• Support PF! Buy your school textbooks, materials and every day products Here!

Contraction, integral, maximum norm

  • Thread starter JulienB
  • Start date
  • #1
408
12

Homework Statement



Hi everybody! Here is another problem about contraction and Banach fixed-point theorem that I don't get:

The function ƒ: C([0,½]) → C([0,½]) is defined by:
[tex]
[f(x)](t) := 1 + \int_{0}^{t} x(s) ds ∀ t∈[0,\frac{1}{2}].
[/tex]

Is ƒ a contraction with respect to the norm || ⋅ ||? If yes, which function is the fixed point of ƒ?

Homework Equations



Contraction mapping theorem, Banach fixed-point theorem

The Attempt at a Solution



Well I could not really get anywhere, because I don't understand really the definition of the function. Here is what I would do anyway:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} = \mbox{max } | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds | \\
= \mbox{max } | \int_{0}^{t} x(s) - y(s) ds |
[/tex]

Then I have no idea how to integrate that since x is a function of s... What is the primitive of x(s)? Also not sure if I did the right thing in the first place as well! Some help would be very appreciated. :)

Thank you in advance for your answers.


Julien.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737

Homework Statement



Hi everybody! Here is another problem about contraction and Banach fixed-point theorem that I don't get:

The function ƒ: C([0,½]) → C([0,½]) is defined by:
[tex]
[f(x)](t) := 1 + \int_{0}^{t} x(s) ds ∀ t∈[0,\frac{1}{2}].
[/tex]

Is ƒ a contraction with respect to the norm || ⋅ ||? If yes, which function is the fixed point of ƒ?

Homework Equations



Contraction mapping theorem, Banach fixed-point theorem

The Attempt at a Solution



Well I could not really get anywhere, because I don't understand really the definition of the function. Here is what I would do anyway:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} = \mbox{max } | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds | \\
= \mbox{max } | \int_{0}^{t} x(s) - y(s) ds |
[/tex]

Then I have no idea how to integrate that since x is a function of s... What is the primitive of x(s)? Also not sure if I did the right thing in the first place as well! Some help would be very appreciated. :)

Thank you in advance for your answers.


Julien.
Looks OK so far. Do some overestimating. How big can that integral be?
 
  • #3
408
12
Looks OK so far. Do some overestimating. How big can that integral be?
@LCKurtz Hi and thx for your answer. I have no idea, really. How should I do some estimation of this integral?


Julien.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
Well, you know you can't actually perform the integration. But remember from calculus$$
\left | \int_a^b f(x)~dx \right | \le \int_a^b |f(x)|~dx$$and other properties of integration. And that integrand looks like it might be useful calculating ##\| \cdot \|_\infty##.
 
  • #5
408
12
@LCKurtz Thank you again for your answer. So I would have:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \cdot \int_{0}^{t} | x(s) - y(s) | ds
[/tex]

and

[tex]
|| x - y ||_{\infty} = max | x(t) - y(t) |
[/tex]

Right? Mm I still don't see how to compare them, the only thing I know is that t is between 0 and 1 and ƒ is continuous between 0 and 1. I'm also not sure about my || x - y ||. Shouldn't it be || x(t) - y(t) || or something?

Maybe I just don't know the inequalities of calculus well enough, I'm gonna take a close look at them during the next days.


Julien.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
@LCKurtz Thank you again for your answer. So I would have:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \cdot \int_{0}^{t} | x(s) - y(s) | ds
[/tex]

and

[tex]
|| x(s) - y(s) ||_{\infty} = max | x(t) - y(t) |
[/tex]

Right? Mm I still don't see how to compare them, the only thing I know is that t is between 0 and 1 and ƒ is continuous between 0 and 1. Maybe I just don't know the inequalities of calculus well enough, I'm gonna take a close look at them during the next days.


Julien.
You know more about ##t## than it is in ##[0,1]##, and it matters.
 
  • #7
408
12
@LCKurtz Yes indeed it is between 0 and ½. Is that what you meant?

PS: I edited my last post while you were writing I believe, sorry for that.
 
  • #8
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
Yes. Now you are close. You just need to continue a couple more inequalities to get your overestimate. It might be easier to see you way through these inequalities if you start with$$
| [f(x)](t) - [f(y)](t)| = | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds |
= | \int_{0}^{t} x(s) - y(s) ds |$$and don't worry about taking the max until you are done overestimating.
 
Last edited:
  • #9
408
12
@LCKurtz What about that:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \int_{0}^{t} | x(s) - y(s) | ds ≤ (t - 0) \cdot max |x(s) - y(s)| \\
[/tex]

Since t is between 0 and ½, ƒ is a contraction. The inequality comes from: if ƒ is bounded on [0,t] and integrable then | ƒ | is integrable and:

[tex]
\frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |
[/tex]

Is that a valid proof? Thx a lot for your help.


Julien.
 
  • #10
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
Yes. Now you are close. You just need to continue a couple more inequalities to get your overestimate.
@LCKurtz What about that:

[tex]
|| [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \int_{0}^{t} | x(s) - y(s) | ds ≤ (t - 0) \cdot max |x(s) - y(s)| \\
[/tex]

Since t is between 0 and ½, ƒ is a contraction. The inequality comes from: if ƒ is bounded on [0,t] and integrable then | ƒ | is integrable and:

[tex]
\frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |
[/tex]

Is that a valid proof? Thx a lot for your help.


Julien.
I think that is a bit sketchy, especially with ##t##'s in your answer. I had edited my previous post while you posted this. You might want to look at my previous post again.
 
  • #11
408
12
@LCKurtz Okay I'm not sure that's exactly what you had in mind but I give it another go:

[tex]
| [f(x)](t) - [f(y)](t) | = | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds | = | \int_{0}^{t} | x(s) - y(s) ds | \\
≤ \int_{0}^{t} | x(s) - y(s) | ds \\
≤ \int_{0}^{\frac{1}{2}} | x(s) - y(s) | ds \\
≤ \frac{1}{2} \cdot max | x(s) - y(s) | \\
\implies || [f(x)](t) - [f(y)](t) ||_{\infty} ≤ \frac{1}{2} \cdot || x(s) - y(s) ||_{\infty} \\
\implies \mbox{f is a contraction}
[/tex]

Very similar to what I've done last post though..


Julien.
 
  • #12
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
Yes, it is similar, but much clearer, leaving no doubt each step is correct. Good. I have to leave for a few hours now. When I return I will check and see how you are doing in finding the fixed point.
 
  • #13
408
12
@LCKurtz Thanks a lot for all this help, I appreciate it. I might search for the fixed point tomorrow as it is close to midnight here in Germany :)


Julien.
 
  • #14
408
12
Okay i couldn't wait so I gave it a go, but then I get an integral equation, which is something I have never done before:

[tex]
1 + \int_{0}^{t} x(s) ds = x(t)
[/tex]

The only thing I got out of it is:

[tex] x(0) = 1 [/tex]

That's not much I know :D I'll sleep over it and go further tomorrow.


Julien.
 
  • #15
408
12
Could it be that the fixed point function is et? I don't really know how to find it, but it seems to work:

[tex]
1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t
[/tex]

Maybe I've misunderstood how to find the fixed point though, I'm not really sure.


Julien.
 
  • #16
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
Okay i couldn't wait so I gave it a go, but then I get an integral equation, which is something I have never done before:

[tex]
1 + \int_{0}^{t} x(s) ds = x(t)
[/tex]

The only thing I got out of it is:

[tex] x(0) = 1 [/tex]

That's not much I know :D I'll sleep over it and go further tomorrow.


Julien.
Could it be that the fixed point function is et? I don't really know how to find it, but it seems to work:

[tex]
1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t
[/tex]

Maybe I've misunderstood how to find the fixed point though, I'm not really sure.


Julien.
Yes, you have found the fixed point by inspection, but your equation in post #14$$
x(t)=1 + \int_{0}^{t} x(s) ds$$will lead you directly to it. Try differentiating both sides with respect to ##t## to change it from an integral equation to a differential equation and see if you can get it from there.
 
  • Like
Likes JulienB
  • #17
408
12
[tex]
1 + \int_{0}^{t} x(s) ds = x(t) \\
\frac{d}{dt} (1 + \int_{0}^{t} x(s) ds) = \frac{d}{dt} (x(t)) \\
0 + \frac{d}{dt} \int_{0}^{t} x(s) ds = \dot{x}(t) \\
\mbox{According to the fundamental theorem of calculus: } \\
x(t) = \dot{x}(t) \\
\implies x_*(t) = e^t
[/tex]

Pretty neat! Thx a lot @LCKurtz, I've learned a lot through this post! Note that x* is the notation my teacher uses for fixed point.

Julien.
 
  • #18
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737
You're welcome. One tiny oversight, you would get ##x_*(t) = Ce^t##, but as you observed earlier, ##x_*(0) = 1##. You need that.
 
  • #19
408
12
@LCKurtz Thanks a lot, that makes sense!
 

Related Threads on Contraction, integral, maximum norm

Replies
9
Views
800
Replies
4
Views
3K
Replies
10
Views
3K
Replies
0
Views
4K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
1
Views
905
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
9
Views
631
  • Last Post
Replies
12
Views
3K
Top