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Geometrically this implies that the images of any two points \(x,y\) are closer together than the points themselves.
**Observation 6**.: Let \(I=[a,b]\) or \([a,b)\) for \(b<\infty\) and \(f:I\to I\). If
\(\alpha=\sup\limits_{x\in I}|f^{\prime}(x)|<1\), then \(f\) is a contraction on \(I\). To see this observe that for any \(x,y\in I\), by the Mean Value Theorem, we have for some \(\zeta\) between \(x\) and \(y\)
\[f(x)-f(y)=f^{\prime}(\zeta)(x-y).\]
Since we know that \(\alpha=\sup\limits_{x\in I}|f^{\prime}(x)|<1\), we have
\[|f(x)-f(y)|=|f^{\prime}(\zeta)||x-y|\leq\alpha|x-y|.\]
### Banach Fixed Point Theorem
The following theorem called the Banach Fixed Point Theorem, Banach Contraction Mapping Theorem, or simply the Contraction Mapping Theorem is one of the jewels of analysis. It is an existence and uniqueness theorem for fixed points for a class of mappings, called contractions. The strength of it comes from the fact that it has a constructive proof. In other words, it gives a **constructive** procedure for obtaining better and better approximations to the fixed point. This procedure, as we mentioned before, is called an iteration process. Iteration procedures are used in nearly every branch of applied mathematics, and convergence proofs and error estimates are very often obtained by an application of Banach’s Fixed Point Theorem. In the following we also present some fundamental applications of this powerful theorem.
**Theorem 83** (Banach Fixed Point Theorem or Contraction Mapping Theorem).: _Let \(X=(X,d)\) be a complete metric space. Then any contraction \(f:X\to X\) 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 \(f\) and \(f\) has no further fixed points. We will do this proof in steps.
* Choose an arbitrary point \(x_{0}\in X\). Define a sequence \(\{x_{n}\}\) inductively as follows: \[x_{n+1}=f(x_{n}).\] So we have \[x_{1} = f(x_{0})\] \[x_{2} = f(x_{1})=f(f(x_{0}))=f^{2}(x_{0})\] \[\vdots\] \[x_{n} = f^{n}(x_{0})\] \[\vdots\] where \(f^{n}\) is the \(n\)th composition of \(f\). * Since \(f\) is continuous (why?), we know from the sequential characterization of continuity that \(f(x_{n})\to f(x^{*})\). So we have \[f(x^{*})=\lim_{n\to\infty}f(x_{n})=\lim_{n\to\infty}x_{n+1}=x^{*}\implies f(x^{*} )=x^{*}.\] Thus we have found a fixed point. **Remark 44** (Error Estimate).: In the proof of the Banach Fixed Point Theorem we are given the method to construct a fixed point as the limit of a sequence \(\{x_{n}\}\). If we take \(n\to\infty\) we have the estimate for the error \[d(x_{m},x)\leq\frac{k^{m}}{1-k}d(x_{1},x_{0}),\] where \(k<1\) is the contraction constant. ### Applications In the following several applications of the Banach Fixed Point Theorem are given. ### Applications to Solutions of the Equation \(f(x)=x\) **1.** Finding a root of the equation \(e^{-x}=x\). We see that a root of \(f(x)=e^{-x}\) is in \([\frac{1}{2},1]\) because for \(x\in[\frac{1}{2},1]\), we have \(|f^{\prime}(x)|=|-e^{-x}|=e^{-x}\leq e^{-\frac{1}{2}}=0.6065<1\). This implies that \(f\) is a contraction on \([\frac{1}{2},1]\) and that the iteration \(x_{n}=f(x_{n-1})\) has a limit. So we will solve this by employing an iterative scheme. Let us start with \(x_{0}=0.5\); then we have that \[x_{1} = f(x_{0})=e^{-0.5}=0.6065\] \[x_{2} = f(x_{1})=e^{-0.6065}=0.5452\] \[\vdots\] Continuing this iteration scheme we obtain,Note that, in the above example, we have \(x=0.567\) is a fixed point correct to three decimal places. Starting with some arbitrary point \(x_{0}\), we can explore this visually as shown by the following Figure 2.10 in which the iterative sequence approaches the fixed point that is the solution to the equation. **2.** Solve the equation \(x^{3}+x-1=0\). We let \(F(x)=x^{3}+x-1\) and observe that \[F\left(\frac{1}{2}\right) = -\frac{3}{8}<0\] \[F\left(\frac{3}{4}\right) = \frac{11}{64}>0,\] which implies that the root is in \([\frac{1}{2},\frac{3}{4}]\). We have some possibilities to solve for \(x\): \[x=1-x^{3}=f(x), \tag{2.1} \]\] \[x=\frac{1}{1+x^{2}}=g(x), \tag{2.2} \]\] \[x=\frac{\sqrt{x}}{\sqrt{1+x^{2}}}=h(x). \tag{2.3} \]\] So, we will check the derivatives of each of these functions. For (2.1), \[|f^{\prime}(x)|=|-3x^{2}|=3x^{2}>1.\] Figure 2.10: Iterations This is close to the solution and the iteration sequence \(\{x_{n}\}\) will not be convergent. Now we take (2.2), \[|g^{\prime}(x)|=\left|-\frac{2x}{(1+x^{2})^{2}}\right|=\frac{2x}{(1+x^{2})^{2}} \leq\frac{2\cdot\frac{3}{4}}{(1+\frac{1}{4})^{2}}=\frac{24}{25}<1.\] This means that \(x_{n}=g(x_{n-1})\) is convergent. Now we look at (2.3), notice that \(|h^{\prime}(x)|<\frac{4}{9}\). This implies that \(x_{n}=h(x_{n-1})\) converges even faster. Note that we have special methods such as Newton's method, the square root method, etc. to solve equations. Thus, in general if we have \(f:[a,b]\to[a,b]\) and \(\alpha\) such that \(\alpha=\sup\limits_{x\in[a,b]}|f^{\prime}(x)|<1\), then the equation \(f(x)=x\) has a unique solution \(\overline{x}\in[a,b]\) which can be approximated by iterations \((x_{0},x_{1},\dots)\) where \(x_{n+1}=f(x_{n})\), and with the error estimate given by \[|x_{n+1}-\overline{x}|\leq\frac{\alpha}{1-\alpha}\,|x_{n+1}-x_{n}|\leq\frac{ \alpha^{n}}{1-\alpha}|x_{1}-x_{0}|.\] ### 2.2 Applications to Systems of Algebraic Linear Equations Let \(T:\mathbb{R}^{n}\to\mathbb{R}^{n}\) be given by \[\vec{y}=A\vec{x}+\vec{b},\] where \(A=(a_{ij})\) and \(B=(b_{i})\) are fixed. We can write this as \[\begin{pmatrix}y_{1}\\ y_{2}\\ \vdots\\ y_{n}\end{pmatrix}=\begin{pmatrix}a_{11}&a_{12}&\dots&a_{1n}\\ a_{21}&a_{22}&\dots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\dots&a_{nn}\end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ \vdots\\ x_{n}\end{pmatrix}+\begin{pmatrix}b_{1}\\ b_{2}\\ \vdots\\ b_{n}\end{pmatrix}. \tag{2.4} \]\] We write this in a more compact form as \[y_{i}=\sum_{j=1}^{n}a_{ij}x_{j}+b_{i},\quad i=1,2,\dots,n.\] To solve the equation \(\vec{y}=A\vec{x}+\vec{b}\) apply the Banach Fixed Point Theorem; but for this theorem, we need a complete metric space and a contraction on it. We already know \(\mathbb{R}^{n}\) is complete metric space, so whether the mapping \[y=Tx=Ax+b\] is a contraction on \((\mathbb{R}^{n},d)\) depends on the metric \(d\) as illustrated in the following. * Suppose for \((\mathbb{R}^{n},d)\) the metric \(d(\cdot,\cdot)\) given by \[d(x,z)=\max_{1\leq i\leq n}|x_{i}-z_{i}|.\]We have that \[d(y^{\prime},y^{\prime\prime}) = \max_{i}|y^{\prime}_{i}-y^{\prime\prime}_{i}|=\max_{i}\left|\sum_{j= 1}^{n}a_{ij}x^{\prime}_{j}+b_{i}-\left(\sum_{j=1}^{n}a_{ij}x^{\prime\prime}_{j} +b_{i}\right)\right|\] \[= \max_{i}\left|\sum_{j=1}^{n}a_{ij}(x^{\prime}_{j}-x^{\prime\prime }_{j})\right|\leq\max_{i}\sum_{j=1}^{n}|a_{ij}||x^{\prime}_{j}-x^{\prime\prime }_{j}|\] \[\leq \max_{i}\sum_{j=1}^{n}|a_{ij}|\max_{j}|x^{\prime}_{j}-x^{\prime \prime}_{j}|=\max_{i}\sum_{j=1}^{n}|a_{ij}|d(x^{\prime},x^{\prime\prime}).\] Thus, using the Banach Fixed Point Theorem we arrive at the following theorem: **Theorem 84** (Row Sum Criterion).: _If a system_
* We will show that this sequence is Cauchy. Observe that \[d(x_{n+1},x_{n} = d(f(x_{n}),f(x_{n-1}))\leq kd(x_{n},x_{n-1})=kd(f(x_{n-1}),f(x_{n -2}))\] \[\leq k^{2}d(x_{n-1},x_{n-2})\leq\cdots\leq k^{n}d(x_{1},x_{0}),\] where we have used the fact that \(f\) is a contraction with contraction constant \(k<1\). Thus if \(n>m\), we have that \[d(x_{m},x_{n}) \leq d(x_{m},x_{m+1})+d(x_{m+1},x_{m+2})+\cdots+d(x_{n-1},x_{n})\] \[\leq(k^{m}+k^{m+1}+\cdots+k^{n-1})d(x_{1},x_{0})\] \[= [k^{m}(1+k+\cdots+k^{n-m-1})]\,d(x_{1},x_{0}).\] We now use a well-known identity for \(r<1\), \[1+r+r^{2}+\cdots+r^{n}=\frac{1-r^{n+1}}{1-r},\] to conclude \[k^{m}(1+k+\cdots+k^{n-m-1})=k^{m}\frac{1-k^{n-m}}{1-k}.\] Since we know that \(n>m\) we have \[0
* Now we will prove uniqueness. Suppose there is another fixed point \(y\) such that \(f(y)=y\). Observe \[d(y,x^{*})=d(f(y),f(x^{*}))\leq k\,d(y,x^{*}),\] \[\Rightarrow(1-k)d(y,x^{*})\leq 0\] since we have a contraction, we must have \(d(y,x^{*})=0\), therefore \(y=x^{*}\).
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