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Unformatted text preview: & m if x = 0 if x 6 = 0 : If f : R m ! C de&ne F ( f ) : c R m ! C by F ( f )( & ) = 1 p m X y 2 R m & ( y ) ± 1 f ( y ) : 1 6) Prove that f ( x ) = 1 p m X & 2 d R m & ( x ) F ( f )( & ) for all f : R m ! C and all x 2 R m . (Hint: Write out the right hand sum as a double sum and show that the double sum can be written 1 m X & 2 d R m X y 2 R m & ( x & y ) f ( y ) : Make the change of variables z = x & y so y = x & z and write the sum 1 m X z 2 R m ( X & 2 d R m & ( z )) f ( x & z ) : Now use the results above.) If f : R m ! C then we can de&ne a new function b f ( j ) = F f ( & j ) mapping R m to C . 7) Write out the formula for b f ( j ) . 8) Prove that b b f ( x ) = f ( & x ) . 2...
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 Winter '09
 N.R.Wallach
 Addition, Category theory, Abelian group, Group isomorphism, Group homomorphism

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