Continuous function and limits

[dancergirlie]

Homework Statement

Consider a function f : R--> R, and assume that there is a c is in (0, 1) so that
|f(x) - f(y)|<= c|x -y|
for all x, y in R.
(a) Show that f is continuous on R.
(b) Given a point y1 in R define a sequence by yn+1 = f(yn). Prove that yn is a Cauchy sequence
(and therefore convergent).
(c) Let y = lim yn. Prove that f(y) = y.
(d) Prove that the limit y is independent of the choice of y1, i.e. any sequence defined by choosing an x1 and defining xn+1 = f(xn) converges to the y from part (c).

Homework Equations

The Attempt at a Solution

a) Well this is what I tried, I'm not too sure if it is right:

Let epsilon>0 and choose delta=epsilon/c so that
0<|x-y|<delta implies that
|f(x)-f(y)|<=c|x-y|<c*delta=epsilon
Since this holds for all x, y in R, that means that the function is continuous on R

b) Not too sure how to do this, but I know that for epsilon > 0 you need to pick an N in N so that for m, n >= N:

|yn-ym|< epsilon

c)No idea what to do,
by definition of a limit of a sequence: for all epsilon>0
|yn-y|< epsilon

d) Need part c to do this one...

 

[lanedance]

Homework Statement

Consider a function f : R--> R, and assume that there is a c is in (0, 1) so that
|f(x) - f(y)|<= c|x -y|
for all x, y in R.
(a) Show that f is continuous on R.
(b) Given a point y1 in R define a sequence by yn+1 = f(yn). Prove that yn is a Cauchy sequence
(and therefore convergent).
(c) Let y = lim yn. Prove that f(y) = y.
(d) Prove that the limit y is independent of the choice of y1, i.e. any sequence defined by choosing an x1 and defining xn+1 = f(xn) converges to the y from part (c).

Homework Equations

The Attempt at a Solution

a) Well this is what I tried, I'm not too sure if it is right:

Let epsilon>0 and choose delta=epsilon/c so that
0<|x-y|<delta implies that
|f(x)-f(y)|<=c|x-y|<c*delta=epsilon
Since this holds for all x, y in R, that means that the function is continuous on R

looks reasonable to me, note the defintion of this function, and i think what you have shown is in fact uniform continuity, which in turn implies pointwise continuity

b) Not too sure how to do this, but I know that for epsilon > 0 you need to pick an N in N so that for m, n >= N:

|yn-ym|< epsilon

i think you're on the right track, try substituting for abitrary m,n & using the inequality

As the real numbers are complete, once you have shown it is Cauchy, you in fact have shown it has a limit within R.

The next part should follow on

c)No idea what to do,
by definition of a limit of a sequence: for all epsilon>0
|yn-y|< epsilon

d) Need part c to do this one...

 

[dancergirlie]

alright, I think I got part b, but I'm still stuck on part c.

for part c: I got that I am assuming that the lim(yn)=y and thus for all epsilon>0 there exists an n in N so that for n>=N

|yn-y|<epsilon
which is equivalent to
y-epsilon<yn<y+epsilon

I know that |f(x)-f(y)|<=c|x-y|
Is that the piece of information that I should use or should I use the fact that
yn+1=f(yn)??

 

[lanedance]

wait so have you shown the sequence is Cauchy for part b? It should pretty much follow on from that for c)

for the Cauchy part, notice that because of the defintion of the function, things cascade... ie

|y_{n} - y_{m}| = |f(y_{n-1}) - f(y_{m-1})| = \leq c|y_{n-1}-y_{m-1}| = \leq c^2|y_{n-2}-y_{m-2}|...

 

[dancergirlie]

That is where I get confused though because we are looking at
|yn-y|. So is y suposed to be y1, because there is no subscript so that is where I'm getting confused. So that would mean:

|yn-y|=|f(yn-1)-f(y0)|?

 

[lanedance]

no, y is the limit of the sequence y = (lim_{n \rightarrow \infty}) y_n

so... if you show the sequence is Cauchy, then you know it has a limit in R as R is complete. As the sequence has a limit then it should follow pretty easily

y= (lim_{n \rightarrow \infty})y_n

f(y) =(lim_{n \rightarrow \infty})f(y_n ) = (lim_{n \rightarrow \infty})y_{n+1}


never quite worked out how to write those limits correctly in tex...