Real Analysis

Lecture 5: Continuity

Manasa Mandava

Indian School of Business, Hyderabad

Term 1, 2015

Limits

Recall:

For a sequence {pn} in a metric space (X, d), we writelimn→∞ pn = p, for some point p ∈ X, if for every ǫ > 0 there isan integer N ≥ 0 such that d(pn, p) < ǫ for all n ≥ N .

Limit of a function:

Definition

Let (X, dX ) and (Y, dY ) be metric spaces. Suppose E ⊂ X, fmaps E into Y , and p is a limit point of E. For q ∈ Y , we sayf(x) → q as x → p, or

limx→p

f(x) = q,

if for every ǫ > 0 there exists a δ > 0 such that dY (f(x), q) < ǫ forall x ∈ E such that 0 < dX(x, p) < δ.

Right and left hand limits

Definition (Right hand limit)

Let f be a real valued function defined on the interval (a, b). Thenfor any p ∈ [a, b), if for every ǫ > 0 there exists a δ > 0 such that|f(x), q| < ǫ for all 0 < x− p < δ, we say

f(p+) := limx→p+

f(x) = q,

Definition (left hand limit)

Let f be a real valued function defined on the interval (a, b). Thenfor any p ∈ (a, b], if for every ǫ > 0 there exists a δ > 0 such that|f(x), q| < ǫ for all 0 < p− x < δ, we say

f(p−) := limx→p−

f(x) = q,

Limits at infinity and infinite limits

Recall:

For any sequence {pn} of real numbers, if for every real M there isan integer N such that pn > M for all n > N , we writelimn→∞ pn = ∞. The neighbourhood of pn: (M,∞), and theneighbourhood of ∞: (N,∞).

Similarly, if for every real M there is an integer N such thatpn < M for all n > N , we write limn→∞ pn = −∞. Theneighbourhood of pn: (−∞,M), and the neighbourhood of ∞:(N,∞).

Definition

For a real sequence {pn}, we write limn→∞ pn = p, if for everyneighbourhood U of p there is a neighbourhood V of ∞ such thatpn ∈ U for all n ∈ V .

Limit of a function at infinity and infinite limits

Definition

For any real c, the set of real numbers x such that x > c is calleda neighbourhood of ∞ and is written (c,∞). Similarly the set(−∞, c) is a neighbourhood of −∞.

Definition

Let f be a real function defined on E ⊂ R̄. For some point q ∈ R̄,we say

limx→p

f(x) = q,

if for every neighbourhood U of q there is a neighbourhood V of psuch that V ∩ E 6= ∅ and such that f(x) ∈ U for all x ∈ V ∩ Eand x 6= p.

Relation between sequences and limits

Theorem

Let X,Y,E, f, and p be as in the definition above. Then,limx→p f(x) = q if and only if limn→∞ f(pn) = q for everysequence {pn} in E such that pn 6= p and limn→∞ pn = p.

Corollary

If f has a limit at p, the limit is unique.

Properties of limits

Theorem

Let X be a metric space and E ⊂ X, p is a limit point of E, fand g are complex functions on E, and

limx→p

f(x) = A and limx→p

g(x) = B.

Then the following properties hold:(a) limx→p(f + g)(x) = A+B;(b) limx→p(fg)(x) = AB;

(c) limx→p(fg)(x) = A

B, if B 6= 0.

Remarks

If the functions f and g take values in R̄, then properties (a) - (c)hold provided the right hand side members are well defined. (ornot equal to ∞−∞, 0 · ∞, ∞/∞, A/0 )

If the functions f and g are Rk valued, then property (a) is true,and property (b) becomes limx→p(f · g)(x) = A ·B.

Continuity

Definition

Let (X, dX ) and (Y, dY ) be metric spaces. Suppose f maps Xinto Y . Then f is said to be continuous at a point p ∈ X if forevery ǫ > 0 there exists a δ > 0 such that dY (f(x), f(p)) < ǫ forall x ∈ X such that 0 < dX(x, p) < δ.

If f is not continuous at point p of X, then f is said to bediscontinuous at p.

If f is continuous at every point of X, then f is said to continuouson X.

Relation between limits and continuity

Theorem

Under assumptions of the above definition, assume also that X isclosed. Then, f is continuous at p if and only if limx→p f(x) = p.

Classification of discontinuities: For a real valued function fdefined on [a, b], if f is discontinuous at a point x,

(a)if f(x+) and f(x−) exist, it is said to have a discontinuity offirst kind.

(b) Otherwise, it is said to have a discontinuity of second kind.

Properties of continuous functions

Continuity of composite functions: Let h(x) := g(f(x)) for allx in the domain f . Then if f is continuous at a point p and if g iscontinuous at the point f(p), then h is continuous at p.

Let f : X → Y . Then f is continuous on X if and only if f−1(V )is open in X for every open set V of Y .

If f and g are continuous functions on X, then f + g, fg, and f/g(provided g(x) 6= 0 for all x ∈ X) are continuous on X.

Continuity and compactness

Theorem

Let f be a continuous mapping of a compact metric space X intoa metric space Y . Then f(X) is compact.

Theorem

If f is continuous mapping of a compact metric space X into Rk,then f(x) is closed and bounded.

Theorem

Suppose f is continuous real function on a compact metric space,and

M = supx∈X

f(x) and m = infx∈X

f(x).

Then there exists points p, q ∈ X such that f(p) = M andf(q) = m.

Special properties of real valued functions defined on R:

Theorem

Let f be a continuous real function on the interval [a, b]. Iff(a) < f(b) and if c is a number such that f(a) < c < f(b), thenthere exists a point x ∈ (a, b) such that f(x) = c.

Theorem

Let f be a monotonic function on (a, b). Then f has nodiscontinuities of the second kind, and the set of points of (a, b) atwhich f is discontinuous is at most countable.

Uniform continuity

Definition (Uniform continuity)

Let f : X → Y . The function f is uniformly continuous on X iffor every ǫ > 0 there exists a δ > 0 such that dY (f(p), f(q)) < ǫfor all p and q in X for which dX(p, q) < δ.

Theorem

A function is continuous if it is uniformly continuous.

Example : Uniformly continuous function

Example

Let f(x) =√x, x ∈ (0,∞). Then for every ǫ > 0 choose δ = ǫ2.

Then for all x, y ∈ R such that |x− y| < ǫ2,

|f(x)− f(y)|2 = |√x−√y|2 ≤ |√x−√

y||√x+√y|

= |x− y| < ǫ2.

Thus, it follows from the above inequality that |f(x)− f(y)| < ǫfor all x, y ∈ R such that |x− y| < ǫ2. Since the choice δ = ǫ2 foreach ǫ does not depend on x, y, the function f(x) =

√x is

uniformly continuous.

Example: Function that is continuous but not uniformly

continuous

Example

Let f(x) = 1/x, x ∈ (0, 2). Clearly the function f is continuous on(0, 2). Assume that f is uniformly continuous. Take ǫ = 1. Fix anarbitrary δ > 0. Then, for x := min(δ, 1) and y = x/2,

|x− y| = |x/2| < δ (1)

and|f(x)− f(y)| = |1/x − 2/x| = |1/x| ≥ 1. (2)

Since (1) and (2) hold for any δ > 0, we have that for ǫ = 1, thereexists no δ > 0 such that |f(x)− f(y)| < ǫ for all |x− y| < δ.Therefore, f(x) = 1/x is not uniformly continuous.