Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 2 / a

For Banach spaces $X,Y$, a nonlinear compact operator $T:X\to Y$ is a continuous map that sends every bounded set to a relatively compact subset of $Y$: the closure of its image is a compact set. For a continuous map, the image condition is equivalent to requiring that every bounded sequence $(x_j)$ has a subsequence along which $T(x_j)$ converges in the norm topology of $Y$. For a linear operator, this is the usual compact operator definition; the continuity requirement is included explicitly here because the subsequent applications are nonlinear.

The compact-domain version of the Schauder fixed-point theorem saysA useful equivalent version replaces $K$ by a nonempty closed bounded convex set $C$ and assumes $T:C\to C$ is a continuous map with relatively compact image. This latter form will be used for the partial differential equation. Both forms assert existence of a fixed point, without asserting uniqueness.

Let $E=\mathrm{aff}K$ be the affine hull and put $k=\dim E\le n$. If $k=0$, $K$ is a singleton and is homeomorphic to ${\mathbb{B}}^0$. Suppose $k>0$. Choose $k+1$ affinely independent points of $K$ spanning $E$. Their simplex is contained in $K$ because $K$ is a convex set, and its barycentre $a$ has a neighbourhood in $E$ contained in that simplex. Identifying $E-a$ with ${\mathbb{R}}^k$, put $D=K-a$. For some $0<r\le R<\infty$,Here the inner inclusion follows from that neighbourhood, and the outer one follows from the compactness of $K$.

Use the Minkowski functionalThe inclusions imply $|z|/R\le p(z)\le |z|/r$. The fact that $D$ is a convex set implies $p(z+w)\le p(z)+p(w)$: if $z\in sD$ and $w\in tD$, then $(z+w)/(s+t)$ is a convex combination of points of $D$. Also $p(tz)=tp(z)$ for $t\ge 0$. ConsequentlyThus $p$ is a continuous function. Since $D$ is closed and contains zero, $D=\{p\le 1\}$: the forward inclusion is immediate, and for the reverse inclusion one takes a limit of $z/t_j\in D$ with $t_j\to p(z)\le 1$, using positive homogeneity and the fact that $D$ is a convex set. For $z\ne 0$, $p(z)>0$, so this limiting argument has no division by zero.

Define a radial continuous map byEach ray meets $D$ in exactly $\{s\theta :0\le s\le 1/p(\theta )\}$. Hence $h$ is a bijection from ${\mathbb{B}}^k$ to $K$. Away from zero its continuity follows from that of $p$, and at zero $|h(t\theta )-a|\le Rt$. A continuous map that is a bijection from a compact space to a Hausdorff space is a homeomorphism. ThereforeUsing the affine hull is essential: $K$ can have empty interior in the ambient ${\mathbb{R}}^n$.

We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.