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.

Definition 2.62 (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 $$\RR$$.

Definition 2.63 (Arithmetic operators)

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

We define $$f+g$$ as:

$(f + g)(x) = f(x) + g(x) \Forall x \in \dom f \cap \dom g.$

We define $$f-g$$ as:

$(f - g)(x) = f(x) - g(x) \Forall x \in \dom f \cap \dom g.$

We define $$f g$$ as:

$(f g)(x) = f(x) g(x) \Forall x \in \dom f \cap \dom g.$

We define the quotient $$f/g$$ as:

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

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

$(cf)(x) = cf(x) \Forall x \in \dom f.$

Example 2.10 (Function sum, product, powers)

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

Then, their sum is defined by:

$(f_1 + f_2 + \dots + f_n)(x) = f_1(x) + f_2(x) + \dots + f_n(x) \Forall x \in \bigcap_{i=1}^n \dom f_i$

and their product is defined by:

$(f_1 f_2 \dots f_n)(x) = f_1(x) f_2(x) \dots f_n(x) \Forall x \in \bigcap_{i=1}^n \dom f_i$

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:

$(f^n)(x) = (f(x))^n \Forall x \in \dom f.$

2.6.1. Limits#

Definition 2.64 (Limit)

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

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

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

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

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

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:

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

Assume, $$\epsilon > 0$$ and let $$\delta = \epsilon$$.

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

$|f(x) - 0| = \left | x \sin \frac{1}{x} \right | = |x| \left |\sin \frac{1}{x} \right | \leq | x | < \delta = \epsilon.$

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

$0 < |x - 0| < \delta \implies |f(x) - 0| < \epsilon.$

Thus,

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

Theorem 2.35 (Limit uniqueness)

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

Proof. Assume that the limit exists and assume that:

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

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:

$|f(x) - a_i | < \epsilon \text{ whenever } 0 < | x - x_0 | < \delta_i, \quad i=1,2.$

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

Then:

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

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).

Theorem 2.36 (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.

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

Subtraction of limits:

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

Multiplication of limits:

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

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

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

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:

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

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

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

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

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

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

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

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

We often write simplify the notation as:

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

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.

$\lim_{x \to a} f(x) = p \iff f(a^-) = f(a^+) = p.$

Definition 2.66 (Limits at infinities)

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

$\lim_{x \to \infty} f(x) = p$

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

$|f(x) - p | < \epsilon \text{ whenever } x > b.$

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

$\lim_{x \to -\infty} f(x) = p$

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

$|f(x) - p | < \epsilon \text{ whenever } x < b.$

We sometimes write:

$\lim_{x \to \infty} f(x) = f(\infty) \text{ and } \lim_{x \to -\infty} f(x) = f(-\infty).$

Definition 2.67 (Infinite limits)

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

$\lim_{x \to a^-} f(x) = \infty$

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 $$\infty$$ as $$x$$ approaches $$a$$ from the right, and write:

$\lim_{x \to a^+} f(x) = \infty$

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 $$-\infty$$ as $$x$$ approaches $$a$$ from the left, and write:

$\lim_{x \to a^-} f(x) = -\infty$

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 $$-\infty$$ as $$x$$ approaches $$a$$ from the right, and write:

$\lim_{x \to a^+} f(x) = -\infty$

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 $$\lim_{x \to a} f(a)$$ 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 $$\lim f = \infty$$ 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#

Definition 2.68 (Monotonic function)

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

$f(x_1) \leq f(x_2) \text{ whenever } x_1 < x_2 \Forall x_1, x_2 \in I.$

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

$f(x_1) \geq f(x_2) \text{ whenever } x_1 < x_2 \Forall x_1, x_2 \in I.$

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

$f(x_1) < f(x_2) \text{ whenever } x_1 < x_2 \Forall x_1, x_2 \in I.$

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

$f(x_1) < f(x_2) \text{ whenever } x_1 < x_2 \Forall x_1, x_2 \in I.$

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:

$f(c^-) \leq f(c) \leq f(c^+)$

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.

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

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 $$\epsilon > 0$$, there exists $$\delta > 0$$ such that:

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

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 $$\epsilon > 0$$, there exists $$\delta > 0$$ such that:

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

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 $$\epsilon > 0$$, there exists $$\delta > 0$$ such that:

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

$$f$$ 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 $$\epsilon-\delta$$ 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 $$\dom f$$ 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.

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 $$\epsilon > 0$$, there exists an open interval $$I_t = (c,d)$$ containing $$t$$ such that

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

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

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

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:

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

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:

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

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 $$[a,b]$$.

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

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

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

$f(a) - \epsilon < f(x) < f(a) + \epsilon \implies \mu < f(x) < 2 f(a) - \mu.$

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

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

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

$f(a) - \epsilon < f(x) < f(a) + \epsilon \implies \mu - 2 f(a) < f(x) < \mu.$

Theorem 2.40 (Continuity and arithmetic)

If $$f$$ and $$g$$ are continuous on a set $$S$$, then so are $$f+g$$, $$f-g$$, and $$f g$$. $$\frac{f}{g}$$ 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 $$\lim_{x\to c} f(x) \neq f(c)$$.

Definition 2.71 (Jump discontinuity)

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

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 $$f(c^-) \neq f(c^+)$$. In this case, $$\lim_{x \to c} f(x)$$ 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 $$f(a) \neq f(a^+)$$. 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 $$f(b) \neq f(b^-)$$. 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 $$a \leq x < b$$;

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

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 $$f(a^+) \neq f(a)$$.

• $$f$$ 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.

Definition 2.73 (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:

$\begin{split} g(x) \triangleq \begin{cases} f(x) & x \in \dom f \setminus \{ a \} \\ \lim_{x \to a} f(x) & x = a \end{cases} \end{split}$

is continuous at $$x=a$$.

Definition 2.74 (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.

Theorem 2.41

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:

1. Let $$\epsilon > 0$$ be arbitrary.

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

$|g(t) - g(f(a))| < \epsilon \text{ whenever } | t - f(a)| < \delta_1.$
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

$|g(f(x)) - g(f(a)) | < \epsilon \text{ whenever } | x - a | < \delta.$
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 $$S \subseteq \dom f$$ if there is a real number $$m$$ such that

$f(x) \geq m \Forall x \in S.$

Then, the set

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

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

$f(x) \leq M \Forall x \in S.$

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 $$\epsilon = 1$$. 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 $$\OOO \triangleq \{I_t \Forall t \in [a,b] \}$$ 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:

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

Then, for each $$t_i$$:

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

Therefore,

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

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

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

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

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

Thus, $$f$$ is bounded with a bound:

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

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

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

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

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

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 $$\OOO = \{ I_t \ST a \leq t \leq b \}$$ 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:

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

Define:

$\alpha_1 = \underset{1 \leq i \leq n}{\min}\frac{f(t_i) + \alpha}{2}.$

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

$f(x) > \alpha_1 \Forall x \in [a,b].$

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 $$f(a) \neq f(b)$$. 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:

$A = \{x \in [a,b] \ST f(x) \leq \mu \}.$
1. The set is bounded since $$a \leq x \leq b$$.

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 $$c = \sup A$$.

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 $$\epsilon > 0$$ such that $$f(x) > \mu$$ whenever $$c - \epsilon < x \leq c$$.

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

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

6. This contradicts the assumption that $$c = \sup A$$.

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

1. $$c < b$$.

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

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

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

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

6. 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#

Definition 2.76 (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

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

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

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

2. 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.

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

Definition 2.77 (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

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

Few observations on this definition:

1. $$\delta$$ depends on $$\epsilon$$.

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 $$\dom f \setminus A$$.

Remark 2.16

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:

$|x - y | < \delta \text{ but } |f(x) - f(y)| \geq \epsilon_0.$

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

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

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

The collection $$\OOO = \{ I_t \ST t \in [a,b] \}$$ 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:

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

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 $$\PPP$$ is a cover for $$[a,b]$$.

Now, from triangle inequality:

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

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

$| f(x) - f(t_r) | < \frac{\epsilon}{2}.$

On the other hand:

$|y - t_r | \leq |y - x | + | x - t_r | < \delta + \delta_{t_r} \leq 2 \delta_{t_r}.$

Thus,

$| f(t_r) - f(y)| < \frac{\epsilon}{2}.$

Together, we have:

$| f(x) - f(y) | < \epsilon.$

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 $$[a,b]$$.

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 $$f(c^-) \neq f(c^+)$$, 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 $$R_f = f([a,b]) = \{f(x) \ST a \leq x \leq b \}$$ is a closed interval with endpoints $$f(a)$$ and $$f(b)$$.

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

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

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

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

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 $$f(a) \leq f(a^+)$$ and $$f(b^+) \leq f(b)$$.

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 $$f(c^-) \leq f(c) \leq f(c^+)$$.

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:

$g (f(x)) = x \Forall a \leq x \leq b$

and

$f(g(y)) = y \Forall c \leq y \leq d.$

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 $$f(x_1) \neq f(x_2)$$ 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.