C3093105 - ) f ( x ) By (a) with a function f and scaler +...

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5. (Statement of the problem.) Proof. We check for A3 and A6. We will use the definition ( αf )( x ) = αf ( x ) ( a ) and ( f + g )( x ) = f ( x ) + g ( x ) ( b ) A3: Take f,g,h C [ a,b ]. So x [ a,b ], (( f + g ) + h )( x ) =( f + g )( x ) + h ( x ) By (b) with functions f + g and h = f ( x ) + g ( x ) + h ( x ) By (b) with functions f and g = f ( x ) + ( g + h )( x ) By (b) with functions g and h =( f + ( g + h ))( x ) By (b) with functions f and g + h Since (( f + g )+ h )( x ) = ( f +( g + h ))( x ) for all x [ a,b ], they must be the same function. So we have ( f + g ) + h = f + ( g + h ). / ( Note: Here we regard ( f + g ) + h as an element of C [ a,b ] . ) A6: Take f C [ a,b ] and scalars α , β . x [ a,b ], (( α + β ) f )( x ) =( α + β
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Unformatted text preview: ) f ( x ) By (a) with a function f and scaler + = f ( x ) + f ( x ) By distributive rule for scalars =( f )( x ) + ( f )( x ) By (a) with functions f and scalars and =( f + f )( x ) By (b) with functions f and f Since (( + ) f )( x ) = ( f + f )( x ) for all x [ a,b ], they must be the same function. So we have ( + ) f = f + f . / 1...
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This note was uploaded on 04/01/2008 for the course MATH 309 taught by Professor Samiabdul during the Spring '08 term at Michigan State University.

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