[PDF] Continuous Functions on Metric Spaces

Continuous Functions

We leave it as an exercise to prove that these definitions are equivalent. Note that c must belong to the domain A of f in order to define the continuity.

Continuity of Functions

A function f is continuous at x0 in its domain if for every sequence (xn) with xn in To show a function is continuous we can do one of three things:.

Continuous Functions on Metric Spaces

Theorem 21. A continuous function on a compact metric space is bounded and uniformly continuous. Proof. If X is a compact metric space and f : 

? ? ? ? ? ? ? ?

(a) Show that f is continuous at. 0. x = (b) For. 0 x ? express ( ). f x. ? as a piecewise-defined function. Find the value of x for which ( ).

Chapter 5. Integration §1. The Riemann Integral Let a and b be two

j=1Fj. The function f is continuous on F which is a finite union of Proof. Let us first show that f2 is integrable on [a

7.17 If f is continuous on [0 1] and if ? 1 f(x)xn dx = 0

1

4.3 Let f be a continuous real function on a metric space X. Let Z(f

zero set of f. Proof Since Z(f) = f?1({0}) and {0} is a closed set in R

Solutions to Assignment-3

x0 /? E. Show that there is an unbounded continuous function f : E ? R. uniform continuity without actually using showing the dependence of ? on ?

MATH 314 Assignment #6 1. Let f be a continuous function from IR

The Intermediate Value Theorem will be used in the following problems. (a) Show that the equation 2x = 3x has a solution c ? (14). Proof . Let g( 

4.20 Assume that f is a continuous real function defined in (a b

Show that the compactness of Y cannot be omitted from the hypotheses even when. X and Z are compact. Proof (a) Suppose f is not uniformly continuous. Then

Lecture 5 : Continuous Functions De nition 1 f a f x f a x a

x where nis a positive integer then f(x) is continuous on the interval [0;1) We can use symmetry of graphs to extend this to show that f(x) is continuous on the interval (1 ;1) when nis odd Hence all n th root functions are continuous on their domains Trigonometric Functions In the appendix we provide a proof of the following Theorem :

Class diary for Math 311:01 spring 2003

First let us show that a continuous extension of thefunctionf to the set is unique (assuming it exists) Suppose ghSince the set ? isdense inE for any c??E0 converging toc Sincec we get g(xn)?g(c) and However g(xn) =h(xn) =f(xn) for all : E are two continuous extensions of f E0 sequence {xn} continuous at ? ? Hence g(c) =h(c) there is a

Solutions to Assignment-3 - University of California Berkeley

Clearly F is continuous on (a;b):To prove continuity at a let fx ngbe a sequence in (a;b) converging to a We need to show that F(x n) = f(x n) !F(a) = A Let ">0 There exists N 1such that for all n>N 1 jA f(a n)j< " 2 : 4 The proof will be complete if we can show that for nlarge enough jf(x n) f(a n)jcan be made smaller than "=2

Assignment-10 - University of California Berkeley

1 Consider the sequence of functions f n(x) = xnon [0;1] (a)Show that each function f nis uniformly continuous on [0;1] Solution: Any continuous function on a compact set is uniformly continuous (b)For a sequence of functions f n on [0;1] write what it means for them to NOT be an equicontinuous sequence

Searches related to show that f is continuous on filetype:pdf

to show that feis continuous at 0 We need to show that if (x n) is a sequence in (0;1] such that x n!0 then fe(x n) !fe(0) which is equivalent to that f(x n) !0 From x n!0 we get x2 n!0 Since jsin(1=x n)j 1 we have jf(x n)j= jx2n sin(1 xn)j jx nj2 By squeeze lemma we have f(x n) !0 as desired Thus feis continuous on [0;1] Thus

How to show that f is not continuous at 0?

That means to show that f is not continuous at 0, it is sufficient to exhibit one sequence (x n) which converges to 0 for which the sequence (f (x n) does not converge to f (0)=1. So the suggestion was take x n =1/n.

Is the function f (x) continuous everywhere?

Consider f:[?1,3]?R, f(x)= ! 2x, ?1? x ?1 3?x,1

Is f uniformly continuous?

If f (x)=x and the domain is R, then f is uniformly continuous. (We can take delta=epsilon.) The range of this f is all of R, an unbounded set. If f (x)=sin (1/x) and the domain in (0,1), then f is not uniformly continuous.

Is F a continuous curve?

If f is continuous on [ 1, 2] (i.e., its graph can be sketched as a continuous curve from ( 1, ? 10) to ( 2, 5)) then we know intuitively that somewhere on [ 1, 2] f must be equal to ? 9, and ? 8, and ? 7, ? 6, …, 0, 1 / 2, etc. In short, f takes on all intermediate values between ? 10 and 5.