MHB Contraction constant in banach contraction principle

[ozkan12]

İn some fixed point theory books, I saw an expression...But I didnt understand what this mean...Please can you help me ?

" It was important in the proof of banach contraction principle that the contraction constant "h" be strictly less than 1. Than gave us control over the rate of convergence of ${f}^{n}\left({x}_{0}\right)$ to the fixed point since ${h}^{n}\to0$ as $n\to\infty$. If we consider f is contractive mapping instead of a contraction, then we lose that control and indeed a fixed point need not exist.

How "h" be strictly less than 1 gave control over the rate of convergence of ${f}^{n}\left({x}_{0}\right)$ to the fixed point ?

How we lose control if we consider f is contractive mapping instead of a contraction ?

Please can you explain these questions ? Thank you so much...Best wishes...

 

[Ackbach]

İn some fixed point theory books, I saw an expression...But I didnt understand what this mean...Please can you help me ?

" It was important in the proof of banach contraction principle that the contraction constant "h" be strictly less than 1. Than gave us control over the rate of convergence of ${f}^{n}\left({x}_{0}\right)$ to the fixed point since ${h}^{n}\to0$ as $n\to\infty$. If we consider f is contractive mapping instead of a contraction, then we lose that control and indeed a fixed point need not exist.

How "h" be strictly less than 1 gave control over the rate of convergence of ${f}^{n}\left({x}_{0}\right)$ to the fixed point ?

In the proof of the Banach Fixed Point Theorem, at least, the fact that $h<1$ is required in a particular step of the proof. If $h=1$, you don't get convergence of the sequence generated, as you mentioned above.

How we lose control if we consider f is contractive mapping instead of a contraction ?

Please can you explain these questions ? Thank you so much...Best wishes...

So, the main difference, as I see it, between a contraction and a contractive mapping, is that with a contraction $A$,
$d(Ax,Ay)$ is strictly bounded away from $d(x,y)$ via the contraction constant, whereas with a contractive mapping $B$, $d(Bx,By)$ can be arbitrarily close to $d(x,y)$. While it's true that $d(Bx,By)<d(x,y)$, there's nothing preventing the LHS from being arbitrarily close to the RHS. Consequently, the convergence of the sequence of iterates can be much slower, and you might lose it altogether. As a result, guaranteeing a fixed point for a contractive mapping requires more assumptions. Here's an illustrative paper proving some fixed point theorems regarding contractive mappings. You can see that you need more hypotheses than with a contraction.

 

[ozkan12]

Dear professor,

What is LHS and RHS ?

Also, how "Consequently, the convergence of the sequence of iterates can be much slower, and you might lose it altogether." ?

And why If h=1, we don't get convergence of the sequence generated ?

 

[Ackbach]

LHS = Left Hand Side, and RHS = Right Hand Side.

From pages 300 to 302 in Introductory Functional Analysis with Applications, by Erwin Kreyszig:

5.1-2 Banach Fixed Point Theorem (Contraction Theorem).
Consider a metric space $X=(X,d),$ where $x\not= \varnothing$. Suppose that $X$ is complete and let $T:X \to X$ be a contraction on $X$. Then $T$ has precisely one fixed point.

Proof. We construct a sequence $(x_n)$ and show that it is Cauchy, so that it converges in the complete space $X$, and then we prove that its limit $x$ is a fixed point of $T$ and $T$ has no further fixed points. This is the idea of the proof.

We choose any $x_0 \in X$ and define the "iterative sequence" $(x_n)$ by
$$(2) \qquad x_0, \quad x_1=Tx_0, \quad x_2=Tx_1=T^2x_0, \quad \cdots, \quad x_n=T^n x_0, \quad \cdots.$$
Clearly, this is the sequence of the images of $x_0$ under repeated application of $T$. We show that $(x_n)$ is Cauchy. By [the definition of contraction with contraction constant $0\le \alpha<1$] and (2),
\begin{align*}
d(x_{m+1},x_m)&=d(Tx_m,Tx_{m-1}) \\
&\le \alpha d(x_m,x_{m-1}) \\
&= \alpha d(Tx_{m-1},Tx_{m-2}) \\
&\le \alpha^2 d(x_{m-1},x_{m-2}) \\
\cdots &\le \alpha^m d(x_1,x_0).
\end{align*}
Hence by the triangle inequality and the formula for the sum of a geometric progression we obtain for $n>m$
\begin{align*}
d(x_m,x_n)&\le d(x_m,x_{m+1})+d(x_{m+1},x_{m+2})+\cdots +d(x_{n-1},x_n) \\
&\le (\alpha^m+\alpha^{m+1}+\cdots+\alpha^{n-1}) \, d(x_0, x_1) \\
&=\alpha^{m} \, \frac{1-\alpha^{n-m}}{1-\alpha} \, d(x_0, x_1).
\end{align*}
Since $0\le \alpha <1$, in the numerator we have $1-\alpha^{n-m}<1$. Consequently,
$$(4) \qquad d(x_m,x_n)\le \frac{\alpha^m}{1-\alpha} \, d(x_0,x_1) \qquad (n>m).$$
On the right, $0\le \alpha<1$ and $d(x_0,x_1)$ is fixed, so that we can make the right-hand side as small as we please by taking $m$ sufficiently large (and $n>m$). This proves that $(x_m)$ is Cauchy. Since $X$ is complete, $(x_m)$ converges, say, $x_m \to x$. We show that this limit $x$ is a fixed point of the mapping $T$.

And the proof continues. As you can see, we use the fact that $\alpha<1$ to obtain the fact that the geometric series converges. It is simply a fact that a geometric series diverges if $\alpha=1$, in this context. As the proof hinges on this fact, not having $d(Tx,Ty)$ strictly bounded away from $d(x,y)$ has important implications for whether the operator $T$ has a fixed point. I suppose it might have a fixed point, anyway, but the Banach Fixed Point Theorem would not guarantee it.