Proving Contractive Homework: Tips & Solutions

[JasMath33]

Homework Statement

I worked on this question and I made it so far, and now I am stuck on how to finish it. Here is the problem and below I will explain what I attempted.

Homework Equations

The Attempt at a Solution

I know looking at the last part about using previous homework, I want to prove that the function is contractive. If I can prove that, I can use a previous assignment which says contractive functions have a fixed point. Here is what I have in my proof so far.

I am stuck now and not completely sure where to go next. Any advice is appreciated.

 

[fresh_42]

I don't understand what your last line in the picture means. Nevertheless, are you allowed to apply Banach's fixed-point theorem?

 

[JasMath33]

I don't understand what your last line in the picture means. Nevertheless, are you allowed to apply Banach's fixed-point theorem?

Yes I could use that theorem. Would I just need to completely erase everything I did and start over?

I was getting the last part from using the previous homework problems below.

 

[fresh_42]

Yes I could use that theorem. Would I just need to completely erase everything I did and start over?

So you are already done. The other exercise (Lipschitz continuity for ##f## on the condition on ##f'##) showed you that ##b## is your Lipschitz constant. Since ##b<1, \; f## is a contraction and the fixed-point theorem applies.

 

[JasMath33]

What other exercise are you talking about? I am lost on that statement. I understand the reasoning afterwards.

 

[JasMath33]

So you are already done. The other exercise (Lipschitz continuity for ##f## on the condition on ##f'##) showed you that ##b## is your Lipschitz constant. Since ##b<1, \; f## is a contraction and the fixed-point theorem applies.

I think you are looking at the proof I found in the book and asked about. I get it now.

 

[fresh_42]

Mean value theorem: ##|f(x) - f(y)| = f'(t) |x-y| < b |x-y|##, i.e. ##f## is a contraction because ##b<1##.

 

[JasMath33]

Mean value theorem: ##|f(x) - f(y)| = f'(t) |x-y| < b |x-y|##, i.e. ##f## is a contraction because ##b<1##.

Thanks I see it now. I forgot I could use that. Thanks.

 

[Ray Vickson]

Thanks I see it now. I forgot I could use that. Thanks.

Even easier: just use very elementary methods. If ##|f'(t)| \leq m## on ##R## (or an an interval ##[a,b]##), that means that ##-m \leq f'(t) \leq m##. Thus, for ##x_1 < x_2## we have
[tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \leq \int_{x_1}^{x_2} m \, dt = m (x_2 - x_1) [/tex]
and
[tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \geq \int_{x_1}^{x_2} (-m) \, dt = -m (x_2 - x_1) [/tex]
Thus, ##|f(x_2) - f(x_1)| \leq m |x_2 - x_1|##.

 

[JasMath33]

Even easier: just use very elementary methods. If ##|f'(t)| \leq m## on ##R## (or an an interval ##[a,b]##), that means that ##-m \leq f'(t) \leq m##. Thus, for ##x_1 < x_2## we have
[tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \leq \int_{x_1}^{x_2} m \, dt = m (x_2 - x_1) [/tex]
and
[tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \geq \int_{x_1}^{x_2} (-m) \, dt = -m (x_2 - x_1) [/tex]
Thus, ##|f(x_2) - f(x_1)| \leq m |x_2 - x_1|##.

That works too. Thanks.