### Archive

Posts Tagged ‘fixed point’

## Shrinkable polygons

Here’s a nice problem inspired from a post on MathOverflow: link

We call a polygon shrinkable if any scaling of itself with a factor ${0<\lambda<1}$ can be translated into itself. Characterize all shrinkable polygons.

It is easy to see that any star-convex polygon is shrinkable. Pick the point ${x_0}$ in the definition of star-convex, and any contraction of the polygon by a homothety of center ${x_0}$ lies inside the polygon.

## Agreg 2012 Analysis Part 4

A Fixed Point Theorem

This parts wishes to extend the next result (which can be used without proof) to infinite dimension.

Theorem. (Browuer) Consider ${F}$ a finite dimensional normed vector space. Consider ${C\subset F}$ a convex, closed, bounded non-void set. If ${f:C\rightarrow C}$ is a continuous application, then ${f}$ has a fixed point in ${C}$.

1. In ${\ell^2(\Bbb{N})}$ endowed with ${\|\cdot \|_2}$ we consider the following application:

$\displaystyle f: B(0,1) \subset \ell^2(\Bbb{N}) \rightarrow \ell^2(\Bbb{N})$

$\displaystyle f(x) =(\sqrt{1-\|x\|_2},x_0,x_1,...).$

Prove that ${f}$ is continuous with values in the unit sphere of ${\ell^2(\Bbb{N})}$, but ${f}$ does not admit any fixed points.

2. Consider ${E}$ a normed vector space, ${B}$ a closed, bounded non-void subset of ${E}$ and ${f:B \rightarrow E}$ a compact application (not necessarily linear). (a compact application maps bounded sets into relatively compact sets)

i) Let ${n \in \Bbb{N}\setminus \{0\}}$. We can cover ${\overline{f(B)}}$ (which is compact) by a finite number ${N_n}$ of open balls of radius ${\frac{1}{n}}$: ${\overline{f(B)} \subset \displaystyle \bigcup_{i=1}^{N_n} \mathring{B}(y_i,\frac{1}{n})}$ with ${y_i \in \overline{f(B)}}$ for every ${i}$. For ${y \in E}$ we define

$\displaystyle \psi(y) =\begin{cases} \frac{1}{n}-\|y-y_i\| & \text{ if } y \in B(y_i,\frac{1}{n}) \\ 0 & \text{ otherwise} \end{cases}$

Prove that ${\Psi : y \in \overline{f(B)} \mapsto \sum_{i=1}^{N_n} \psi_i(y)}$ is continuous and that there exists ${\delta>0}$ such that for ${y \in \overline{f(B)}}$ we have ${\Psi(y)\geq \delta}$.

ii) We introduce the application ${f_n : B \rightarrow E}$ defined by

$\displaystyle f_n(x)= \left( \sum_{i=1}^{N_n} \psi_i(f(x))\right)^{-1} \sum_{i=1}^{N_n} \psi_i(f(x))y_i.$

Prove that for every ${x \in B}$ we have ${\|f(x)-f_n(x) \|\leq \frac{1}{n}}$.

Categories: Analysis, Fixed Point

## Darboux Functions with no Iterate Fixed Points

It is well known that there exist functions ${f:[0,1] \rightarrow [0,1]}$ which have Darboux property and they have no fixed points. An example can be found in an earlier post of mine. Here is a generalization of that result.

There exist functions ${f:[0,1] \rightarrow [0,1]}$ which have Darboux property and for which none of its iterates has a fixed point, i.e. ${\underbrace{f\circ ..\circ f}_{n \text{ times}}(x)\neq x}$ for every ${x \in [0,1]}$ and for every ${n \geq 1}$.

## Brouwer fixed point theorem

Let $D$ be an open subset of $\Bbb{R}^n$ such that $\overline{D}$ is homeomorphic to the closed unit ball $\overline{B}$ ( in $\Bbb{R}^n$). If $\phi \in C(\overline{D})$ (continuous function on $\overline{D}$) and $\phi(\overline{D})\subset\overline{D}$, then $\phi$ has a fixed point in $\overline{D}$.

Categories: Analysis, Fixed Point Tags: ,

## The Existence of a Triangle with Prescribed Angle Bisector Lengths

August 9, 2010 1 comment

Prove that for any $m,n,p>0$ there exists a unique triangle (up to an isometry) such that $m,n,p$ are the lengths of bisectors of this triangle
Solved by L. Panaitopol & P. Mironescu in 1994, AMM 101, 58-60

## Fixed point for an operator

Suppose $X$ is a Hilbert space and $T$ is a linear operator on $X$ with $\| T\| \leq 1$. Prove that $Tx=x$ if and only if $T^* x =x$.
$\| x \| ^2 =$ $\langle x, x \rangle = \langle Tx,x \rangle= \langle x,T^*x \rangle \leq \|x \| \| T^* x\| \leq \|x\|^2$. Since this implies equality in the C-B inequality, we must have $T^* x= \lambda x,\ \lambda\geq 0$. We easily find that $\lambda=1$. The converse is equivalent to the implication above.
Suppose $K \subset \mathbb{R}^n$ is a compact convex set and $f: K \to K$ is a function satisfying $\| f(x)-f(y) \| \leq \|x -y\|,\ \forall x,y \in K$. Prove that $f$ has a fixed point ( i.e. $\exists x \in K$ with $f(x)=x$). Read more…