# HW10 - MATH185 Solutions to Assignment 10 1 Recall that a...

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MATH185: Solutions to Assignment 10 1. Recall that a sequence of functions f n : U C is uniformly Cauchy on U iﬀ lim N →∞ ( sup {| f n ( w ) - f m ( w ) | : n,m N and w U } ) = 0 (a) Show that a sequence of functions f n is Uniformly Cauchy on U if and only if there is a function g : U C such that f n converges uniformly to g on U . Sol: Assume that ( f n ) n is a uniformly Cauchy sequence of func- tions. For any w U , we have ( * ) | f n ( w ) - f m ( w ) | ≤ sup {| f n ( z ) - f m ( z ) | : z U } as a result, the sequence of complex numbers ( f n ( w )) is a Cauchy sequence and hence converges to a complex number which we will call g ( w ). We will now show that the sequence of functions ( f n ) converges uniformly to the function g on U . Given ± > 0 there is a number N such that, for all n,m N sup {| f n ( z ) - f m ( z ) | : z U } ≤ ± so, by ( * ) we know that for all n,m N and all w U | f n ( w ) - f m ( w ) | ≤ ± taking limit when m → ∞ we have that for all w U | f n ( w ) - g ( w ) | ≤ ± and thus sup {| f n ( w ) - g ( w ) | : w U } ≤ ± Since ± was arbitrary it follows that lim n →∞ sup {| f n ( z ) - g ( z ) | : z U } = 0 so that ( f n ) n converges uniformly to g on U . Conversely, suppose that ( f n ) n converges uniformly to g on U and let ± > 0 be given. Then there is an N N such that for 1

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n N sup {| f n ( z ) - g ( z ) | : z U } < ± 2 . If m,k N we see that for all w U | f n ( w ) - f m ( w ) | = | f n ( w ) - g ( w )+ g ( w ) - f m ( w ) | ≤ | f n ( w ) - g ( w ) | + | g ( w ) - f m ( w ) | ≤ ± so sup {| f n ( w ) - f m ( w ) | : n,m N ± . Since ± was arbitrary it follows that the sequence of functions ( f n ) n is uniformly Cauchy on U . (b) Use part ( a ) to show that, if the series n =1 | f n ( z ) | converges uniformly on U then the series n =1 f n ( z ) converges uniformly on U (Recall that a series converges unformly if it converges to and moreover the sequence of partial sums converges uniformly to the limit). Sol: Let F N ( z ) = N k =1 f n ( z ) and let A N ( z ) = N k =1 | f n ( z ) | . Saying that the series n =1 | f n ( z ) | converges uniformly on U means that the sequence of partial sums ( A N ) N converges uni- formly on U and thus, by part (a), that this sequence is uniformly Cauchy on U . We will now show that the sequence ( F N ) N is uniformly Cauchy on U . Assume m n , and let w U | F m ( w ) - F n ( w ) | = | m X k = n +1 f k ( w ) | ≤ m X k = n +1 | f k ( w ) | = | A m ( w ) - A n ( w ) | In turn, for all w U | A m ( w ) - A n ( w ) | ≤ sup {| A m ( z ) - A n ( z ) | : z U } So putting both inequalities together we get sup {| F m ( z ) - F n ( z ) | : z U } ≤ sup
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HW10 - MATH185 Solutions to Assignment 10 1 Recall that a...

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