# Contraction, integral, maximum norm

1. Apr 29, 2016

### JulienB

1. The problem statement, all variables and given/known data

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:
$$[f(x)](t) := 1 + \int_{0}^{t} x(s) ds ∀ t∈[0,\frac{1}{2}].$$

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

2. Relevant equations

Contraction mapping theorem, Banach fixed-point theorem

3. 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:

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

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. :)

Julien.

2. Apr 29, 2016

### LCKurtz

Looks OK so far. Do some overestimating. How big can that integral be?

3. Apr 29, 2016

### JulienB

@LCKurtz Hi and thx for your answer. I have no idea, really. How should I do some estimation of this integral?

Julien.

4. Apr 29, 2016

### LCKurtz

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. Apr 29, 2016

### JulienB

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

and

$$|| x - y ||_{\infty} = max | x(t) - y(t) |$$

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. Apr 29, 2016

### LCKurtz

You know more about $t$ than it is in $[0,1]$, and it matters.

7. Apr 29, 2016

### JulienB

@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. Apr 29, 2016

### LCKurtz

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: Apr 29, 2016
9. Apr 29, 2016

### JulienB

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

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:

$$\frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |$$

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

Julien.

10. Apr 29, 2016

### LCKurtz

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. Apr 29, 2016

### JulienB

@LCKurtz Okay I'm not sure that's exactly what you had in mind but I give it another go:

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

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

Julien.

12. Apr 29, 2016

### LCKurtz

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. Apr 29, 2016

### JulienB

@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. Apr 29, 2016

### JulienB

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:

$$1 + \int_{0}^{t} x(s) ds = x(t)$$

The only thing I got out of it is:

$$x(0) = 1$$

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

Julien.

15. Apr 29, 2016

### JulienB

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

$$1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t$$

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

Julien.

16. Apr 29, 2016

### LCKurtz

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.

17. Apr 30, 2016

### JulienB

$$1 + \int_{0}^{t} x(s) ds = x(t) \\ \frac{d}{dt} (1 + \int_{0}^{t} x(s) ds) = \frac{d}{dt} (x(t)) \\ 0 + \frac{d}{dt} \int_{0}^{t} x(s) ds = \dot{x}(t) \\ \mbox{According to the fundamental theorem of calculus: } \\ x(t) = \dot{x}(t) \\ \implies x_*(t) = e^t$$

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. Apr 30, 2016

### LCKurtz

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. May 1, 2016

### JulienB

@LCKurtz Thanks a lot, that makes sense!