Unformatted text preview: g ( a ) = 0, then g ( x ) = g ( x-a ) g ( a ) = 0. For the former, assuming that g is continuous at x = 0, we need to show that g ( y ) → g ( x ) as y → x for every x ∈ R . But g ( y ) = g ( x ) g ( y-x ). As y → x , y-x → 0, and since g is continuous at 0, it follows that g ( y ) = g ( x ) g ( y-x ) → g ( x ) g (0). Moreover since 0 = 0 + 0, we have g (0) = g (0) g (0) and thus g (0) is either 0 and g ( x ) ≡ 0 and is trivially continuous or g (0) = 1. In this latter case, we have g ( y ) → g ( x ) as desired. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
- Spring '10