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Chapter 3
Continuous Functions
In this chapter, we define continuous functions and study their properties.
3.1. Continuity
According to the definition introduced by Cauchy, and developed by Weierstrass, continuous functions are functions that take nearby values at nearby points.
Denition 3.1.
Letf:A→R, whereA⊂R, and suppose thatc∈A. Thenfis continuous atcif for everyϵ >0 there exists aδ >0 such that |x-c|< δandx∈Aimplies that|f(x)-f(c)|< ϵ. A functionf:A→Ris continuous on a setB⊂Aif it is continuous at every point inB, and continuous if it is continuous at every point of its domainA. The definition of continuity at a point may be stated in terms of neighborhoods as follows.
Denition 3.2.
A functionf:A→R, whereA⊂R, is continuous atc∈Aif for every neighborhoodVoff(c) there is a neighborhoodUofcsuch that x∈A∩Uimplies thatf(x)∈V . Theϵ-δdefinition corresponds to the case whenVis anϵ-neighborhood off(c) andUis aδ-neighborhood ofc. We leave it as an exercise to prove that these definitions are equivalent. Note thatcmust belong to the domainAoffin order to define the continuity offatc. Ifcis an isolated point ofA, then the continuity condition holds auto- matically since, for sufficiently smallδ >0, the only pointx∈Awith|x-c|< δ isx=c, and then 0 =|f(x)-f(c)|< ϵ. Thus, a function is continuous at every isolated point of its domain, and isolated points are not of much interest. 21
223. Continuous Functions
Ifc∈Ais an accumulation point ofA, then continuity offatcis equivalent to the condition that lim x!cf(x) =f(c), meaning that the limit offasx→cexists and is equal to the value offatc.
Example 3.3.
Iff: (a,b)→Ris defined on an open interval, thenfis continuous on (a,b) if and only if lim x!cf(x) =f(c) for everya < c < b since every point of (a,b) is an accumulation point.
Example 3.4.
Iff: [a,b]→Ris defined on a closed, bounded interval, thenfis continuous on [a,b] if and only if lim x!cf(x) =f(c) for everya < c < b, lim x!a+f(x) =f(a),limx!bf(x) =f(b).
Example 3.5.
Suppose that
A={ 0,1,1 2 ,1 3 ,...,1 n andf:A→Ris defined by f(0) =y0, f(1 n =yn for some valuesy0,yn∈R. Then 1/nis an isolated point ofAfor everyn∈N, sofis continuous at 1/nfor every choice ofyn. The remaining point 0∈Ais an accumulation point ofA, and the condition forfto be continuous at 0 is that lim n!1yn=y0. As for limits, we can give an equivalent sequential definition of continuity, which follows immediately from Theorem 2.4
Theorem 3.6.
Iff:A→Randc∈Ais an accumulation point ofA, thenfis continuous atcif and only if lim n!1f(xn) =f(c) for every sequence (xn) inAsuch thatxn→casn→ ∞. In particular,fis discontinuous atc∈Aif there is sequence (xn) in the domain
Aoffsuch thatxn→cbutf(xn)̸→f(c).
Let's consider some examples of continuous and discontinuous functions to illustrate the definition.
Example 3.7.
xis continuous c =x-c c c |x-c|,
3.1. Continuity23
cϵ >0 in the definition of continuity. To prove x < ϵ.
Example 3.8.
The function sin :R→Ris continuous onR. To prove this, we use |sinx-sinc|=2cos(x+c 2 sin(x-c 2 2 It follows that we can takeδ=ϵin the definition of continuity for everyc∈R.
Example 3.9.
The sign function sgn :R→R, defined by
sgnx= 1 ifx >0,
0 ifx= 0,
-1 ifx <0, is not continuous at 0 since lim x!0sgnxdoes not exist (see Example 2.6 ). The left and right limits of sgn at 0, lim x!0f(x) =-1,limx!0+f(x) = 1, do exist, but they are unequal. We say thatfhas a jump discontinuity at 0.
Example 3.10.
The functionf:R→Rdefined by
f(x) ={
1/xifx̸= 0,
0 ifx= 0,
is not continuous at 0 since lim x!0f(x) does not exist (see Example 2.7 ). Neither the left or right limits offat 0 exist either, and we say thatfhas an essential discontinuity at 0.
Example 3.11.
The functionf:R→R, defined by
f(x) ={ sin(1/x) ifx̸= 0,
0ifx= 0.
is continuous atc̸= 0 (see Example 3.20 below) but discontinuous at 0 because lim x!0f(x) does not exist (see Example 2.8
Example 3.12.
The functionf:R→Rdefined by
f(x) ={ xsin(1/x) ifx̸= 0,
0ifx= 0.
is continuous at every point ofR. (See Figure 1 . The continuity atc̸= 0 is proved in Example 3.21 below. To prove continuity at 0, note that forx̸= 0,
243. Continuous Functions-3-2-10123-0.4
-0.2 0 0.2 0.4 0.6 0.8 1 -0.4-0.3-0.2-0.100.10.20.30.4-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4
Figure 1.
A plot of the functiony=xsin(1=x) and a detail near the origin with the linesy=xshown in red. sof(x)→f(0) asx→0. If we had definedf(0) to be any value other than
0, thenfwould not be continuous at 0. In that case,fwould have a removable
discontinuity at 0.
Example 3.13.
The Dirichlet functionf:R→Rdefined by
f(x) ={
1 ifx∈Q,
0 ifx /∈Q
is discontinuous at everyc∈R. Ifc /∈Q, choose a sequence (xn) of rational numbers such thatxn→c(possible sinceQis dense inR). Thenxn→cand f(xn)→1 butf(c) = 0. Ifc∈Q, choose a sequence (xn) of irrational numbers such thatxn→c; for example ifc=p/q, we can take x n=p q 2 n
2∈Q. Thenxn→candf(xn)→0 but
f(c) = 1. In fact, taking a rational sequence (xn) and an irrational sequence (˜xn) that converge toc, we see that limx!cf(x) does not exist for anyc∈R.
Example 3.14.
The Thomae functionf:R→Rdefined by
f(x) ={
1/qifx=p/qwherepandq >0 are relatively prime,
0 ifx /∈Qorx= 0
is continuous at 0 and every irrational number and discontinuous at every nonzero rational number. See Figure 2 for a plot. We can give a rough classification of a discontinuity of a functionf:A→Rat an accumulation pointc∈Aas follows. (1) Removable discontinuity: limx!cf(x) =Lexists butL̸=f(c), in which case we can makefcontinuous atcby redefiningf(c) =L(see Example 3.12 (2) Jump discontinuity: limx!cf(x) doesn't exist, but both the left and right limits lim x!cf(x), limx!c+f(x) exist and are different (see Example 3.9
3.2. Properties of continuous functions25
Figure 2.
A plot of the Thomae function on [0;1]
(3) Essential discontinuity: limx!cf(x) doesn't exist and at least one of the left or right limits lim x!cf(x), limx!c+f(x) doesn't exist (see Examples 3.10 3.11 3.13
3.2. Properties of continuous functions
The basic properties of continuous functions follow from those of limits.
Theorem 3.15.
Iff,g:A→Rare continuous atc∈Aandk∈R, thenkf,f+g, andfgare continuous atc. Moreover, ifg(c)̸= 0 thenf/gis continuous atc.
Proof.
This result follows immediately Theorem
2.22 A polynomial function is a functionP:R→Rof the form
P(x) =a0+a1x+a2x2+···+anxn
wherea0,a1,a2,...,anare real coefficients. A rational functionRis a ratio of polynomialsP,Q
R(x) =P(x)
Q(x). The domain ofRis the set of points inRsuch thatQ̸= 0.
Corollary 3.16.
Every polynomial function is continuous onRand every rational function is continuous on its domain.
Proof.
The constant functionf(x) = 1 and the identity functiong(x) =xare continuous onR. Repeated application of Theorem 3.15 for scalar multiples, sums, and products implies that every polynomial is continuous onR. It also follows that a rational functionR=P/Qis continuous at every point whereQ̸= 0.
263. Continuous Functions
Example 3.17.
The functionf:R→Rgiven by
f(x) =x+ 3x3+ 5x5
1 +x2+x4
is continuous onRsince it is a rational function whose denominator never vanishes. In addition to forming sums, products and quotients, another way to build up more complicated functions from simpler functions is by composition. We recall that iff:A→Randg:B→Rwheref(A)⊂B, meaning that the domain ofgcontains the range off, then we define the compositiong◦f:A→R by (g◦f)(x) =g(f(x)). The next theorem states that the composition of continuous functions is continuous. Note carefully the points at which we assumefandgare continuous.
Theorem 3.18.
Letf:A→Randg:B→Rwheref(A)⊂B. Iffis continuous atc∈Aandgis continuous atf(c)∈B, theng◦f:A→Ris continuous atc.
Proof.
Letϵ >0 be given. Sincegis continuous atf(c), there existsη >0 such that |y-f(c)|< ηandy∈Bimplies that|g(y)-g(f(c))|< ϵ. Next, sincefis continuous atc, there existsδ >0 such that |x-c|< δandx∈Aimplies that|f(x)-f(c)|< η.
Combing these inequalities, we get that
|x-c|< δandx∈Aimplies that|g(f(x))-g(f(c))|< ϵ, which proves thatg◦fis continuous atc.
Corollary 3.19.
Letf:A→Randg:B→Rwheref(A)⊂B. Iffis continuous onAandgis continuous onf(A), theng◦fis continuous onA.
Example 3.20.
The function
f(x) ={ sin(1/x) ifx̸= 0,
0ifx= 0.
is continuous onR\{0}, since it is the composition ofx7→1/x, which is continuous onR\ {0}, andy7→siny, which is continuous onR.
Example 3.21.
The function
f(x) ={ xsin(1/x) ifx̸= 0,
0ifx= 0.
is continuous onR\ {0}since it is a product of functions that are continuous on
R\ {0}. As shown in Example
3.12 ,fis also continuous at 0, sofis continuous onR.
3.3. Uniform continuity27
3.3. Uniform continuity
Uniform continuity is a subtle but powerful strengthening of continuity.
Denition 3.22.
Letf:A→R, whereA⊂R. Thenfis uniformly continuous onAif for everyϵ >0 there exists aδ >0 such that |x-y|< δandx,y∈Aimplies that|f(x)-f(y)|< ϵ. The key point of this definition is thatδdepends only onϵ, not onx,y. A uniformly continuous function onAis continuous at every point ofA, but the converse is not true, as we explain next. If a function is continuous onA, then givenϵ >0 there existsδ(c)>0 for every c∈Asuch that |x-c|< δ(c) andx∈Aimplies that|f(x)-f(c)|< ϵ. In general,δ(c) depends on bothϵandc, but we don't show theϵ-dependence explicitly since we're thinking ofϵas fixed. If inf c2Aδ(c) = 0 however we chooseδ(c), then noδ0>0 depending only onϵworks simultaneously for allc∈A. In that case, the function is continuous onAbut not uniformly continuous. Before giving examples, we state a sequential condition for uniform continuity to fail.
Proposition 3.23.
A functionf:A→Ris not uniformly continuous onAif and only if there existsϵ0>0 and sequences (xn), (yn) inAsuch that lim n!1|xn-yn|= 0 and|f(xn)-f(yn)| ≥ϵ0for alln∈N.
Proof.
Iffis not uniformly continuous, then there existsϵ0>0 such that for every δ >0 there are pointsx,y∈Awith|x-y|< δand|f(x)-f(y)| ≥ϵ0. Choosing x n,yn∈Ato be any such points forδ= 1/n, we get the required sequences. Conversely, if the sequential condition holds, then for everyδ >0 there exists n∈Nsuch that|xn-yn|< δand|f(xn)-f(yn)| ≥ϵ0. It follows that the uniform continuity condition in Definition 3.22 cannot hold for anyδ >0 ifϵ=ϵ0, sofis not uniformly continuous.
Example 3.24.
Example
3.8 shows that the sine function is uniformly continuous onR, since we can takeδ=ϵfor everyx,y∈R.
Example 3.25.
Definef: [0,1]→Rbyf(x) =x2. Thenfis uniformly continuous on [0,1]. To prove this, note that for allx,y∈[0,1] we have so we can takeδ=ϵ/2 in the definition of uniform continuity. Similarly,f(x) =x2 is uniformly continuous on any bounded set.
283. Continuous Functions
Example 3.26.
The functionf(x) =x2is continuous but not uniformly continuous onR. We have already proved thatfis continuous onR(it's a polynomial). To prove thatfis not uniformly continuous, let x n=n, yn=n+1 n Then lim n!1|xn-yn|= limn!11 n = 0, but |f(xn)-f(yn)|=( n+1 n 2 -n2= 2 +1 n
2≥2 for everyn∈N.
It follows from Proposition
3.23 thatfis not uniformly continuous onR. The problem here is that, for givenϵ >0, we need to makeδ(c) smaller ascgets larger to prove the continuity offatc, andδ(c)→0 asc→ ∞.
Example 3.27.
The functionf: (0,1]→Rdefined by
f(x) =1 x is continuous but not uniformly continuous on (0,1]. We have already proved that fis continuous on (0,1] (it's a rational function whose denominatorxis nonzero in (0,1]). To prove thatfis not uniformly continuous, definexn,yn∈(0,1] forn∈N by x n=1 n , yn=1 n+ 1. Thenxn→0,yn→0, and|xn-yn| →0 asn→ ∞, but |f(xn)-f(yn)|= (n+ 1)-n= 1 for everyn∈N.
It follows from Proposition
3.23 thatfis not uniformly continuous on (0,1]. The problem here is that, for givenϵ >0, we need to makeδ(c) smaller ascgets closer to 0 to prove the continuity offatc, andδ(c)→0 asc→0+. The non-uniformly continuous functions in the last two examples were un- bounded. However, even bounded continuous functions can fail to be uniformly continuous if they oscillate arbitrarily quickly.
Example 3.28.
Definef: (0,1]→Rby
f(x) = sin(1 x Thenfis continuous on (0,1] but it isn't uniformly continuous on (0,1]. To prove this, definexn,yn∈(0,1] forn∈Nby xquotesdbs_dbs15.pdfusesText_21
Math 312, Intro to Real Analysis: Homework Solutions