201A_F10_hw5_sol

201A_F10_hw5_sol - 201A Fall 10 Thomases Homework 5...

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201A, Fall ’10, Thomases Homework 5 Solutions Mihaela Ifrim 1. If f is continuous on [0 , 1] and if Z 1 0 f ( x ) x n dx = 0 , n N * , then f ( x ) = 0 on [0 , 1]. Hint. The integral of the product of f with any polynomial is zero. Use the Weierstrass theorem to show that R 1 0 f 2 ( x ) dx = 0 . Proof. Since f is continuous on the compact set [0 , 1], it is bounded, i.e., | f ( x ) | ≤ M x . The complex conjugate f ( x ) is continuous also, so by the Weierstrass Approximation Theorem, there is a sequence of complex polynomials P n , such that P n f uniformly on [0 , 1]. Since f is bounded, then fP n f f = | f | 2 , as n → ∞ , uniformly as well. More precisely, this is obvious from the following fact: k fP n - f f k M k P n - f k 0 . By problem no 6 from HW3, we can conclude that: 0 = lim n →∞ Z 1 0 fP n dx = Z 1 0 lim n →∞ fP n dx = Z 1 0 | f | 2 dx. (1) By the continuity of f on [0 , 1] and by 1, it follows that f 0 on [0 , 1]. 2. Consider the following equation for an unknown function f : [0 , 1] R : f ( x ) = g ( x ) + λ Z 1 0 ( x - y ) 2 f ( y ) dy + 1 2 sin( f ( x )) . (2) Prove that there exists a number λ 0 > 0 such that for all λ [0 0 ), and all continuous functions g on [0 , 1], the equation (2) has a unique continuous solution.
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Proof. This problem is solvable using Contraction mapping theorem . Let’s denote by Tf ( x ) the follwing : Tf = g ( x ) + λ Z 1 0 ( x - y ) 2 f ( y ) dy + 1 2 sin( f ( x )) . Now, we need to find the space on which T is defined. It is obvious that the space on which T is defined is C ([0 , 1]). Also, we still need to prove that the rage of T is indeed included in C ([0 , 1]). Let’s first show that T is a contraction. For this we need to prove that: k Tf 1 - Tf 2 k c k f 1 - f 2 k , for c < 1. So, computting we get: k Tf 1 - Tf 2 k = k λ Z 1 0 ( x - y ) 2 ( f 1 ( y ) - f 2 ( y )) dy + 1 2 [sin( f 1 ( x )) - sin( f 1 ( x ))] k . The first part can be recognized as a Fredholm operator. This can be bounded as follows: sup 0 x 1 | Z 1 0 ( x - y ) 2 ( f 1 ( y ) - f 2 ( y )) dy | ≤ sup 0 x 1 Z 1 0 | ( x - y ) | 2 | ( f 1 ( y ) - f 2 ( y )) | dy ≤k f 1 - f 2 k sup 0 x 1 Z 1 0 | ( x - y ) | 2 | dy 1 3 k f 1 - f 2 k . But now we still need to find a bound for the remaider part, the one with sin . First recall that by the Mean Value Theorem we have that sin( f 1 ( x ) - sin( f 2 ( x )) f 1 ( x ) - f 2 ( x ) = cos( α ) 1 (3) for α [0 , 1]. Note: The fact that cos( ) [ - 1 , 1] for all ♥ ∈ R . But 3 tell us that sin f 1 ( x ) - sin f 2 ( x ) f 1 ( x ) - f 2 ( x ) for all x [0 , 1] . Therefore 1 2 [sin f 1 ( x ) - sin f 2 ( x )] 1 2 [ f 1 ( x ) - f 2 ( x )] , x.
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Hence 1 2 [sin f 1 ( x ) - sin f 2 ( x )] 1 2 k f 1 - f 2 k . Now, we can put everything together and see that we got: k Tf 1 - Tf 2 k λ 3 k f 1 - f 2 k + 1 2 k f 1 - f 2 k ( λ 3 + 1 2 ) k f 1 - f 2 k . In oder to apply the Contraction Mapping Theorem , we need T to be a contraction, fact that translates in impossing that λ 3 + 1 2 < 1 i.e., λ < 3 2 . Let’s pick the critical value λ 0 = 3 2 . This value gives us indeed that T is a contraction!
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This note was uploaded on 01/29/2012 for the course MATH 34 taught by Professor Wiley during the Spring '11 term at UC Merced.

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201A_F10_hw5_sol - 201A Fall 10 Thomases Homework 5...

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