21-356 Lecture 2

# 21-356 Lecture 2 - Wednesday 2 Elementary Geometry of RN A...

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2E l e m e n t a r yG e o m e t r y o f R N Avector v ! R N of norm one is called a direction .G iv enapo in t x 0 ! R N and ad irect ion v ,the line through x 0 in the direction v is given by L := ! x ! R N : x = x 0 + t v ,t ! R " . Given two points y , z ! R N with y " = z line through y and z is given by L := ! x ! R N : x = t y +(1 # t ) z ! R " , while the segment joining y and z is given by S := ! x ! R N : x = t y # t ) z ! [0 , 1] " . Aset E \$ R N is convex if for every x , y ! E ,thesegmentjo ing x and y is contained in E . Remark 3 Given a point x 0 ! R N and r> 0 ball centered at x 0 and of radius r is the set B ( x 0 ,r ):= ! x ! R N : % x # x 0 % <r " . Let’s prove that the ball B ( x 0 ) is convex. Consider two points x , y ! B ( x 0 ) and let t ! [0 , 1] . We want to prove that t x # t ) y belongs to B ( x 0 ) .W r i te x 0 = t x 0 # t ) x 0 . Then we have % t x # t ) y # x 0 % = % t x # t ) y # ( t x 0 # t ) x 0 ) % = % t ( x # x 0 )+(1 # t )( y # x 0 ) % &% t ( x # x 0 ) % + % (1 # t y # x 0 ) % = t % x # x 0 % # t ) % y # x 0 % <tr # t ) r = r, where we have used the triangle inequality and the positive homogeneity of the norm. This shows that the segment joining x and y is contained in B ( x 0 ) .

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2-bis Continuous functions in R N Let E R n ,let f : E R and let x 0
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21-356 Lecture 2 - Wednesday 2 Elementary Geometry of RN A...

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