201A_F10_hw4_sol

# 201A_F10_hw4_sol - 201A Fall 10 Thomases Homework 4...

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201A, Fall ’10, Thomases Homework 4 - Solutions Mihaela Ifrim 1. Suppose f n C ([0 , 1]) is a monotone decreasing sequence that con- verges pointwise to f C ([0 , 1]). Prove that f n converges uniformly to f . This result is called Dini’ s monotone convergence theorem . Proof. Put g n ( x ) := f n ( x ) - f ( x ). The reason I am doing this is because it will be easier for me to prove that g n converges uniformly to zero rather than prove that f n converges uniformly to f (note that this is just a rewriting of the our problem). The new sequence we have constructed g n is in C ([0 , 1]) since f n C ([0 , 1]) and f C ([0 , 1]). Also, note that g n 0 pointwise and also g n g n +1 . This last inequality is true since we know that f n f n +1 , which im- plies that f n - f f n +1 - f i.e., g n g n +1 . So, our goal is to prove that g n 0 uniformly on [0 , 1]. Let ε > 0. For each x [0 , 1] there is an integer N x such that 0 g N x ( x ) ε 2 . By the continuity and by the monotocity of the sequence { g n } n (we said above that a monotone decreasing sequence), there exists an open set I ( x ) that contains x , such that 0 g n ( t ) ε (1) if t I ( x ) and if n N x . Since [0 , 1] is compact, we know that we can ﬁnd a ﬁnite open cover of it; in particular there exists a ﬁnite set of points x 1 , x 2 ··· ,x m such that [0 , 1] I ( x 1 ) I ( x 2 ) I ( x 2 ) ∪ ··· ∪ I ( x m ) . (2) Choosing N := max { N x 1 ,N x 2 ,N x 3 , ··· ,N x m }

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it follows from (1) and (2) that 0 g n ( t ) ε, if t I ( x ) and if n N . This prove the uniformity of convergence. Note The compactness of [0 , 1] is necessarily ! For instance look at f n ( x ) = 1 nx + 1 , where x (0 , 1) and n N * . Then f n ( x ) 0 monotonically in (0 , 1), but the convergence is not uniform. 2. Consider the space ofcontinuous diﬀerentiable functions C 1 ([ a,b ]) = n f : [ a,b ] R | f,f 0 are continous } with the C 1 - norm, k f k C 1 = sup a x b | f ( x ) | + sup a x b | f 0 ( x ) | . Proove that C 1 ([ a,b ]) is a Banach space with respect to the given norm. Proof. Set I = [ a,b ]. Since we are asked to prove that C 1 ([ a,b ]) is a Banach space, let { f n } n be a Cauchy sequence in C 1 ([ a,b ]) . Since k f n - f m k ≤k f n - f m k C 1 the sequence { f n } n is Cauchy in C ( I ) such that f n f uniformly. Next set g n = f 0 n . Then k g n - g m k = k f 0 n - f 0 m k ≤k f n - f m k C 1 , so { g n } n is Cauchy in C ( I ). Therefore, there exists a function g C ( I ) such that g n g
uniformly. It remains to prove that f C 1 ( I ) and that f n f in C 1 ( I ). Fix any x I and any h R such that x + h I .

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201A_F10_hw4_sol - 201A Fall 10 Thomases Homework 4...

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