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Continuity

Module by: Lawrence Baggett. E-mail the author

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

Definition 1:

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

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

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

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

Theorem 1

Let SS and TT be subsets of C,C, let ff and gg be functions from SS into T,T, and suppose that ff and gg are both continuous at a point cc of S.S. Then

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

Proof

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

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

To prove part (5), we first make use of part 1. Let δ1,M1δ1,M1 and δ2,M2δ2,M2 be chosen so that if |y-c|<δ1|y-c|<δ1 and ySyS then

| f ( y ) | < M 1 | f ( y ) | < M 1
(1)

and if |y-c|<δ2|y-c|<δ2 and ySyS then

| g ( y ) | < M 2 | g ( y ) | < M 2
(2)

Next, let ϵ'ϵ' be the positive number |g(c)|/2.|g(c)|/2. Then, there exists a δ'>0δ'>0 such that if |y-c|<δ'|y-c|<δ' and ySyS then |g(y)-g(c)|<ϵ'=|g(c)|/2.|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 ) | | g ( y ) | > ϵ ' = | g ( c ) | / 2 so that | 1 / g ( y ) | < 2 / | g ( c ) |
(3)

Now, to finish the proof of part (5), let ϵ>0ϵ>0 be given. If |y-c|<min(δ1,δ2,δ')|y-c|<min(δ1,δ2,δ') and yS,yS, 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 . | 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 .
(4)

Finally, using the continuity of both ff and gg applied to the positive numbers ϵ1=ϵ/(4M2|g(c)|2)ϵ1=ϵ/(4M2|g(c)|2) and ϵ2=ϵ/(4M1|g(c)|2),ϵ2=ϵ/(4M1|g(c)|2), choose δ>0,δ>0, with δ<min(δ1,δ2,δ'),δ<min(δ1,δ2,δ'), and such that if |y-c|<δ|y-c|<δ and ySyS then |f(y)-f(c)|<ϵ4M2/|g(c)|2|f(y)-f(c)|<ϵ4M2/|g(c)|2 and |g(c)-g(y)|<ϵ4M1/|g(c)|2.|g(c)-g(y)|<ϵ4M1/|g(c)|2. Then, if |y-c|<δ|y-c|<δ and ySyS we have that

| f ( y ) g ( y ) - f ( c ) g ( c ) | < ϵ | f ( y ) g ( y ) - f ( c ) g ( c ) | < ϵ
(5)

as desired.

Exercise 1

  1. Prove part (2) of the preceding theorem. (It's an ϵ/2ϵ/2 argument.)
  2. Prove part (3) of the preceding theorem. (It's similar to the proof of part (5) only easier.)
  3. Prove part (4) of the preceding theorem.
  4. Prove part (6) of the preceding theorem.
  5. Suppose SS is a subset of R.R. Verify the above theorem replacing “ continuity” with left continuity and right continuity.
  6. If SS is a subset of R,R, show that ff is continuous at a point cScS if and only if it is both right continuous and left continuous at c.c.

Theorem 2: The composition of continuous functions is continuous.

Let S,T,S,T, and UU be subsets of C,C, and let f:STf:ST and g:TUg:TU be functions. Suppose ff is continuous at a point cScS and that gg is continuous at the point f(c)T.f(c)T. Then the composition gfgf is continuous at c.c.

Proof

Let ϵ>0ϵ>0 be given. Because gg is continuous at the point f(c),f(c), there exists an α>0α>0 such that |g(t)-g(f(c))|<ϵ|g(t)-g(f(c))|<ϵ if |t-f(c)|<α.|t-f(c)|<α. Now, using this positive number α,α, and using the fact that ff is continuous at the point c,c, there exists a δ>0δ>0 so that |f(s)-f(c)|<α|f(s)-f(c)|<α if |s-c|<δ.|s-c|<δ. Therefore, if |s-c|<δ,|s-c|<δ, then |f(s)-f(c)|<α,|f(s)-f(c)|<α, and hence |g(f(s))-g(f(c))|=|gf(s)-gf(c)|<ϵ,|g(f(s))-g(f(c))|=|gf(s)-gf(c)|<ϵ, which completes the proof.

Exercise 2

  1. If f:CCf:CC is the function defined by f(z)=z,f(z)=z, prove that ff is continuous at each point of C.C.
  2. Use part (a) and Theorem 3.2 to conclude that every rational function is continuous on its domain.
  3. Prove that a step function h:[a,b]Ch:[a,b]C is continuous everywhere on [a,b][a,b] except possibly at the points of the partition PP that determines h.h.

Exercise 3

  1. Let SS be the set of nonnegative real numbers, and define f:SSf:SS by f(x)=x.f(x)=x. Prove that ff is continuous at each point of S.S. HINT: For c=0,c=0, use δ=ϵ2.δ=ϵ2. For c0,c0, use the identity
    y-c=(y-c)y+cy+c=y-cy+cy-cc.y-c=(y-c)y+cy+c=y-cy+cy-cc.
    (6)
  2. If f:CRf:CR is the function defined by f(z)=|z|,f(z)=|z|, show that ff is continuous at every point of its domain.

Exercise 4

Using the previous theorems and exercises, explain why the following functions ff are continuous on their domains. Describe the domains as well.

  1. f(z)=(1-z2)/(1+z2).f(z)=(1-z2)/(1+z2).
  2. f(z)=|1+z+z2+z3-(1/z)|.f(z)=|1+z+z2+z3-(1/z)|.
  3. f(z)=1+1-|z|2.f(z)=1+1-|z|2.

Exercise 5

  1. If cc and dd are real numbers, show that max(c,d)=(c+d)/2+|c-d|/2.max(c,d)=(c+d)/2+|c-d|/2.
  2. If ff and gg are functions from SS into R,R, show that max(f,g)=(f+g)/2+|f-g|/2.max(f,g)=(f+g)/2+|f-g|/2.
  3. If ff and gg are real-valued functions that are both continuous at a point c,c, show that max(f,g)max(f,g) and min(f,g)min(f,g) are both continuous at c.c.

Exercise 6

Let NN be the set of natural numbers, let PP be the set of positive real numbers, and define f:NPf:NP by f(n)=1+n.f(n)=1+n. Prove that ff is continuous at each point of N.N. Show in fact that every function f:NCf:NC is continuous on this domain N.N.

HINT: Show that for any ϵ>0,ϵ>0, the choice of δ=1δ=1 will work.

Exercise 7: Negations

  1. Negate the statement: “For every ϵ>0,ϵ>0,|x|<ϵ.''|x|<ϵ.''
  2. Negate the statement: “For every ϵ>0,ϵ>0, there exists an xx for which |x|<ϵ.''|x|<ϵ.''
  3. Negate the statement that “ ff is continuous at c.''c.''

The next result establishes an equivalence between the basic ϵ,δϵ,δ definition of continuity and a sequential formulation. In many cases, maybe most, this sequential version of continuity is easier to work with than the ϵ,δϵ,δ version.

Theorem 3

Let f:SCf:SC be a complex-valued function on S,S, and let cc be a point in S.S. Then ff is continuous at cc if and only if the following condition holds: For every sequence {xn}{xn} of elements of SS that converges to c,c, the sequence {f(xn)}{f(xn)} converges to f(c).f(c). Or, said a different way, if {xn}{xn} converges to c,c, then {f(xn)}{f(xn)} converges to f(c).f(c). And, said yet a third (somewhat less precise) way, the function ff converts convergent sequences to convergent sequences.

Proof

Suppose first that ff is continuous at c,c, and let {xn}{xn} be a sequence of elements of SS that converges to c.c. Let ϵ>0ϵ>0 be given. We must find a natural number NN such that if nNnN then |f(xn)-f(c)|<ϵ.|f(xn)-f(c)|<ϵ. First, choose δ>0δ>0 so that |f(y)-f(c)|<ϵ|f(y)-f(c)|<ϵ whenever ySyS and |y-c|<δ.|y-c|<δ. Now, choose NN so that |xn-c|<δ|xn-c|<δ whenever nN.nN. Then if nN,nN, we have that |xn-c|<δ,|xn-c|<δ, whence |f(xn)-f(c)|<ϵ.|f(xn)-f(c)|<ϵ. This shows that the sequence {f(xn)}{f(xn)} converges to f(c),f(c), as desired.

We prove the converse by proving the contrapositive statement; i.e., we will show that if ff is not continuous at c,c, then there does exist a sequence {xn}{xn} that converges to cc but for which the sequence {f(xn)}{f(xn)} does not converge to f(c).f(c). Thus, suppose ff is not continuous at c.c. Then there exists an ϵ0>0ϵ0>0 such that for every δ>0δ>0 there is a ySyS such that |y-c|<δ|y-c|<δ but |f(y)-f(c)|ϵ0.|f(y)-f(c)|ϵ0. To obtain a sequence, we apply this statement to δδ's of the form δ=1/n.δ=1/n. Hence, for every natural number nn there exists a point xnSxnS such that |xn-c|<1/n|xn-c|<1/n but |f(xn)-f(c)|ϵ0.|f(xn)-f(c)|ϵ0. Clearly, the sequence {xn}{xn} converges to cc since |xn-c|<1/n.|xn-c|<1/n. On the other hand, the sequence {f(xn)}{f(xn)} cannot be converging to f(c),f(c), because |f(xn)-f(c)||f(xn)-f(c)| is always ϵ0.ϵ0.

This completes the proof of the theorem.

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