]$ if and only if its range $R_f=f([a,b])=\{f(x)|a\le x\le b\}$ is a closed interval with endpoints $f(a)$ and $f(b)$.

In other words, $f$ is continuous on $[a,b]$ if and only if:

$$f([a,b])=[f(a),f(b)]\text{ if }f(a)\le f(b)\text{ else }[f(b),f(a)].$$

Proof. If $f$ is constant over $[a,b]$, there is nothing to prove. Hence, we shall restrict our attention to the case where $f$ is non-constant. Then $f(a)\ne f(b)$ since $f$ is monotonic. Without loss of generality, assume that $f(a)<f(b)$ ($f$ is increasing). If $f(a)>f(b)$, we can replace $f$ by $-f$ and proceed.

Consider the set

$$S_f=\{f((a,b))\}=\{f(x)|a<x<b\}.$$

Due to Theorem 2.38:

$$S_f\subseteq [f(a^{+}),f(b^{-})].$$

Thus,

$$R_f=\{f(a)\}\cup S_f\cup \{f(b)\}\subseteq \{f(a)\}\cup [f(a^{+}),f(b^{-})]\cup \{f(b)\}.$$

If $f$ is continuous on $[a,b]$, then $f(a)=f(a^{+})$ and $f(b)=f(b^{-})$. Thus, we have: $R_f\subseteq [f(a),f(b)]$.

Further, due to intermediate value theorem, for every $f(a)<\mu <f(b)$, there exists $x\in (a,b)$, such that $\mu =f(x)$. Thus, $R_f=[f(a),f(b)]$.

We now assume that $R_f=[f(a),f(b)]$ and show that $f$ must be continuous.

1. Since $f$ is increasing, hence $f(a)\le f(a^{+})$ and $f(b^{+})\le f(b)$.

2. We also have: $[f(a),f(b)]\subseteq \{f(a)\}\cup [f(a^{+}),f(b^{-})]\cup \{f(b)$.

3. If $f(a)<f(a^{+})$ were true, then the subset relationship above will be invalid. Similar case with $f(b^{-})<f(b)$.

4. Thus, we both $f(a)=f(a+)$ and $f(b)=f(b^{-})$ must be true.

5. $f$ is continuous from the right at $a$ and from the left at $b$.

6. Also, since $f$ is increasing, hence for any $c\in (a,b)$, we have $f(c^{-})\le f(c)\le f(c^{+})$.

7. If $f(c^{-})<f(c)$, then $(f(c^{-}),f(c))$ cannot be part of $R_f$. Thus, $f(c^{-})=f(c)$

For strictly monotonic functions which are continuous on an interval $[a,b]$, it is possible to find an inverse function on the same interval.

Theorem 2.47

Let $f$ be a strictly increasing and continuous function on an interval $[a,b]$. Let $f(a)=c$ and $f(b)=d$. Then, there is a unique function $g$ defined on $[c,d]$ such that:

$$g(f(x))=x\mathrm{\forall }a\le x\le b$$

and

$$f(g(y))=y\mathrm{\forall }c\le y\le d.$$

Moreover, $g$ is continuous and strictly increasing on $[c,d]$.

Proof. We first show that such a function $g$ can be defined:

1. Since $f$ is (strictly) monotone, hence due to Theorem 2.46, $f([a,b])=[c,d]$.

2. In other words, $f$ restricted to $[a,b]$ as $f:[a,b]\to [c,d]$ is total and surjective.

3. Thus, for each $y\in [c,d]$, there exists $x\in [a,b]$ such that $y=f(x)$.

4. Since $f$ is strictly increasing, hence $f(x_1)\ne f(x_2)$ for any $x_1,x_2\in [a,b]$.

5. Hence, $f:[a,b]\to [c,d]$ is injective.

6. Thus, $f:[a,b]\to [c,d]$ is bijective.

7. Thus, we can introduce an inverse function $g:[c,d]\to [a,b]$ with the rule $g(y)=x$ whenever $f(x)=y$.

Next, we show that $g$ is strictly increasing.

1. Let $y_1,y_2\in [c,d]$ such that $y_1<y_2$.

2. Let $x_1,x_2\in [a,b]$ such that $y_1=f(x_1)$ and $y_2=f(x_2)$.

3. Thus, $x_1=g(y_1)$ and $x_2=g(y_2)$.

4. Since $f$ is strictly increasing, hence $f(x_1)<f(x_2)$ implies $x_1<x_2$.

5. Thus, $y_1<y_2$ implies $g(y_1)=x_1<g(y_2)=x_2$.

6. Thus, $g$ is strictly increasing.

Finally, notice that $g$ is monotonic and the range $g([c,d])=[a,b]$ is a closed interval with the end points $a=g(c)$ and $b=g(d)$. Thus, due to Theorem 2.46, $g$ is continuous.