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Posts Tagged ‘fixed point’

Shrinkable polygons

October 7, 2014 Leave a comment

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.

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Agreg 2012 Analysis Part 4

October 20, 2013 Leave a comment

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}}.

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Darboux Functions with no Iterate Fixed Points

June 3, 2012 Leave a comment

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}.

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Brouwer fixed point theorem

February 19, 2011 Leave a comment

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}.

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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
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Fixed point for an operator

January 12, 2010 Leave a comment

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.
Solution: Using the Cauchy-Buniakovski inequality, we get
\| 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.

Fixed point theorem

January 12, 2010 Leave a comment

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…

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