Please help me to prove the problem! and show all your work A function f : D rig

Question

Please help me to prove the problem! and show all your work

A function f : D rightarrow R. D R. is called monotone increasing if f(x) f(y) whenever x,y D and x y. Now let f :R rightarrow R be monotone increasing and continuous. Choose some r R and define a sequence (xn) inductively by x1 = r and xn+1 = f(xn) for n N. Show that one of the following occurs: (xn) diverges to +infinity. (xn) diverges to - infinity. (xn) converges to some c R with f(c) = c. Remark: The element c in (c) is called a fixed point of f. Hint: Study the cases f(r) r and f(r) r. Show that (xn) is monotone. Let f : R rightarrow R be continuous and D be a dense subset of R. Assume that f(x) = 1 for all x D. Show that f(x) = 1 for all x R.

Explanation / Answer

f(x) is monotonically increasing

x1=r

x2=f(x1)

x3=f(x2)

There are 3 possibilities:

1. x2>x1

=> f(x2)>=f(x1) //reason is f is monotonically increasing function

=> x3>=x2

2. x2<x1

=> f(x2)<=f(x1) //reason is f is monotonically increasing function

=> x3<=x2

3. x2=x1

=> f(x2)=f(x1)

=> x3=x2

Suppose we are given:

xk

x(k+1)=f(xk)

x(k+2)=f(x(k+1)

There are 3 possibilities:

1. x(k+1)>xk

=> f(x(k+1))>=f(xk) //reason is f is monotonically increasing function

=> x(k+2)>=x(k+1)

2. x(k+1)<xk

=> f(x(k+1))<=f(xk) //reason is f is monotonically increasing function

=> x(k+2)<=x(k+1)

3. x(k+1)=xk

=> f(x(k+1))=f(xk)

=> x(k+2)=x(k+1)

So if x2> x1 => x3>x2 => x4>x3 .........=>x(k+1) > xk ....

As k->infinity, xk->infinity

So if x2<x1 => x3<x2 => x4<x3 .........=>x(k+1) < xk ....

As k->infinity, xk->negative infinity

So if x2= x1 => x3=x2 => x4=x3 .........=>x(k+1) =xk....

As k->infinity, xk->x1

SO xn either converges to positive infinity or negative infinity or a constant c = x1=r

A set D is called dense subset of R if for any x in R, within every neighbourhood of x there exists an element of D.

Suppose f(x) is not 1 for all x in R.

Suppose f(x)=some k not equal to 1

Within a small neighbourhood of x (x-a,x+a), there exists a y such that y is in D.

we know that f(y)=1

As a->0, (x-a,x+a)->x

Even in such a small neighbourhood, there exists a y such that y is in D and f(y)=1

But f(x)=k not equal to 1

This means that the function is not continuous at x

Which is a contradiction

So f(x) is 1 for all x in R

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---