Paper 2, Section II, F

Topics in Analysis | Part II, 2009

(a) State Brouwer's fixed point theorem in the plane.

(b) Let a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c} be unit vectors in R2\mathbb{R}^{2} making 120120^{\circ} angles with one another. Let TT be the triangle with vertices given by the points a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} and let I,J,KI, J, K be the three sides of TT. Prove that the following two statements are equivalent:

(1) There exists no continuous function f:TTf: T \rightarrow \partial T with f(I)I,f(J)Jf(I) \subseteq I, f(J) \subseteq J and f(K)Kf(K) \subseteq K.

(2) If A,B,CA, B, C are closed subsets of R2\mathbb{R}^{2} such that TABC,IA,JBT \subseteq A \cup B \cup C, I \subseteq A, J \subseteq B and KCK \subseteq C, then ABCA \cap B \cap C \neq \emptyset.

(c) Let f,g:R2Rf, g: \mathbb{R}^{2} \rightarrow \mathbb{R} be continuous positive functions. Show that the system of equations

(1x2)f2(x,y)x2g2(x,y)=0(1y2)g2(x,y)y2f2(x,y)=0\begin{aligned} &\left(1-x^{2}\right) f^{2}(x, y)-x^{2} g^{2}(x, y)=0 \\ &\left(1-y^{2}\right) g^{2}(x, y)-y^{2} f^{2}(x, y)=0 \end{aligned}

has four distinct solutions on the unit circle S1={(x,y)R2:x2+y2=1}\mathbb{S}^{1}=\left\{(x, y) \in \mathbb{R}^{2}: x^{2}+y^{2}=1\right\}.

Typos? Please submit corrections to this page on GitHub.