Proof 3 moreras theorem states that if f is

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cannot be continued analytically past the unit disc. Proof. 3. Morera’s theorem states that if f is continuous in C , and R T f ( z ) dz = 0 for all triangles T , then f is holomorphic in C . Naturally, we may ask if the conclusion still holds if we replace triangles by other sets. (a)Suppose that f is continuous on C , and R C f ( z ) dz = 0 for every circle C . Prove that f is holomorphic. (b) More generally, let Γ be any toy contour, and f the collection of all translates and dilates of Γ . Show that if f is continuous on C , and Z γ f ( z ) dz = 0 for all γ f 8
then f is holomorphic. In particular, Morera’s theorem holds under the weaker assumption that R T f ( z ) dz = 0 for all equilateral triangles. [Hint: As a first step, assume that f is twice real differentiable, and write f ( z ) = f ( z 0 )+ a ( zz 0 )+ b ( zz 0 ) + O ( | zz 0 | 2 ) for z near z 0 . Integrating this expansion over small circles around z 0 yields ∂f z = b = 0 at z 0 . Alternatively, suppose only that f is differentiable and apply Greens theorem to conclude that the real and imaginary parts of f satisfy the Cauchy-Riemann equations. In general, let ϕ ( w ) = ϕ ( x, y ) (when w = x + iy ) denote a smooth function with 0 ϕ ( w ) 1, and R R 2 ϕ ( w ) dV ( w ) = 1, where dV ( w ) = dxdy , and R denotes the usual integral of a function of two variables in R 2 . For each > 0, let ϕ ( z ) = 2 ϕ ( 1 z ), as well as f ( z ) = Z R 2 f ( z - w ) ϕ ( w ) dV ( w ) , where the integral denotes the usual integral of functions of two variables, with dV ( w ) the area element of R 2 . Then f is smooth, satisfies condition for circle in (a), and f f uniformly on any compact subset of C .] Proof. 4. Prove the converse to Runge’s theorem: if K is a compact set whose complement is not connected, then there exists a function f holomorphic in a neighborhood of K which cannot be approximated uniformly by polynomial on K . [Hint: Pick a point z 0 in a bounded component of K c , and let f ( z ) = 1 / ( z - z 0) . If f can be approximated uniformly by polynomials on K , show that there exists a polynomial p such that | ( z - z 0 ) p ( z ) - 1 | < 1 . Use the maximum modulus principle (Chapter 3) to show that this inequality continues to hold for all z in the component of K c that contains z 0 .] Proof. 5. There exists an entire function F with the following universal property: given any entire function h , there is an increasing sequence { N k } k =1 of positive integers, so that lim k →∞ F ( z + N k ) = h ( z ) uniformly on every compact subset of C . 9
(a) Let p 1 , p 2 , · · · denote an enumeration of the collection of polynomials whose coefficients have rational real and imaginary parts. Show that it suffices to find an entire function F and an increasing sequence { M n } of positive integers, such that | F ( z ) - p n ( z - M n ) | < 1 n whenever z D n , (1) where D n denotes the disc centered at M n and of radius n . [Hint: Given h entire, there exists a sequence { n k } such that lim k →∞ p n k ( z ) = h ( z ) uniformly on every compact subset of C .] (b) Construct F satisfying (1) as an infinite series F ( z ) = X n =1 u n ( z ) where u n ( z ) = p n ( z - M n ) e - c n ( z - M n ) 2 and the quantities c n > 0 and M n > 0 are chosen appropriately with c n 0 and M n → ∞ . [Hint: The function e z 2 vanishes rapidly as | z | → ∞ in the sectors {| arg z | < π/ 4 - δ } and {| π - arg z | < π/ 4 - δ } .] In the same spirit, there exists an alternate universal entire function G with the following property:

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