Contraction, integral, maximum norm

[JulienB]

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

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:

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

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.

 

[LCKurtz]

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

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:

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

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?

 

[JulienB]

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

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

 

[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 $\| \cdot \|_\infty$.

 

[JulienB]

@LCKurtz Thank you again for your answer. So I would have:

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

and

<br /> || x - y ||_{\infty} = max | x(t) - y(t) |<br />

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 going to take a close look at them during the next days.Julien.

 

[LCKurtz]

@LCKurtz Thank you again for your answer. So I would have:

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

and

<br /> || x(s) - y(s) ||_{\infty} = max | x(t) - y(t) |<br />

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 going to 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.

 

[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.

 

[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.

 

[JulienB]

@LCKurtz What about that:

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

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:

<br /> \frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |<br />

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

 

[LCKurtz]

Yes. Now you are close. You just need to continue a couple more inequalities to get your overestimate.

@LCKurtz What about that:

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

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:

<br /> \frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |<br />

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.

 

[JulienB]

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

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

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

 

[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.

 

[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.

 

[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:

<br /> 1 + \int_{0}^{t} x(s) ds = x(t)<br />

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.

 

[JulienB]

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

<br /> 1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t<br />

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

 

[LCKurtz]

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:

<br /> 1 + \int_{0}^{t} x(s) ds = x(t)<br />

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.

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

<br /> 1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t<br />

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.

 

[JulienB]

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

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.

 

[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.

 

[JulienB]

@LCKurtz Thanks a lot, that makes sense!