# C3093105 - 5(Statement of the problem Proof We check for A3...

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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
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Unformatted text preview: ) f ( x ) By (a) with a function f and scaler α + β = αf ( x ) + βf ( x ) By distributive rule for scalars =( αf )( x ) + ( βf )( x ) By (a) with functions f and scalars α and β =( αf + βf )( x ) By (b) with functions αf and βf Since (( α + β ) f )( x ) = ( αf + βf )( x ) for all x ∈ [ a,b ], they must be the same function. So we have ( α + β ) f = αf + βf . / 1...
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