Unformatted text preview: < c ≤ 2 is sufﬁciently small, then their is a continuous function u ( x ) that satisﬁes u ( x ) = f ( x )+ Z c h ( x , y ) u ( y ) dy . (*) b) In the special case where h ( x , y ) ≡ 1 and f ( x ) ≡ 1, solve equation (*) explicitly. [This is easy. Let α = R 1 u ( y ) dy and then use (*) to solve for α ]. From this, show that indeed for some value of c a solution may not exist. 6. Let f ( x ) and h ( x , y ) be as in the previous problem. Show that if λ > 0 is sufﬁciently small, the equation u ( x ) = f ( x )+ λ Z 2 h ( x , y ) u ( y ) dy . (1) has a unique continuous solution u ( x ) . 7. Let f be an even continuous function on [-1 , 1 ] with R 1-1 f ( x ) x n dx = 0 for all even n ≥ 0. Show that f ≡ 0. [Last revised: December 8, 2010] 1...
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- Fall '10
- Derivative, Continuous function, uniform norm, Jerry L. Kazdan