Bounded Functions

A schematic illustration of a bounded function (red) and an unbounded one (blue). Intuitively, the graph of a bounded function stays within a horizontal band, while the graph of an unbounded function does not.

A function f defined on some set X with real or complex values is called bounded if the set of its values is bounded. In other words, there exists a real number M such that

$|f(x)|\leq M$

for all x in X. A function that is not bounded is said to be unbounded.

If f is real-valued and f(x) ≤ A for all x in X, then the function is said to be bounded (from) above by A. If f(x) ≥ B for all x in X, then the function is said to be bounded (from) below by B. A real-valued function is bounded if and only if it is bounded from above and below.

An important special case is a bounded sequence, where X is taken to be the set N of natural numbers. Thus a sequence f = (a₀, a₁, a₂, ...) is bounded if there exists a real number M such that

$|a_{n}|\leq M$

for every natural number n. The set of all bounded sequences forms the sequence space $l^{\infty }$.

The definition of boundedness can be generalized to functions f : X → Y taking values in a more general space Y by requiring that the image f(X) is a bounded set in Y.

Related notions

Weaker than boundedness is local boundedness. A family of bounded functions may be uniformly bounded.

A bounded operator T : X → Y is not a bounded function in the sense of this page's definition (unless T = 0), but has the weaker property of preserving boundedness: Bounded sets M ⊆ X are mapped to bounded sets T(M) ⊆ Y. This definition can be extended to any function f : X → Y if X and Y allow for the concept of a bounded set. Boundedness can also be determined by looking at a graph.

Examples

• The sine function sin : R → R is bounded since $|\sin(x)|\leq 1$ for all $x\in \mathbf {R}$.

• The function $f(x)=(x^{2}-1)^{-1}$, defined for all real x except for −1 and 1, is unbounded. As x approaches −1 or 1, the values of this function get larger and larger in magnitude. This function can be made bounded if one considers its domain to be, for example, [2, ∞) or (−∞, −2].

• The function ${\textstyle f(x)=(x^{2}+1)^{-1}}$, defined for all real x, is bounded.

• The inverse trigonometric function arctangent defined as: y = arctan(x) or x = tan(y) is increasing for all real numbers x and bounded with −Template:Sfrac < y < Template:Sfrac radians.
• By the boundedness theorem, every continuous function on a closed interval, such as f : [0, 1] → R, is bounded. More generally, any continuous function from a compact space into a metric space is bounded.

• All complex-valued functions f : C → C which are entire are either unbounded or constant as a consequence of Liouville's theorem. In particular, the complex sin : C → C must be unbounded since it is entire.

• The function f which takes the value 0 for x rational number and 1 for x irrational number (cf. Dirichlet function) is bounded. Thus, a function does not need to be "nice" in order to be bounded. The set of all bounded functions defined on [0, 1] is much larger than the set of continuous functions on that interval. Moreover, continuous functions need not be bounded; for example, the functions $g:\mathbb {R} ^{2}\to \mathbb {R}$ and $h:(0,1)^{2}\to \mathbb {R}$ defined by $g(x,y):=x+y$ and $h(x,y):={\frac {1}{x+y}}$ are both continuous, but neither is bounded. (However, a continuous function must be bounded if its domain is both closed and bounded.)

Boundedness Theorem

Recall 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 $I$.

<img src="http://mathonline.wdfiles.com/local--files/boundedness-theorem/Screen%20Shot%202014-11-27%20at%2011.29.09%20PM.png" alt="Screen%20Shot%202014-11-27%20at%2011.29.09%20PM.png" class="image" />

• 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 <a href="/the-bolzano-weierstrass-theorem">The Bolzano Weierstrass Theorem</a> 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 <a href="/sequential-criterion-for-the-continuity-of-a-function">Sequential Criterion for the Continuity of a Function</a> 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 <a href="/proof-that-convergent-sequences-of-real-numbers-are-bounded">we know that this sequence is also bounded</a>.

• But this is a contradiction. Notice that Failed to parse (syntax error): {\displaystyle \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 Failed to parse (syntax error): {\displaystyle \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.

Licensing

Content obtained and/or adapted from:

• Bounded function, Wikipedia under a CC BY-SA license