Is a Continuous and Periodic Function Bounded and Uniformly Continuous on R?

[steven187]

hello all,

i have been working on problems with continuity and i have come across some question in which i understand generally what i have to do but i just don't know where to start and how to put it together

a function f:R->R is said to be periodic if there exists a number k>0 such that
f(x+k)=f(x) for all x an element of R. suppose that f:R->R is continuous and periodic. Prove that f is bounded and uniformly continuous on R.

also

let f:R-->R be a function which satisfies the conditions
f(x+y)=F(x)+f(y)
and
f(-x)=-f(x) for al x znd y an element of R
suppose that f is continuous at 0 show that f is continuous at every point in R

please help

Steven

 

[HallsofIvy]

Smells like homework. First problem is peculiar in that proving "bounded" is far simpler than proving "uniformly continuous". If you are working with "uniformly continuous" you should long ago have seen the proof that any continuous function (you don't need "periodic") is bounded. In order to prove uniformly continuous you will need to use the fact that a function continuous on a compact set (here, closed and bounded) is uniformly continuous. Do you already know that? How does "periodic" help you there?

For the second problem, I don't see why they include "f(-x)=-f(x) for all x and y an element of R". That follows from "f(x+ y)= f(x)+ f(y)". Any way: What is the DEFINITION of "continuous at 0" ?? Now what conditions must be met to prove "continuous at a"?

 

[steven187]

continuity

your right every continuous function is bounded i just thought that there would be more to it to actually show it, and in terms of the uniform continuity so you are saying anythin on a closed interval and bounded and continuous will prove that f is uniformly continuous ,i can see that but we havnt actually used anythin about the periodic function, without the periodic function concept the question does not sound that complicated if i had to use the concept of the periodic function then from my understanding so far is that if i prove that it is bounded and uniformly continuous on an interval [0,k] which is the length of the period then i would be able to prove it for all intervals say [nk,nk+k] (where n is a positive intiger) since it is a periodic function does that sound right,

about the second problem, do you mean : if we choose epsilon>0 then there will be delta such that !x!<delta then !f(x)!<epsilon but i don't know how that follows on to prove that it is continuous at a also

 

[quasar987]

$$\forall \epsilon >0 , \ \exists \delta \ \ s.t. \ \ |x|<\delta \Rightarrow |f(x) - f(0)|< \epsilon$$

This is the hypothesis. You got to use that to show that

$$\forall \epsilon >0 , \ \exists \delta \ \ s.t. \ \ |x-a|<\delta \Rightarrow |f(x) - f(a)|< \epsilon$$

I suggest making the change of variable x - a = y. Then what we want to show now is that

$$\forall \epsilon >0 , \ \exists \delta \ \ s.t. \ \ |y|<\delta \Rightarrow |f(y+a) - f(a)|< \epsilon$$

but f(y+a) = f(y)+f(a), and f(a) = f(a+0) = f(a) + f(0), so

$$|f(y+a) - f(a)|< \epsilon \Leftrightarrow |f(y) + f(a) - f(a) - f(0)|< \epsilon \Leftrightarrow |f(y) - f(0)|< \epsilon$$

But this is precisely the statement of the hypothesis. So it's true and we won.