# Real Functions

## Contents

# 2.6. Real Functions#

In this section, we will deal with functions of type \(f : \RR \to \RR\). 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.

(Real function)

A (partial) function of type \(f: \RR \to \RR\) mapping
real values to real values is called a *real function*.

The domain and range of a real function are both subsets of \(\RR\).

(Arithmetic operators)

Let \(f,g\) be real functions with \(\dom f \cap \dom g \neq \EmptySet\).

We define \(f+g\) as:

We define \(f-g\) as:

We define \(f g\) as:

We define the quotient \(f/g\) as:

For some \(c \in \RR\), we define \(cf\) as:

(Function sum, product, powers)

Let \(f_1, f_2, \dots, f_n\) be real functions.

Then, their sum is defined by:

and their product is defined by:

provided the domain \(D = \bigcap_{i=1}^n \dom f_i\) is nonempty.

If \(f = f_1 = f_2 = \dots = f_n\), then the n-th power of \(f\) is defined as:

## 2.6.1. Limits#

(Limit)

We say that \(f\) approaches the limit \(a\) as \(x\) approaches \(x_0\) and write:

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

\(f\) 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 \(a\).

(Limit outside domain points)

Let

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

Assume, \(\epsilon > 0\) and let \(\delta = \epsilon\).

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

Thus, for every \(\epsilon > 0\), there exists \(\delta > 0\), given by \(\delta = \epsilon\), such that:

Thus,

(Limit uniqueness)

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

Proof. Assume that the limit exists and assume that:

hold true. We will show that \(a_1=a_2\) must be true.

Let \(\epsilon > 0\). Since, the limit exists, hence there are positive numbers \(\delta_1\) and \(\delta_2\) such that:

Choose \(\delta = \min(\delta_1, \delta_2)\).

Then:

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

(Arithmetic of limits)

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

Then, we have the following rules.

Addition of limits:

Subtraction of limits:

Multiplication of limits:

Division of limits. If \(b \neq 0\), then:

(One sided limits)

We that that \(f\) approaches the *left hand limit* \(p\) as \(x\) approaches \(a\)
from the left and write:

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

We that that \(f\) approaches the *right hand limit* \(p\) as \(x\) approaches \(a\)
from the right and write:

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

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

We often write simplify the notation as:

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.

(Limits at infinities)

We say that \(f\) approaches the limit \(p\) as \(x\) approaches \(\infty\) and write:

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

We say that \(f\) approaches the limit \(p\) as \(x\) approaches \(-\infty\) and write:

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

We sometimes write:

(Infinite limits)

We say that \(f\) approaches \(\infty\) as \(x\) approaches \(a\) from the left, and write:

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

We say that \(f\) approaches \(\infty\) as \(x\) approaches \(a\) from the right, and write:

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

We say that \(f\) approaches \(-\infty\) as \(x\) approaches \(a\) from the left, and write:

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

We say that \(f\) approaches \(-\infty\) as \(x\) approaches \(a\) from the right, and write:

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

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

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 \(\lim f = \infty\) and \(\lim g = \infty\), then \(\lim (f - g) = \lim f - \lim g\) doesn’t make sense as \(\infty - \infty\) is indeterminate.

Division rule for limits remains valid if \(\lim f / \lim g\) is not indeterminate and \(\lim g \neq 0\).

## 2.6.2. Monotonicity#

(Monotonic function)

A function \(f\) is *increasing* on an interval \(I\) if

A function \(f\) is *decreasing* on an interval \(I\) if

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

A function \(f\) is *strictly decreasing* on an interval \(I\) if

If \(f\) is *strictly increasing* or
*strictly decreasing* on \(I\), then we say that
\(f\) is *strictly monotonic* on \(I\).

(Monotonicity and one sided limits)

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

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

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

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

\[ f(c^-) \leq f(c) \leq f(c^+) \]and, if \(f\) is decreasing, then:

\[ f(c^-) \geq f(c) \geq f(c^+). \]

## 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.

(Continuity)

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 \(\lim_{x \to c} f(x) = f(c)\).We say that \(f\) is

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

*continuous from the right*at \(x=c\) if \(f\) is defined on an interval \([c,b)\) and \(f(c^+) = f(c)\).

(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 \(\epsilon > 0\), there exists \(\delta > 0\) such that:

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 \(\epsilon > 0\), there exists \(\delta > 0\) such that:

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 \(\epsilon > 0\), there exists \(\delta > 0\) such that:

\(f\) is continuous at \(x=c\) if and only if

This theorem is a restatement of continuity definition in the form of \(\epsilon-\delta\) rules.

(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 \(\dom f\) 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 \(S \subseteq \dom f\), we mean that \(S\) is a union of finitely or infinitely many disjoint intervals.

\([a,b]\))

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

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

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:

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:

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

The set \(\OOO \triangleq \{I_t \Forall t \in [a,b] \}\) forms an
*open cover* of \([a,b]\).

Let \(f\) be continuous at \(x=a\)

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

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

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

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

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

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

(Continuity and arithmetic)

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

## 2.6.4. Discontinuities#

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

(Jump discontinuity)

Let \(f : \RR \to \RR\) be a real function and let \([a,b] \subseteq \dom f\).

\(f\) 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 \(f(c^-) \neq f(c^+)\). In this case, \(\lim_{x \to c} f(x)\) doesn’t exist and \(f\) is not continuous at \(x=c\).\(f\) has a

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

*jump discontinuity*at \(x=b\) if the left hand limit \(f(b^-)\) exists but \(f(b) \neq f(b^-)\). In this case, \(f\) is not continuous from the left at \(x=b\).

(Piecewise continuity)

A function \(f\) is *piecewise-continuous* on \([a,b]\) if

\(f(x^+)\) exists for all \(a \leq x < b\);

\(f(x^-)\) exists for all \(a < x \leq b\);

\(f(x^+) = f(x^-) = f(x)\) 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 \(f(a^+) \neq f(a)\).

\(f\) has a jump discontinuity at \(b\) if \(f(b^-) \neq f(b)\).

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.

(Removable discontinuity)

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

is continuous at \(x=a\).

(Essential discontinuity)

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

If \(x \in \dom f\) 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.

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

Proof. We proceed as follows:

Let \(\epsilon > 0\) be arbitrary.

Since \(g\) is continuous at \(f(a)\), there is \(\delta_1 > 0\) such that

\[ |g(t) - g(f(a))| < \epsilon \text{ whenever } | t - f(a)| < \delta_1. \]Since \(f\) is continuous at \(a\), hence, there exists \(\delta > 0\) such that

\[ |f(x) - f(a)| < \delta_1 \text{ whenever } | x - a | < \delta. \]Together, they imply that

\[ |g(f(x)) - g(f(a)) | < \epsilon \text{ whenever } | x - a | < \delta. \]Therefore, \(g \circ f\) is continuous at \(x=a\).

## 2.6.6. Boundedness#

(Bounded function)

A real function \(f\) is *bounded below* on a set
\(S \subseteq \dom f\) if
there is a real number \(m\) such that

Then, the set

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

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
\(S \subseteq \dom f\) if
there is a real number \(M\) such that

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

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 \(S\).

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 \(\epsilon = 1\). Then, due to Proposition 2.22, there exists an interval \(I_t\) such that

The set \(\OOO \triangleq \{I_t \Forall t \in [a,b] \}\) forms an
*open cover* of \([a,b]\).

Due to Heine-Borel theorem, \([a,b]\) has a finite subcover contained in \(\OOO\).

Let \(t_1 < t_2 < \dots < t_n\) index the finite open subcover. We have:

Then, for each \(t_i\):

Therefore,

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

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

Thus, \(f\) is bounded with a bound:

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

Since \(f\) is bounded, hence \(V\) is bounded, hence \(V\) has an infimum as well as a supremum.

Let \(f\) be continuous on a finite closed interval \([a,b]\). Let

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:

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,

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

The set \(\OOO = \{ I_t \ST a \leq t \leq b \}\) is an open cover of \([a,b]\).

Due to Heine-Borel theorem, \([a,b]\) has a finite subcover contained in \(\OOO\). Thus, there are finitely many points \(t_1, t_2, \dots, t_n\) such that:

Define:

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, \(f(x) > \frac{f(t_i) + \alpha}{2} \geq \alpha_1\). Thus,

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 \(x_2 \in [a,b]\).

(Intermediate value theorem)

Let \(f\) be continuous on a finite closed interval \([a,b]\). Assume that \(f(a) \neq f(b)\). 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:

The set is bounded since \(a \leq x \leq b\).

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.

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

Let \(c = \sup A\).

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

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

\(c > a\).

\(f\) is continuous at \(x=c\).

Thus, there exists an \(\epsilon > 0\) such that \(f(x) > \mu\) whenever \(c - \epsilon < x \leq c\).

Therefore, the interval \((c - \epsilon, c]\) is not included in \(A\).

Therefore, \(c-\epsilon\) is an upper bound for \(A\).

This contradicts the assumption that \(c = \sup A\).

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

\(c < b\).

\(f\) is continuous at \(x=c\).

Thus, there exists an \(\epsilon > 0\) such that \(f(x) < \mu\) whenever \(c \leq x < c + \epsilon\).

Therefore, the interval \([c, c + \epsilon)\) is included in \(A\).

Therefore, \(c\) is not an upper bound for \(A\).

This also contradicts the assumption that \(c = \sup A\).

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#

(Continuity over a set)

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

The clause \(x \in \dom f\) in the definition is important.

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

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

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

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 \(\epsilon\) as well as the point \(x_0\). This is not always desirable. Uniform continuity addresses this concern.

(Uniform continuity)

Let \(f: \RR \to \RR\) be a real function.
Let \(A \subseteq \dom f\).
We say that \(f\) is *uniformly continuous*
over the set \(A\) if for every \(\epsilon > 0\),
there exists \(\delta > 0\) (depending on \(\epsilon\)) such that

Few observations on this definition:

\(\delta\) depends on \(\epsilon\).

\(\delta\) is independent of the choice of \(x\) and \(y\).

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

The definition is restricted to points in \(A\). It doesn’t consider points in \(\dom f \setminus A\).

If \(f\) is not uniformly continuous on a set \(A\), then there is an \(\epsilon_0 > 0\) such that for any \(\delta > 0\), there are points \(x,y\) in \(A\) such that:

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

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

Proof. Let \(\epsilon > 0\) be arbitrary. Since \(f\) is continuous on \([a,b]\), for every \(t \in [a,b]\), there exists \(\delta_t > 0\) such that

Choose an open interval \(I_t = (t - \delta_t, t + \delta_t)\) for every \(t \in [a,b]\).

The collection \(\OOO = \{ I_t \ST t \in [a,b] \}\) 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 \(\PPP = \{ I_{t_1}, I_{t_2}, \dots, I_{t_n} \}\) form a finite open cover of \([a,b]\).

Define:

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 \(\PPP\) is a cover for \([a,b]\).

Now, from triangle inequality:

Since \(x \in I_{t_r}\), hence \(|x - t_r | < \delta_{t_r}\), hence:

On the other hand:

Thus,

Together, we have:

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

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#

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 \(c \in (a,b)\).

If \(f(c^-) = f(c^+)\), then by monotonicity, \(f(c^-) = f(c) = f(c^+)\), thus \(f\) is continuous at \(c\).

If \(f(c^-) \neq f(c^+)\), then we have a jump discontinuity.

Next, the boundary points. Consider \(x=a\).

The right hand limit \(f(a^+)\) exists due to monotonicity.

If \(f(a) = f(a^+)\), then \(f\) is continuous from the right at \(x=a\).

Otherwise, we have a jump discontinuity at \(x=a\).

Similarly, for \(x=b\):

The left hand limit \(f(b^-)\) exists due to monotonicity.

If \(f(b) = f(b^-)\), then \(f\) is continuous from the left at \(x=b\).

Otherwise, we have a jump discontinuity at \(x=b\).

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

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

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 \(f(a) \neq f(b)\) 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

Due to Theorem 2.38:

Thus,

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.

Since \(f\) is increasing, hence \(f(a) \leq f(a^+)\) and \(f(b^+) \leq f(b)\).

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

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

Thus, we both \(f(a) = f(a+)\) and \(f(b) = f(b^-)\) must be true.

\(f\) is continuous from the right at \(a\) and from the left at \(b\).

Also, since \(f\) is increasing, hence for any \(c \in (a,b)\), we have \(f(c^-) \leq f(c) \leq f(c^+)\).

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.

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:

and

Moreover, \(g\) is continuous and strictly increasing on \([c,d]\).

Proof. We first show that such a function \(g\) can be defined:

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

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

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

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

Hence, \(f : [a,b] \to [c,d]\) is injective.

Thus, \(f : [a,b] \to [c,d]\) is bijective.

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.

Let \(y_1, y_2 \in [c,d]\) such that \(y_1 < y_2\).

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

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

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

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

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.