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Next, we come to the definition of continuity. Unlike the preceding discussion, which can beviewed as being related primarily to the algebraic properties of functions, this one is an analytic notion.

Next, we come to the definition of continuity. Unlike the preceding discussion, which can beviewed as being related primarily to the algebraic properties of functions, this one is an analytic notion.

Let S and T be sets of complex numbers, and let f : S T . Then f is said to be continuous at a point c of S if for every positive ϵ , there exists a positive δ such that if x S satisfies | x - c | < δ , then | f ( x ) - f ( c ) | < ϵ . The function f is called continuous on S if it is continuous at every point c of S .

If the domain S of f consists of real numbers, then the function f is called right continuous at c if for every ϵ > 0 there exists a δ > 0 such that | f ( x ) - f ( c ) | < ϵ whenever x S and 0 x - c < δ , and is called left continuous at c if for every ϵ > 0 there exists a δ > 0 such that | f ( x ) - f ( c ) | < ϵ whenever x S and 0 x - c > - δ .

REMARK If f is continuous at a point c , then the positive number δ of the preceding definition is not unique (any smaller number would work as well), but it does depend both on the number ϵ and on the point c . Sometimes we will write δ ( ϵ , c ) to make this dependence explicit. Later, we will introduce a notion of uniform continuityin which δ only depends on the number ϵ and not on the particular point c .

The next theorem indicates the interaction between the algebraic properties of functions and continuity.

Let S and T be subsets of C , let f and g be functions from S into T , and suppose that f and g are both continuous at a point c of S . Then

  1. There exists a δ > 0 and a positive number M such that if | y - c | < δ and y S then | f ( y ) | M . That is, if f is continuous at c , then it is bounded near c .
  2. f + g is continuous at c .
  3. f g is continuous at c .
  4. | f | is continuous at c .
  5. If g ( c ) 0 , then f / g is continuous at c .
  6. If f is a complex-valued function, and u and v are the real and imaginary parts of f , then f is continuous at c if and only if u and v are continuous at c .

We prove parts (1) and (5), and leave the remaining parts to the exercise that follows.

To see part (1), let ϵ = 1 . Then, since f is continuous at c , there exists a δ > 0 such that if | y - c | < δ and y S then | f ( y ) - f ( c ) | < 1 . Since | z - w | | | z | - | w | | for any two complex numbers z and w (backwards Triangle Inequality), it then follows that | | f ( y ) | - | f ( c ) | | < 1 , from which it follows that if | y - c | < δ then | f ( y ) | < | f ( c ) | + 1 . Hence, setting M = | f ( c ) | + 1 , we have that if | y - c | < δ and y S , then | f ( y ) | M as desired.

To prove part (5), we first make use of part 1. Let δ 1 , M 1 and δ 2 , M 2 be chosen so that if | y - c | < δ 1 and y S then

| f ( y ) | < M 1

and if | y - c | < δ 2 and y S then

| g ( y ) | < M 2

Next, let ϵ ' be the positive number | g ( c ) | / 2 . Then, there exists a δ ' > 0 such that if | y - c | < δ ' and y S then | g ( y ) - g ( c ) | < ϵ ' = | g ( c ) | / 2 . It then follows from the backwards triangle inequality that

| g ( y ) | > ϵ ' = | g ( c ) | / 2 so that | 1 / g ( y ) | < 2 / | g ( c ) |

Now, to finish the proof of part (5), let ϵ > 0 be given. If | y - c | < min ( δ 1 , δ 2 , δ ' ) and y S , then from Inequalities (3.1), (3.2), and (3.3) we obtain

| f ( y ) g ( y ) - f ( c ) g ( c ) | = | f ( y ) g ( c ) - f ( c ) g ( y ) | | g ( y ) g ( c ) | = | f ( y ) g ( c ) - f ( c ) g ( c ) + f ( c ) g ( c ) - f ( c ) g ( y ) | | g ( y ) | | g ( c ) | | f ( y ) - f ( c ) | | g ( c ) | + | f ( c ) | | g ( c ) - g ( y ) | | g ( y ) | | g ( c ) | < ( | f ( y ) - f ( c ) | M 2 + M 1 | g ( c ) - g ( y ) | ) × 2 | g ( c ) | 2 .

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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