Difference between revisions of "Bounded Functions"

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* 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 <math>g:\mathbb{R}^2\to\mathbb{R}</math> and <math>h: (0, 1)^2\to\mathbb{R}</math> defined by <math>g(x, y) := x + y</math> and <math>h(x, y) := \frac{1}{x+y}</math> are both continuous, but neither is bounded. (However, a continuous function must be bounded if its domain is both closed and bounded.)
 
* 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 <math>g:\mathbb{R}^2\to\mathbb{R}</math> and <math>h: (0, 1)^2\to\mathbb{R}</math> defined by <math>g(x, y) := x + y</math> and <math>h(x, y) := \frac{1}{x+y}</math> are both continuous, but neither is bounded. (However, a continuous function must be bounded if its domain is both closed and bounded.)
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==Boundedness Theorem==
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<p>Recall that a function <span class="math-inline"><math>f</math></span> is bounded on a set <span class="math-inline"><math>A</math></span> if for every <span class="math-inline"><math>M \in \mathbb{R}</math></span>, <span class="math-inline"><math>M > 0</math></span>, then <span class="math-inline"><math>\forall x \in A</math></span>, we have that <span class="math-inline"><math>\mid f(x) \mid < M</math></span>.</p>
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<p>We will now look at an important theorem known as the boundedness theorem which says that if <span class="math-inline"><math>f</math></span> is a continuous function over the closed and bounded interval <span class="math-inline"><math>I</math></span>, then <span class="math-inline"><math>f</math></span> is a bounded function over the set <span class="math-inline"><math>I</math></span>.</p>
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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<td><strong>Theorem 1 (Boundedness):</strong> If <span class="math-inline"><math>I = [a, b]</math></span> is a closed and bounded interval, and <span class="math-inline"><math>f : I \to \mathbb{R}</math></span> is a continuous function on <span class="math-inline"><math>I</math></span>, then <span class="math-inline"><math>f</math></span> is bounded on <span class="math-inline"><math>I</math></span>.
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[[File:Continuous function bounded on interval.png|Continuous function bounded on interval]]
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</td>
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</blockquote>
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<ul>
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<li><strong>Proof:</strong> We will carry out this proof by contradiction. Let <span class="math-inline"><math>I = [a, b]</math></span> be a closed and bounded interval, and let <span class="math-inline"><math>f : I \to \mathbb{R}</math></span> be a continuous function on <span class="math-inline"><math>I</math></span>.</li>
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</ul>
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<ul>
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<li>Now suppose that <span class="math-inline"><math>f</math></span> is NOT bounded on the interval <span class="math-inline"><math>I</math></span>. Then for any <span class="math-inline"><math>n \in \mathbb{N}</math></span> there exists an element <span class="math-inline"><math>x_n \in I</math></span> such that <span class="math-inline"><math>\mid f(x_n) \mid > n</math></span>. Now consider the sequence <span class="math-inline"><math>(x_n)</math></span>. Since <span class="math-inline"><math>I</math></span> is a bounded interval, this implies that the sequence <span class="math-inline"><math>(x_n)</math></span> which contains elements from <span class="math-inline"><math>I</math></span> is also bounded. Therefore, by <a href="/the-bolzano-weierstrass-theorem">The Bolzano Weierstrass Theorem</a> there exists a subsequence <span class="math-inline"><math>(x_{n_k})</math></span> that converges to <span class="math-inline"><math>L</math></span>, that is <span class="math-inline"><math>\lim_{k \to \infty} x_{n_k} = L</math></span>.</li>
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</ul>
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<ul>
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<li>Now since <span class="math-inline"><math>I</math></span> is also a closed interval, we have that the elements of the sequence <span class="math-inline"><math>(x_{n_k})</math></span> are contained within <span class="math-inline"><math>I</math></span>, and so we have that <span class="math-inline"><math>L \in I</math></span>.</li>
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</ul>
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<ul>
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<li>Now since <span class="math-inline"><math>f : I \to \mathbb{R}</math></span> is a continuous function at <span class="math-inline"><math>L</math></span> (since <span class="math-inline"><math>L \in I</math></span> and <span class="math-inline"><math>f</math></span> is continuous on <span class="math-inline"><math>I</math></span>, 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 <span class="math-inline"><math>(f(x_{n_k}))</math></span> converges to <span class="math-inline"><math>f(L)</math></span>, that is <span class="math-inline"><math>\lim_{k \to \infty} f(x_{n_k}) = f(L)</math></span>. Since the sequence <span class="math-inline"><math>(f(x_{n_k}))</math></span> is convergent, then <a href="/proof-that-convergent-sequences-of-real-numbers-are-bounded">we know that this sequence is also bounded</a>.</li>
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</ul>
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<ul>
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<li>But this is a contradiction. Notice that <span class="math-inline"><math>\mid f(x_{n_k}) \mid > n_{k} \geq k</math></span> for all <span class="math-inline"><math>k \in \mathbb{N}</math></span>, and so our supposition that <span class="math-inline"><math>f</math></span> was not bounded on <span class="math-inline"><math>I</math></span> was false. Therefore <span class="math-inline"><math>f</math></span> is bounded on <span class="math-inline"><math>I</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
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</ul>
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<p>We should make special note that the conclusion to the boundedness theorem is guaranteed to hold provided that:</p>
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<ul>
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<li><strong>(1)</strong> <span class="math-inline"><math>f: I \to \mathbb{R}</math></span> is a continuous function on <span class="math-inline"><math>I</math></span>.</li>
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</ul>
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<ul>
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<li><strong>(2)</strong> <span class="math-inline"><math>I</math></span> is a closed interval.</li>
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</ul>
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<ul>
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<li><strong>(3)</strong> <span class="math-inline"><math>I</math></span> is a bounded interval.</li>
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</ul>
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<p>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.</p>
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===Example 1===
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<p><strong>Verify that the function <span class="math-inline"><math>f : [0, 2] \to \mathbb{R}</math></span> defined by <span class="math-inline"><math>f(x) = x^2</math></span> is bounded.</strong></p>
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<p>We first note that <span class="math-inline"><math>f</math></span> defined by <span class="math-inline"><math>f(x) = x^2</math></span> is continuous on all of <span class="math-inline"><math>\mathbb{R}</math></span> and so <span class="math-inline"><math>f</math></span> is also continuous on the interval <span class="math-inline"><math>[0, 2]</math></span>. Furthermore, <span class="math-inline"><math>I = [0, 2]</math></span> is a closed bounded interval. Since <span class="math-inline"><math>f</math></span> is an increasing function on the interval <span class="math-inline"><math>[0, 2]</math></span> as <span class="math-inline"><math>f' > 0</math></span> on <span class="math-inline"><math>[0, 2]</math></span> we conclude that <span class="math-inline"><math>\mid f(x) \mid \leq 4</math></span> for all <span class="math-inline"><math>x \in [0, 2]</math></span>.</p>
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===Example 2===
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<p><strong>Verify that the function <span class="math-inline"><math>f : [-1, 1] \to \mathbb{R}</math></span> defined by <span class="math-inline"><math>f(x) = \frac{1}{x}</math></span> does not satisfy the conditions of the boundedness theorem.</strong></p>
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<p>Note that <span class="math-inline"><math>f</math></span> is not continuous on all of <span class="math-inline"><math>I = [-1, 1]</math></span>. In fact, <span class="math-inline"><math>f</math></span> is not continuous at <span class="math-inline"><math>x = 0</math></span> and so we are not guaranteed that <span class="math-inline"><math>f</math></span> is to be bounded on <span class="math-inline"><math>I</math></span>. Precisely, <span class="math-inline"><math>f</math></span> is not bounded on <span class="math-inline"><math>[-1, 1]</math></span>, since for all <span class="math-inline"><math>n \in \mathbb{N}</math></span> there exists <span class="math-inline"><math>\frac{1}{n} \in [-1, 1]</math></span> such that <span class="math-inline"><math>f(\frac{1}{n}) = n</math></span>, and we know that the set of natural numbers is not bounded, that is there DOES NOT exist an <span class="math-inline"><math>M \in \mathbb{R}</math></span>, <span class="math-inline"><math>M > 0</math></span> such that <span class="math-inline"><math>\mid f(\frac{1}{n}) \mid = n < M</math></span> for all <span class="math-inline"><math>n \in \mathbb{N}</math></span>.</p>
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===Example 3===
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<p><strong>Verify that the function <span class="math-inline"><math>f : (0, 2) \to \mathbb{R}</math></span> defined by <span class="math-inline"><math>f(x) = \frac{1}{x}</math></span> does not satisfy the conditions of the boundedness theorem.</strong></p>
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<p>In this example, <span class="math-inline"><math>f</math></span> is continuous on all of <span class="math-inline"><math>I = (0, 2)</math></span>, however, this interval is not closed. We can use the same argument in example 2 to show that hence <span class="math-inline"><math>f</math></span> is not bounded, or we can use limits to show that as <span class="math-inline"><math>x \to 0^+</math></span>, then <span class="math-inline"><math>f(x) \to \infty</math></span>.</p>
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===Example 4===
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<p><strong>Verify that the function <span class="math-inline"><math>f : [0, \infty) \to \infty</math></span> defined by <span class="math-inline"><math>f(x) = x</math></span> does not satisfy the conditions of the boundedness theorem.</strong></p>
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<p>In this example, <span class="math-inline"><math>f</math></span> is continuous on all of <span class="math-inline"><math>[0, \infty)</math></span>, and this interval is also closed. However, this interval is not bounded, and using the Archimedean property we can show that <span class="math-inline"><math>f</math></span> is not bounded as a result.</p>
  
 
== Licensing ==  
 
== Licensing ==  
 
Content obtained and/or adapted from:
 
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* [https://en.wikipedia.org/wiki/Bounded_function Bounded function, Wikipedia] under a CC BY-SA license
 
* [https://en.wikipedia.org/wiki/Bounded_function Bounded function, Wikipedia] under a CC BY-SA license

Latest revision as of 15:56, 17 November 2021

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

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

for every natural number n. The set of all bounded sequences forms the sequence space .

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 : XY 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 : RR is bounded since for all .
  • The function , 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 , 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 : CC which are entire are either unbounded or constant as a consequence of Liouville's theorem. In particular, the complex sin : CC 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 and defined by and 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 is bounded on a set if for every , , then , we have that .

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

Theorem 1 (Boundedness): If is a closed and bounded interval, and is a continuous function on , then is bounded on .

Continuous function bounded on interval

  • Proof: We will carry out this proof by contradiction. Let be a closed and bounded interval, and let be a continuous function on .
  • Now suppose that is NOT bounded on the interval . Then for any there exists an element such that . Now consider the sequence . Since is a bounded interval, this implies that the sequence which contains elements from is also bounded. Therefore, by <a href="/the-bolzano-weierstrass-theorem">The Bolzano Weierstrass Theorem</a> there exists a subsequence that converges to , that is .
  • Now since is also a closed interval, we have that the elements of the sequence are contained within , and so we have that .
  • Now since is a continuous function at (since and is continuous on , 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 converges to , that is . Since the sequence 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 for all , and so our supposition that was not bounded on was false. Therefore is bounded on .

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

  • (1) is a continuous function on .
  • (2) is a closed interval.
  • (3) 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 defined by is bounded.

We first note that defined by is continuous on all of and so is also continuous on the interval . Furthermore, is a closed bounded interval. Since is an increasing function on the interval as on we conclude that for all .

Example 2

Verify that the function defined by does not satisfy the conditions of the boundedness theorem.

Note that is not continuous on all of . In fact, is not continuous at and so we are not guaranteed that is to be bounded on . Precisely, is not bounded on , since for all there exists such that , and we know that the set of natural numbers is not bounded, that is there DOES NOT exist an , such that for all .

Example 3

Verify that the function defined by does not satisfy the conditions of the boundedness theorem.

In this example, is continuous on all of , however, this interval is not closed. We can use the same argument in example 2 to show that hence is not bounded, or we can use limits to show that as , then .

Example 4

Verify that the function defined by does not satisfy the conditions of the boundedness theorem.

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

Licensing

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