Proof. This is Brouwer’s fixed point theorem, and it is quoted rather than proved here: every known proof needs machinery from algebraic topology — degree theory, homology, or Sperner’s lemma — which is outside the tools of this course. It is stated because the next theorem and Picard’s theorem both rest on it, and it would be dishonest to present a proof that does not exist at this level.

The idea, at least, can be given. Suppose a continuous \(f:S\rightarrow S\) had no fixed point. Then \(f(x)\neq x\) for every \(x\), so the ray from \(f(x)\) through \(x\) meets the boundary sphere in a well-defined point \(r(x)\), and \(r\) is continuous. On the boundary \(r(x)=x\). So \(r\) would be a continuous retraction of the ball onto its boundary sphere — and the content of the topology is precisely that no such retraction exists, because the ball can be continuously shrunk to a point and the sphere cannot.

For \(n=1\) the theorem is elementary: a continuous \(f:[-1,1]\rightarrow [-1,1]\) has \(g(x)=f(x)-x\) with \(g(-1)\geq 0\) and \(g(1)\leq 0\), so \(g\) vanishes somewhere by the intermediate value theorem. □

Proof. Let \(A\) be a compact convex subset of a Banach space \(X\) and \(f:A\rightarrow A\) continuous. The argument reduces the statement to the finite dimensional case, where Brouwer’s theorem applies.

Fix \(\varepsilon >0\). Since \(A\) is compact it is totally bounded, so there are finitely many points \(a_1,\dots ,a_m\in A\) whose \(\varepsilon \)-balls cover \(A\). Let \(A_\varepsilon \) be the convex hull of \(\{a_1,\dots ,a_m\}\): it lies in \(A\), because \(A\) is convex, and it sits inside the finite dimensional subspace spanned by those points.

Define the Schauder projection \(P_\varepsilon :A\rightarrow A_\varepsilon \) by \[P_\varepsilon (x)=\frac {\sum _{i=1}^{m}\lambda _i(x)\,a_i}{\sum _{i=1}^{m}\lambda _i(x)}, \qquad \lambda _i(x)=\max \big (0,\ \varepsilon -||x-a_i||\big ).\] The denominator never vanishes, since every \(x\) is within \(\varepsilon \) of some \(a_i\), and \(P_\varepsilon \) is continuous. Moreover \(||P_\varepsilon (x)-x||\leq \varepsilon \), because \(P_\varepsilon (x)\) is a convex combination of points within \(\varepsilon \) of \(x\).

Now \(P_\varepsilon \circ f\) maps the compact convex finite dimensional set \(A_\varepsilon \) continuously into itself, so by Brouwer’s theorem it has a fixed point \(x_\varepsilon \). Then \[||f(x_\varepsilon )-x_\varepsilon || =||f(x_\varepsilon )-P_\varepsilon \big (f(x_\varepsilon )\big )||\leq \varepsilon .\]

Take \(\varepsilon =\frac 1n\) and let \(x_n\) be the corresponding points. By compactness of \(A\) a subsequence converges, say \(x_{n_k}\rightarrow x\in A\). Continuity of \(f\) and \(||f(x_{n_k})-x_{n_k}||\leq \frac {1}{n_k}\rightarrow 0\) give \(f(x)=x\). □

Proof. Let \(x_0\in X\) and define \(x_1=f(x_0)\), \(x_2=f^2(x_0)=f(x_1)\) and in general \(x_n=f^n(x_0)=f(x_{n-1})\). If \(m<n\) then \begin {align*} d(x_m,x_n) &=d(f^m(x_0),f^n(x_0))\\ &=d(f^m(x_0),f^mf^{n-m}(x_0))\\ &\leq r^md(x_0,f^{n-m}(x_0))\\ &=r^md(x_0,x_{n-m})\\ &\leq r^m[d(x_0,x_1)+d(x_1,x_2)+\cdots +d(x_{n-m-1},x_{n-m})]\\ &\leq r^md(x_0,x_1)[1+r+\cdots +r^{n-m-1}]\\ &<\frac {r^md(x_0,x_1)}{1-r}\rightarrow 0\hspace {0.2cm}as\hspace {0.2cm} m\rightarrow \infty \hspace {0.2cm}\text {since}\hspace {0.2cm} r<1. \end {align*}

Thus \(\{x_n\}^{\infty }_{n=1}\) is Cauchy in \(X\) and since \(X\) is complete, \(\{x_n\}^{\infty }_{n=1}\) converges to \(x\in X\). Now \begin {align*} f(x) &=f\Big (\lim _{n\rightarrow \infty }x_n\Big )\\ &=\lim _{n\rightarrow \infty }f(x_n), \hspace {0.2cm}\text {since it is continuous.}\\ &=\lim _{n\rightarrow \infty }(x_n+1)\\ &=x \end {align*}

i.e \(f(x)=x\).

Now, suppose that \(\exists y\in X\) such that \(f(y)=y\), then \(d(x,y)=d(f(x),f(y))\leq rd(x,y)\), (\(f\) is a contraction). Since \(r<1\implies d(x,y)=0\) or \(x=y\). □

Proof. Since \(f(x,y)\) and \(\dfrac {\partial f}{\partial y}\) are continuous in a closed rectangle \(R\), they are bounded. Thus,

\(\forall (x,y)\in R,\exists \) constants \(K\) and \(M\) such that \(|f(x,y)|\leq K\) and \(|\partial f(x,y)/\partial y|\leq M\).

Let \((x,y_1), (x,y_2)\in R\). Then by the mean value theorem, for some \(\theta \in (0,1)\) we have \[|f(x,y_1)-f(x,y_2)|=|y_1-y_2|\Big |\frac {\partial }{\partial y}f(x,y_1+\theta (y_2-y_1))\Big |\leq M|y_1-y_2|\qquad (2)\] Now if \(y=g(x)\) satisfies equation \((1)\) and has the property that \(g(x_0)=y_0\), then integrating equation \((1)\) from \(x_0\) to \(x\) gives \(\displaystyle {g(x)-g(x_0)=\int ^x_{x_0}f(t,g(t))dt}\) \[g(x)=y_0+\int ^x_{x_0}f(t,g(t))dt\qquad (3)\] Conversely, if \(y=g(x)\) satisfies \((3)\), then \(g(x_0)=y_0\), and differentiating \((3)\) gives back \((1)\).
It therefore suffices to show that an equivalent problem, the integral equation \((3)\) has a unique solution.

Now choose \(a>0\ni Ma<1\) and \(R'=\{(x,y):|x-x_0|\leq a\hspace {0.3cm} and\hspace {0.3cm} |y-y_0|\leq Ka\}\in R\). Let \(X\) be the set of all continuous real functions \(y=g(x)\) defined on \(|x-x_0|\leq a\) such that \(|g(x)-y_0|\leq Ka\). Then \(X\) is a complete metric space being a closed subspace of the complete metric space \(C[x_0-a,x_0+a]\).
Next, define \(T:X\rightarrow X\) by \(Tg=h\) where \(\displaystyle {h(x)=y_0+\int ^x_{x_0}f(t,g(t))dt}.\) It then follows from equation \((3)\) and \((2)\) that \[\Big |h_1(x)-h_2(x)\Big |=\Bigg |\int ^x_{x_0}[f(t,g_1(t))-f(t,g_2(t))]dt\Bigg |\leq Ma\sup |g_1(x)-g_2(x)|,\] and since \(Ma<1\), this implies that \(T\) is a contraction. Therefore, by the contraction mapping theorem, \(Tg=g\) has a unique solution, that is, the integral equation \((3)\) has a unique solution.
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