Proving $T:X\to X$ is not a Contraction on $X=[1, \infty)$

[ozkan12]

Let $X=[1,\infty)$ and $T:X\to X$. Define $T=x+\frac{1}{x}$...Please show that T is not a contraction

 

[Ackbach]

I'm a little confused.
$$T'(x)=1-\frac{1}{x^2} <1,$$
showing that $T$ is a contraction. Are you using a different metric? Something exotic?

 

[ozkan12]

Dear,

How this shows that T is a contraction ? I don't understand...We use usual metric..

Also, I want to show that T is not a contraction

 

[Ackbach]

If the derivative is strictly smaller than 1 everywhere in the space, this suggests that the mapping is a contraction. The criteria for a contraction mapping is that $|T(x)-T(y)| \le k |x-y|$, where $0\le k <1$. Now, if we assume $x\not=y$, this is like saying that
$$\frac{|T(x)-T(y)|}{|x-y|} \le k.$$
The LHS, you can see, looks a lot like a derivative. By the Mean Value Theorem, if the derivative is always strictly less than 1, then the LHS (slope of a secant line) must always be less than 1. Now I will say this: the value of $k$, it seems to me, is not constant on the entire space. I'm not sure you could have a particular $k$ that works everywhere. But my intuition tells me that you're never going to be able to produce a counterexample for the basic inequality of $|T(x)-T(y)| \le k |x-y|$. That's what you'd have to do to show that $T$ is not a contraction.

[EDIT] See below for a correction.

 

[ozkan12]

Dear,

https://books.google.com.tr/books?i...HeoJAxE#v=onepage&q=x+1/x contraction&f=false

in this link, after lemma, there is an example...Can you examine this ?

 

[Ackbach]

Yes, I see now where you're going with this. I was too hasty in my previous post. The mapping $T$ isn't a contraction, even though it always maps two elements closer together than they were before. And that's because there isn't any one $0\le k<1$ such that the slopes of secant lines are always less than or equal to $k$. But you have to go to the limit to get a counterexample, which makes sense. $T$ is about as close to a contraction as you can get without actually being one.

 

[ozkan12]

Dear,

I didnt understand...How can you explain this by limit ?

 

[johng1]

Here's a link to information on the contraction mapping principle -- https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.
By this theorem, if your map were a contraction, there would be a unique fixed point. But of course the equation $x=x+1/x$ clearly has no solutions. If you read the information in the link, your function is actually briefly discussed.