Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 1

The Banach fixed-point theorem states that a contraction mapping $T:X\to X$ of a nonempty complete metric space $(X,d)$ has a unique fixed point, and iterations of a map $x_{n+1}=T(x_n)$ converge to it from every $x_0\in X$. Precisely, assume a single constant $q\in [0,1)$ satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y\in X$.

Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and the geometric series therefore give, for $m>n$,Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x_{\ast }\in X$. The contraction mapping is a continuous map, so $T{x}_{\ast }={\lim }_{n}T{x}_n={\lim }_n{x}_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing $x_{\ast }=y_{\ast }$. Letting $m\to \infty$ in the estimate also gives the contraction iteration error bound:If $q=0$, $T$ is constant and the sequence reaches its unique fixed point after the first step; this also removes any ambiguity about $0^0$ in the displayed estimate at $n=0$.

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

The retraction of a Hilbert ball onto its sphere exists. In the real Hilbert space ${\ell }^2$, let $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$ be the unilateral shift operator, and let $e_0=(1,0,\dots )$. For $x$ in the closed unit ball $\mathbb{B}$, defineThe orthogonality of $e_0$ and $Sx$ gives $\Vert F(x){\Vert }^2=1$. Thus $F$ is a continuous map into the unit sphere. It has no fixed point: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\ge 0$. A constant sequence in ${\ell }^2$ must be zero, whereas $F(0)=e_0\ne 0$.

Put $d=x-F(x)$, which is nonzero everywhere, and defineThis is a continuous map on the whole closed unit ball, including zero. To check its range, observe that $⟨F(x),d⟩=⟨F(x),x⟩-1$, soIf $\Vert x\Vert =1$, then $\Vert x-F(x){\Vert }^2=2(1-⟨F(x),x⟩)$, hence $\alpha (x)=1$ and $r(x)=x$. Therefore $r:\mathbb{B}\to \partial \mathbb{B}$ is a retraction.

Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the unit sphere. In factThe construction illustrates why the finite-dimensional hypothesis in the no-retraction theorem matters. The closed unit ball of ${\ell }^2$ is not a compact set in the norm topology.

Let $S_{+}^{n-1}$ be the nonnegative portion of the unit sphere specified in the paper. If $Ax=0$ for some $x\in S_{+}^{n-1}$, then this nonzero vector is an eigenvector with eigenvalue zero, which already proves the assertion.

Otherwise $Ax\ne 0$ for every $x\in S_{+}^{n-1}$, andis a continuous map from $S_{+}^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a nonnegative matrix and every entry of $x$ is nonnegative. By the supplied homeomorphism $h:{\mathbb{B}}^{n-1}\to S_{+}^{n-1}$, the continuous map $h^{-1}Th$ maps the finite-dimensional closed unit ball into itself. The Brouwer fixed-point theorem supplies a fixed point, hence $T(x_{\ast })=x_{\ast }$ for some $x_{\ast }\in S_{+}^{n-1}$. ThereforeTogether with the zero case this proves that $A$ has a nonnegative eigenvalue, with a nonnegative eigenvector. For $n=1$, the auxiliary closed unit ball is the one-point space ${\mathbb{B}}^0$ and the argument remains valid. No claim of strict positivity of every entry of the eigenvector is needed.