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Unformatted text preview: Analysis II Tutorial Solution sketches Alexandre Tomberg * 1 January 20, 2011 1.1 Let a,b be real numbers, and define C 1 ([ a,b ]) = { f : [ a,b ] R  f exists and is continuous } . On this space of all continuously differentiable functions, we define a norm k f k = max a t b (  f ( t )  +  f ( t )  ) . Prove that the pair ( C 1 ([ a,b ]) , kk ) is a real normed vector space. Proof. For a space to be a vector space, it needs to satisfy 8 axioms. However, if you remember the fact that the space of all functions f : [ a,b ] R is a vector space, we only need to check the three axioms of a subspace, namely that 1. f,g C 1 ([ a,b ]), f + g C 1 ([ a,b ]); 2. f C 1 ([ a,b ]) and R , f C 1 ([ a,b ]); These are practically obvious. Now, we have k f k = max a t b (  f ( t )  +  f ( t )  ) =   max a t b (  f ( t )  +  f ( t )  ) ; k f + g k = max a t b (  f ( t ) + g ( t )  +  f ( t ) + g ( t )  ) max a t b (  f ( t )  +  f ( t )  ) + max a t b (  g ( t )  +  g ( t )  ) ; These relations prove two of the axioms of a norm. As k f k = max a t b (  f ( t )  +  f ( t )  ) max a t b (  f ( t )  ), k f k = 0 forces f = 0, which completes the proof. * Please report any errors and typos to atomberg@math.mcgill.ca. All tutorial solutions in one pdf can be found at http://www.cs.mcgill.ca/ ~ atombe2/AnalysisTutorial.pdf . 1 1.2 Let X = R n , equipped with the usual inner product h i . Consider the unit sphere in X , S = { x X  k x k = 1 } , where k x k = p h x  x i . Set d ( x,y ) = arccos( h x  y i ), taken in [0 , ]. Prove that d is a metric on S . Proof. Let us check the axioms: 1. By definition since we take arccos in [0 , ], d ( x,y ) 0. 2. On S , d ( x,x ) = arccos( h x  x i ) = arccos( k x k 2 ) = arccos(1) = 0. Conversely, if d ( x,y ) = 0, h x  y i = 1. Thus, k x y k 2 = k x k + k y k 2 2 h x  y i = 2 2 = 0, and so, x = y by the Euclidean norm axiom....
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This document was uploaded on 02/22/2011.
 Spring '09
 Real Numbers

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