This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Boundedness Theorem

Recall from the Functions Bounded on a Set page that a function $f$ is bounded on a set $A$ if for every $M \in \mathbb{R}$, $M > 0$, then $\forall x \in A$, we have that $\mid f(x) \mid < M$.

We will now look at an important theorem known as the boundedness theorem which says that if $f$ is a continuous function over the closed and bounded interval $I$, then $f$ is a bounded function over the set $I$.

Theorem 1 (Boundedness): If $I = [a, b]$ is a closed and bounded interval, and $f : I \to \mathbb{R}$ is a continuous function on $I$, then $f$ is bounded on $I$.

• Proof: We will carry out this proof by contradiction. Let $I = [a, b]$ be a closed and bounded interval, and let $f : I \to \mathbb{R}$ be a continuous function on $I$.

• Now suppose that $f$ is NOT bounded on the interval $I$. Then for any $n \in \mathbb{N}$ there exists an element $x_n \in I$ such that $\mid f(x_n) \mid > n$. Now consider the sequence $(x_n)$. Since $I$ is a bounded interval, this implies that the sequence $(x_n)$ which contains elements from $I$ is also bounded. Therefore, by The Bolzano Weierstrass Theorem there exists a subsequence $(x_{n_k})$ that converges to $L$, that is $\lim_{k \to \infty} x_{n_k} = L$.

• Now since $I$ is also a closed interval, we have that the elements of the sequence $(x_{n_k})$ are contained within $I$, and so we have that $L \in I$.

• Now since $f : I \to \mathbb{R}$ is a continuous function at $L$ (since $L \in I$ and $f$ is continuous on $I$, then we have that by the Sequential Criterion for the Continuity of a Function we have that $(f(x_{n_k}))$ converges to $f(L)$, that is $\lim_{k \to \infty} f(x_{n_k}) = f(L)$. Since the sequence $(f(x_{n_k}))$ is convergent, then we know that this sequence is also bounded.

• But this is a contradiction. Notice that $\mid f(x_{n_k}) \mid > n_{k} ≥ k$ for all $k \in \mathbb{N}$, and so our supposition that $f$ was not bounded on $I$ was false. Therefore $f$ is bounded on $I$. $\blacksquare$

We should make special note that the conclusion to the boundedness theorem is guaranteed to hold provided that:

• (1) $f: I \to \mathbb{R}$ is a continuous function on $I$.

• (2) $I$ is a closed interval.

• (3) $I$ is a bounded interval.

We will now look at an example of where the conclusion to the boundedness theorem holds provided these three conditions are met, and some examples of where the conclusion does not hold when some of the conditions are NOT met.

Example 1

Verify that the function $f : [0, 2] \to \mathbb{R}$ defined by $f(x) = x^2$ is bounded.

We first note that $f$ defined by $f(x) = x^2$ is continuous on all of $\mathbb{R}$ and so $f$ is also continuous on the interval $[0, 2]$. Furthermore, $I = [0, 2]$ is a closed bounded interval. Since $f$ is an increasing function on the interval $[0, 2]$ as $f' > 0$ on $[0, 2]$ we conclude that $\mid f(x) \mid ≤ 4$ for all $x \in [0, 2]$.

Example 2

Verify that the function $f : [-1, 1] \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ does not satisfy the conditions of the boundedness theorem.

Note that $f$ is not continuous on all of $I = [-1, 1]$. In fact, $f$ is not continuous at $x = 0$ and so we are not guaranteed that $f$ is to be bounded on $I$. Precisely, $f$ is not bounded on $[-1, 1]$, since for all $n \in \mathbb{N}$ there exists $\frac{1}{n} \in [-1, 1]$ such that $f(\frac{1}{n}) = n$, and we know that the set of natural numbers is not bounded, that is there DOES NOT exist an $M \in \mathbb{R}$, $M > 0$ such that $\mid f(\frac{1}{n}) \mid = n < M$ for all $n \in \mathbb{N}$.

Example 3

Verify that the function $f : (0, 2) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ does not satisfy the conditions of the boundedness theorem.

In this example, $f$ is continuous on all of $I = (0, 2)$, however, this interval is not closed. We can use the same argument in example 2 to show that hence $f$ is not bounded, or we can use limits to show that as $x \to 0^+$, then $f(x) \to \infty$.

Example 4

Verify that the function $f : [0, \infty) \to \infty$ defined by $f(x) = x$ does not satisfy the conditions of the boundedness theorem.

In this example, $f$ is continuous on all of $[0, \infty)$, and this interval is also closed. However, this interval is not bounded, and using the Archimedean property we can show that $f$ is not bounded as a result.