Schauder fixed-point theorem ($T(K)\subset K⟹\exists u\in K:T(u)=u$)

Every continuous map from a nonempty compact convex set $K$ in a Banach space to itself has a fixed point. A useful equivalent form takes a nonempty closed bounded convex set $C$ and a continuous map $T:C\to C$ whose image is relatively compact. To obtain this version from the compact-domain version, restrict $T$ to $\overline{\mathrm{conv}}T(C)\subset C$ and use closed convex hull of a compact set.

For a proof on $K$, the Schauder approximation lemma gives continuous maps $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with finite-dimensional $L_{\epsilon }$ and $\Vert T-T_{\epsilon }{\Vert }_{\infty }\le \epsilon$. The Brouwer fixed-point theorem on the nonempty compact convex set $K\cap L_{\epsilon }$ gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, so $\Vert T{x}_{\epsilon }-x_{\epsilon }\Vert \le \epsilon$. A convergent subsequence as $\epsilon \downarrow 0$ and the continuity of $T$ give an exact fixed point.

Let $G:[0,1{]}^2\to \mathbb{R}$ and $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be continuous. For a fixed $R>0$, put $K=\max |G|$ and $M={\max }_{y\in [0,1],|s|\le R}|f(y,s)|$. The integral equationhas a continuous solution with $\Vert u{\Vert }_{\infty }\le R$ whenever $|\mu |KM\le R$. The integral kernel is $G$; it is not the null space of a linear map.

Define the right side as $Tu$. Its supremum norm is at most $|\mu |KM$, so it maps the closed radius-$R$ ball in $C([0,1])$ into itself. Uniform continuity of $f$ on its compact box gives continuity of $T$. Uniform continuity of $G$ makes the image equicontinuous as well as bounded. The Arzelà-Ascoli theorem makes the image relatively compact, and the Schauder fixed-point theorem supplies a fixed point.

No uniqueness follows. For $G=1$, $f(y,s)=1+s^2$, and $0<\mu <1/2$, every solution is constant and its value satisfies $c=\mu (1+c^2)$, which has two real roots. A suitable ball captures the small root while the other lies outside; the existence proof does not rule it out.

For a smooth bounded domain and a bounded continuous scalar function $f$, this problem has a weak solution. Let $S$ be the zero-Dirichlet inverse of $-\Delta$, defined by the Lax-Milgram theorem. The Poincaré inequality gives $\Vert Sh{\Vert }_2\le C_P^2\Vert h{\Vert }_2$. The Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. The Nemytskii operator $N_f$ is continuous and has bounded image, so $T=S\circ N_f$ is a nonlinear compact operator mapping a sufficiently large closed $L^2$ ball into itself. The Schauder fixed-point theorem gives a fixed point, necessarily in $H_0^1$, and its defining identity is the desired weak formulation. This proves existence; uniqueness requires additional assumptions.

If $f$ is continuous, bounded and nonnegative on $[0,\infty )$, extend it to a bounded continuous function on the real line by the displayed formula. The bounded semilinear Dirichlet existence theorem gives $-\Delta u=F(u)$ with $u\in H_0^1$. Test with the negative part $u_{-}=\max \{-u,0\}$ to obtain $-\Vert \nabla u_{-}{\Vert }_2^2=\int F(u)u_{-}\ge 0$. The Poincaré inequality gives $u_{-}=0$, so $u\ge 0$ and $F(u)=f(u)$. Thus singularities or uncontrolled growth of the original formula at negative arguments need not prevent a nonnegative weak solution.

Let $K$ be a nonempty compact convex set in a Banach space and $T:K\to K$ a continuous map. For every $\epsilon >0$, there is a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ that approximates $T$ uniformly to within $\epsilon$ in norm.

Choose $y_1,\dots ,y_N\in T(K)$ whose open radius-$\epsilon$ balls cover the compact image $T(K)$. Set $w_i(x)=\max \{0,\epsilon -\Vert Tx-y_i\Vert \}$ andThe denominator is positive. This convex combination belongs to $K\cap \mathrm{span}\{y_1,\dots ,y_N\}$ and depends continuously on $x$. Every contributing $y_i$ has distance less than $\epsilon$ from $Tx$, so the triangle inequality gives $\Vert T_{\epsilon }(x)-Tx\Vert <\epsilon$.