1617-257/TUT-R-8: Difference between revisions

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'''Problem 2'''. Let <math>f: B_1(0) \to \mathbb{R}^2</math> be a function which is "jelly-rigid": for all <math>x,y \in B_1(0)</math>: <math>|f(x) - f(y) - (x - y)| \leq 0.1 |x - y|</math>. Prove that <math>f</math> maps onto <math>B_{0.4}(0)</math>.
'''Problem 2'''. Let <math>f: B_1(0) \to \mathbb{R}^2</math> be a function which is "jelly-rigid": for all <math>x,y \in B_1(0)</math>: <math>|f(x) - f(y) - (x - y)| \leq 0.1 |x - y|</math>. Prove that <math>f</math> maps onto <math>B_{0.4}(0)</math>.


''Proof''. Since <math>f</math> is Lipschitz, it has a unique extension to its boundary and so we regard it as a function on <math>B:= \overline{B_1(0)}</math>. A simple estimate shows if <math>f(x) \in B_{0.4}(0)</math>, then <math>x \in B_1(0)</math>. Suppose now that there is some point <math>z \in B_{0.4}(0)</math> which is not in the image of <math>f</math>. Let <math>x_0 \in B</math> be a closest element to <math>z</math> in the image of <math>f</math>. Consider now the point <math>x_1 := x_0 + \delta (z - f(x_0))</math> (jelly-rigidity says that the function is almost like the identity, so moving closer to the point <math>z</math> from <math>f(x_0)</math> in the codomain side can be obtained by moving in that same direction on the domain side first) where <math>\delta > 0</math> is chosen to be small enough so that <math>x_1 \in B_1(0)</math> and <math>0< \delta < 1</math>.
''Proof''. Since <math>f</math> is Lipschitz, it has a unique extension to its boundary and so we regard it as a function on <math>B:= \overline{B_1(0)}</math>. A simple estimate shows if <math>f(x) \in B_{0.4}(0)</math>, then <math>x \in B_1(0)</math>. Suppose now that there is some point <math>z \in B_{0.4}(0)</math> which is not in the image of <math>f</math>. Let <math>x_0 \in B_1(0)</math> be a closest element to <math>z</math> in the image of <math>f</math>. Consider now the point <math>x_1 := x_0 + \delta (z - f(x_0))</math> (jelly-rigidity says that the function is almost like the identity, so moving closer to the point <math>z</math> from <math>f(x_0)</math> in the codomain side can be obtained by moving in that same direction on the domain side first) where <math>\delta > 0</math> is chosen to be small enough so that <math>x_1 \in B_1(0)</math> and <math>0< \delta < 1</math>. Then <math>f(x_1)</math> is closer to <math>z</math> than is <math>f(x_0)</math>.

Latest revision as of 15:36, 7 November 2016

On 11/3/16, we discussed some questions from the exam:

Problem 1. Let be an infinite subset of a compact metric space . Show that has a limit point.

Proof. If has no limit points, then is a closed subset of a compact space and is therefore compact in itself. Since each point of is isolated, we may find for each point a neighborhood such that . The collection is an open cover of E which clearly has no finite subcover.

Problem 2. Let be a function which is "jelly-rigid": for all : . Prove that maps onto .

Proof. Since is Lipschitz, it has a unique extension to its boundary and so we regard it as a function on . A simple estimate shows if , then . Suppose now that there is some point which is not in the image of . Let be a closest element to in the image of . Consider now the point (jelly-rigidity says that the function is almost like the identity, so moving closer to the point from in the codomain side can be obtained by moving in that same direction on the domain side first) where is chosen to be small enough so that and . Then is closer to than is .