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>. |
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''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 |
''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>. |
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Latest revision as of 16:36, 7 November 2016
On 11/3/16, we discussed some questions from the exam:
Problem 1. Let [math]\displaystyle{ E }[/math] be an infinite subset of a compact metric space [math]\displaystyle{ X }[/math]. Show that [math]\displaystyle{ E }[/math] has a limit point.
Proof. If [math]\displaystyle{ E }[/math] has no limit points, then [math]\displaystyle{ E }[/math] is a closed subset of a compact space and is therefore compact in itself. Since each point of [math]\displaystyle{ E }[/math] is isolated, we may find for each point [math]\displaystyle{ e \in E }[/math] a neighborhood [math]\displaystyle{ U_e }[/math] such that [math]\displaystyle{ E \cap U_e = \{ e \} }[/math]. The collection [math]\displaystyle{ \{ U_e\}_{e \in E} }[/math] is an open cover of E which clearly has no finite subcover.
Problem 2. Let [math]\displaystyle{ f: B_1(0) \to \mathbb{R}^2 }[/math] be a function which is "jelly-rigid": for all [math]\displaystyle{ x,y \in B_1(0) }[/math]: [math]\displaystyle{ |f(x) - f(y) - (x - y)| \leq 0.1 |x - y| }[/math]. Prove that [math]\displaystyle{ f }[/math] maps onto [math]\displaystyle{ B_{0.4}(0) }[/math].
Proof. Since [math]\displaystyle{ f }[/math] is Lipschitz, it has a unique extension to its boundary and so we regard it as a function on [math]\displaystyle{ B:= \overline{B_1(0)} }[/math]. A simple estimate shows if [math]\displaystyle{ f(x) \in B_{0.4}(0) }[/math], then [math]\displaystyle{ x \in B_1(0) }[/math]. Suppose now that there is some point [math]\displaystyle{ z \in B_{0.4}(0) }[/math] which is not in the image of [math]\displaystyle{ f }[/math]. Let [math]\displaystyle{ x_0 \in B_1(0) }[/math] be a closest element to [math]\displaystyle{ z }[/math] in the image of [math]\displaystyle{ f }[/math]. Consider now the point [math]\displaystyle{ 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]\displaystyle{ z }[/math] from [math]\displaystyle{ f(x_0) }[/math] in the codomain side can be obtained by moving in that same direction on the domain side first) where [math]\displaystyle{ \delta \gt 0 }[/math] is chosen to be small enough so that [math]\displaystyle{ x_1 \in B_1(0) }[/math] and [math]\displaystyle{ 0\lt \delta \lt 1 }[/math]. Then [math]\displaystyle{ f(x_1) }[/math] is closer to [math]\displaystyle{ z }[/math] than is [math]\displaystyle{ f(x_0) }[/math].
