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
f(x_2) - f(x_1) = \int_{x_1}^{x_2} f&#039;(t) \, dt \leq \int_{x_1}^{x_2} m \, dt = m (x_2 - x_1)
and
f(x_2) - f(x_1) = \int_{x_1}^{x_2} f&#039;(t) \, dt \geq \int_{x_1}^{x_2} (-m) \, dt = -m (x_2 - x_1)
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
f(x_2) - f(x_1) = \int_{x_1}^{x_2} f&#039;(t) \, dt \leq \int_{x_1}^{x_2} m \, dt = m (x_2 - x_1)
and
f(x_2) - f(x_1) = \int_{x_1}^{x_2} f&#039;(t) \, dt \geq \int_{x_1}^{x_2} (-m) \, dt = -m (x_2 - x_1)
Thus, $|f(x_2) - f(x_1)| \leq m |x_2 - x_1|$.

That works too. Thanks.