




























































































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Given two sets A and B a function f from A to B assigns an element of B to each element of A. Thus, every function comes “equipped,”.
Typology: Lecture notes
1 / 244
This page cannot be seen from the preview
Don't miss anything!
Keith Ball
Chapter 1. Continuous Functions
Continuity
Given two sets A and B a function f from A to B assigns an element of B to each element of A. Thus, every function comes “equipped,” with two sets: A, the set of points where the function is defined, and B, a set to which the values of the function belong.
We can draw attention to these sets by writing
f : A → B.
if x < y < z and x and z both belong to the subset then so does y. So an interval contains all the points between its ends but the ends themselves may or may not be included.
Examples are
{x : a ≤ x ≤ b} = [a, b] a closed interval {x : a < x < b} = (a, b) an open interval {x : a ≤ x < b} = [a, b) a half-open interval {x : a ≤ x} = [a, ∞) a half-infinite interval
For this course we will be considering real valued functions whose domains are intervals of the real line. We want to say what it means for such a function to be continuous.
At each point of the domain there are essentially three types of problem that can occur. In each of the cases below, the function is discontin- uous at 0 (but continuous elsewhere).
0
In the third situation f (x) oscillates infinitely often over a wide range as x approaches 0.
We shall define what it means for a function to be continuous at a point in such a way as to rule out these problems. Then we say that a function is continuous on an interval if it is continuous at each point of the interval.
I containing the number c is said to be continuous at c if for every ε > 0 , there is a δ > 0 so that if x ∈ I and |x − c| < δ then
|f (x) − f (c)| < ε.
The function is continuous on I if it is continuous at each point of I.
Example (Continuity of x 7 → x). The function x 7 → x is continuous at every point of the real line.
Proof If f (x) = x then we can always take δ = ε since if |x − c| < ε then
|f (x) − f (c)| = |x − c| < ε.
As an exercise, you can check that constant functions are continuous.
Example (A discontinuity). The function f given by
f (x) =
{ 1 if x > 0 0 if x ≤ 0 is discontinuous at 0.
Proof We need to find a band around f (0) = 0 which we cannot guarantee to land in, merely by starting near 0. A band of width ε = 1/2 will do because if x is positive, however close to zero, f (x) = 1 and this is not within 1/2 of f (0).
We will do one more example which will also follow from our later arguments but will help to fix the ideas: the function x 7 → x^2.
The aim will be to show that if x − c is small then so is x^2 − c^2. Now x^2 − c^2 = (x − c)(x + c) so if |x − c| < δ then |x^2 − c^2 | < δ|x + c| which looks good because the factor δ is small. But the second factor |x + c| might be large (if c = 1, 000 , 000 say).
However, if x is close to c then x + c is close to 2c which may be large but is a fixed number. So we shall cook up δ to compensate: roughly we want δ = ε/(2|c|). However we need to be a bit careful because x + c isn’t actually equal to 2c.
Example (Continuity of x 7 → x^2 ). The function x 7 → x^2 is continuous at every point of the real line.
|x^2 − c^2 | < ε
by choosing x close enough to c. We know that |x^2 −c^2 | = |x−c|.|x+c|.
If we choose |x − c| < 1 then |x| cannot be larger than |c| + 1 and so |x + c| cannot be larger than 2|c| + 1.
We wish to check the continuity of functions such as polynomials, more complicated than x 7 → x^2. To do this we want a machine which allows us to build continuous functions from simpler ones.
We want to know that when you add or multiply continuous functions the result is still continuous and also when you compose two continuous functions.
There are a number of ways to do this but in view of what you already know about limits there is a particularly simple approach which depends upon rewriting the continuity property in terms of limits of sequences.
interval I and suppose that c ∈ I. Then f is continuous at c if and only if for every sequence (xn) of points in I which converges to c,
f (xn) → f (c) as n → ∞.
Proof Suppose first that f is continuous at c and xn → c. Then, given ε > 0 we can find δ > 0 so that if |x − c| < δ then |f (x) − f (c)| < ε.
Now choose N so that if n > N , |xn − c| < δ.
Then if n > N we have |f (xn) − f (c)| < ε. So f (xn) → f (c).
The sequential continuity theorem immediately gives us the algebra of continuous functions.
defined on the interval I and continuous at c ∈ I. Then
Proof The proofs of all 3 are essentially the same. We shall do the first.
We wish to show that if xn → c then (f + g)(xn) = f (xn) + g(xn) → f (c) + g(c) as n → ∞.
But we know that f (xn) → f (c) and g(xn) → g(c) so we can apply the properties of limits of sequences to conclude that f (xn) + g(xn) → f (c) + g(c).
From this we can immediately conclude that polynomials are contin- uous and that rational functions are continuous except where the de- nominator is zero.