Contraction Question Is the alpha in the contraction formula necessary:
$|T(x)-T(y)|\leq\alpha|x-y|$ where $0< \alpha<1$
 what is the problem  if written like this:
$|T(x)-T(y)|<|x-y|$ why this expression is not a contraction?
I cannot think of an example.Please help.
 A: Let $T : [1, \infty) \to [1, \infty)$ be defined by
$$T(x) = x + \frac 1 x$$
Then we have, given $x > y$ (I'll drop the absolute values since everything is positive)
$$T(x) - T(y) = x+\frac 1 x - y - \frac 1 y = (x-y) - \left(\frac 1 y - \frac 1 x\right) < x-y$$
However, we wouldn't get good results by saying that this map is a contraction mapping - for instance, it has no fixed point.
Edit to add some random rambling: A function $T : X \to X$ (where $X$ is a complete metric space) that satisfies $d(f(x), f(y)) < d(x, y)$ without a constant $\alpha$ is not called a contraction, but it's called a strictly metric map. As the example above shows, a strictly metric map does not necessarily have a fixed point. However, if $X$ is a compact space then one can show a strictly metric map has a fixed point. I believe I've seen this answered in the past here on MSE - but I'll leave this unanswered here, as it's a pretty nice exercise.
A: $T$ is a contraction if and only if it satisfies the definition of contraction. Your second example doesn't and is hence not a contraction.
More conceptually, the $\alpha$ is necessary to prevent the existence of a sequence of pairs $(x_n,y_n)$ such that
$$
\frac{|T(x_n)-T(y_n)|}{|x_n-y_n|}
$$
gets arbitrarily close to $1$. The definition with the $\alpha$ ensures that this quotient is uniformly bounded away from one.
A: We want to ensure that $|T^n(x)-T^n(y)| \le \alpha^n\,|x-y|$ tends to $0$ as $n\to\infty$.
And this is ensured by $0<\alpha<1$.
