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Tutorial-2

# Tutorial-2 - Analysis II Tutorial Solution sketches...

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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 0 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 0 ( t ) | ) . Prove that the pair ( C 1 ([ a, b ]) , k·k ) 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 α R , αf C 1 ([ a, b ]); These are practically obvious. Now, we have k αf k = max a t b ( | αf ( t ) | + | αf 0 ( t ) | ) = | α | · max a t b ( | f ( t ) | + | f 0 ( t ) | ) ; k f + g k = max a t b ( | f ( t ) + g ( t ) | + | f 0 ( t ) + g 0 ( t ) | ) max a t b ( | f ( t ) | + | f 0 ( t ) | ) + max a t b ( | g ( t ) | + | g 0 ( t ) | ) ; These relations prove two of the axioms of a norm. As k f k = max a t b ( | f ( t ) | + | f 0 ( 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 [email protected] All tutorial solutions in one pdf can be found at http://www.cs.mcgill.ca/ ~ atombe2/AnalysisTutorial.pdf . 1

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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 , π ]. Prove that d is a metric on S . Proof. Let us check the axioms: 1. By definition since we take arccos in [0 , π ], 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. 3. The symmetry follows directly from the symmetry of the inner product. The triangle inequality is a little harder to prove. Pick x, y, z S , and define ( θ = d ( x, y ) = arccos( h x | y i ) , φ = d ( y, z ) = arccos( h y | z i ) .
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