Math 655 Notes

Theorem 1 Suppose USC is open, overline{U}) Suppose \[f_{n}\Rightarrow f~\cdot nU\] Then On \[\overline{f}(z)=f(z)z\in U,\] \[\overline{f}(z)=cin_{n}f_{n}(z)\] -continuous where Moreover, Fis mif. cont. on J. Proof. Step 1: fu Suce founded and closed, it is compact. Continuous on [\overline{\nu}] uniform ly cantiumaus [\overline{\nu}] Show that fos ucrif. cont. on U. Let Eso be given. [\forall z_{1},z_{2}\in U] [|f(z_{1})-f(z_{2})|\le|f(z_{1})-f_{n}(z_{1})|] Choose 2 [+|f_{n}(z)-f_{n}(z)|+|f_{n}(z_{2})-f(z);] v2tv for this n choose S so small so that that \[(z_{1}-z_{2})<\delta\Rightarrow f_{\mu}(z_{1})-f_{\mu}(z_{2})f<\frac{\epsilon}{3}\] Then [|f_{n}(z)-f(z)|\le\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon] ⇒ f is unif. cont. on U. ⇒ the sequence [f_{n}(z_{0})] is Cauchy Step 2: ⇒ has a limit on C Show that moreover the limit does not depend on the choice of the subsequence [lim f_{n}(z_{0})] exists, and fn is equal to the limit of f(z) as z→z_{0}, z∈U. Let z_{k}→z_{0}, z_{k}∈U. [|f_{n}(z_{k})-f_{m}(z_{k})|\le|f_{n}(z_{k})-f_{n}(z_{0})|+] [|f_{n}(z_{0})-f_{m}(z_{0})|+|f_{m}(z_{0})-f_{m}(z_{k})|] Given ε>0. Choose n, m so large that ∀z∈U [|f_{n}(z)-f(z)|<\frac{\epsilon}{3}] Choose k, l so large that [|z_{k}-z_{0}|<\delta] and [|f_{n}(z_{k})-f_{n}(z_{0})|<\frac{\epsilon}{3}]. Then [|f_{n}(z_{k})-f_{m}(z_{k})|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon] Further, since ∀n we can choose kn so large that [|f_{n}(z_{k_{n}})-f_{n}(z_{0})|\le\frac{1}{n}] then [\lim_{n\to\infty}f_{n}(z_{0})] exists and is equal to [\lim_{n\to\infty}f_{n}(z_{k_{n}})] for this subsequence (z_{k_{n}}) and therefore for any other as well Step 3. Show that [|f_{n}(z_{0})-f(z_{0})|\to 0.] [\sup_{z_{0}\in U}|f_{n}(z_{0})-f(z_{0})|\to 0] Given ε>0 ∀z_{0}∈U choose n_{m} such that \[|f_{n}(z_{0})-f_{n}(z_{k_{m}})|<\epsilon\] and \[|f(z_{0})-f_{n}(z_{k_{m}})|<\epsilon\] Then \[|f_{n}(z_{0})-f(z_{0})|\le|f_{n}(z_{0})-f_{n}(z_{k_{m}})|+|f_{n}(z_{k_{m}})-f(z_{0})|<\epsilon+\epsilon=2\epsilon\] Since this is true for an arbitrary z_{0}∈U the inequality ≤2ε is true for the supremum. choose Step 4: Since fn ⇒ f on U and each fn is uniformly continuous on U then f is uniformly continuous on U. Theorem 2: Suppose the series Σan z^n converges uniformly for |z|≤R. Then the same series must converge A