From: spamless@nil.nil Subject: Re: Complex Analysis Date: 14 Sep 1999 22:09:49 -0400 Newsgroups: sci.math Keywords: No bounded analytic functions on the half-plane with enough zeros. lena wrote: > 1) h is an analytic and bounded function on the half-plane D = > {z:Re(z)>0}. And h(k)=0 for k=1,2,3,4...... > Show that h vanishes identically. Well, I changed the name of the function to f (and used h later for something else) ... but ... f analytic and bounded in the half plane: Re(z)>0 having zeros at z=1,2,3,... implies that f is identically zero. ==== Consider g(z)=f(1/(1+z)-1/2) for |z|<1 (1/(1+z)-1/2 maps |z|<1 onto the right half plane ... it is nothing but a Moebius transformation) g is analytic in the (open) unit disk and bounded there. g(z)=0 whenever 1/(1+z)-1/2=n for n a positive integer, or: z=-1+1/(n+1/2) Let me call -1+1/(n+1/2)=z_n and note that the product PROD(|z_n|) diverges to zero! That should just about do it, as one would now expect a Blashke product argument. (Spoiler below) Spoiler: We want to show that g(z) is identically zero. As it is analytic at zero, we will show that it has a zero of "infinite mutliplicity" there (heck, if g is analytic at zero and is not identically zero, it has a zero of some finite order ... that is what we will contradict). Let B_k(z)=PROD[(1-z*z_n')/(z-z_n):n=1 to k] where '=conjugate (of course, z_n is real, so I don't really need the prime). g(z)*B_k(z) is analytic (as g has zeros at z_n) and by the maximum modulus principal we have: |g(z)*B_k(z)|<=max_value_of_this_for_(|z|=r) whenever |z|1. |B_k(z)| has a limit of 1 as |z|-->1 and |g| is bounded by, let's say, M (the original bound we have for the original function on the half plane). We have |g(z)|<=M/|B_k(z)| for all |z|<1 and all k. Put in z=0 |g(z)|<=M*PROD{|z_n|:n<=k} and take the limit as k-->inf. The product diverges to zero, so g(0)=0! Well, then, g(z)/z is analytic so do this again, replacing g by g/z |(g(z)/z)*B_k(z)|<=M (since we use maximum modulus and as |z|-->1, both |B_k(z)| AND |z| tend to one). Again, since the product of |z_n| diverges to zero, g(z)/z has a "value" (value of h(z)=g(z)/z, the analytic function with the first zero at z=0 removed) of 0 ... or g has a double zero at z=0. Guess what ... a triple zero! Etc. (if g is not identically zero, there exists h(z) with h(0)<>0, g(z)=z^a*h(z) for some positive integer a ... replace g by h in the argument above to show that h(0)=0, a contradiction). (QUESTION: What if the original problem indicated a function with zeros at 2, 4, 9, 16, 25, ... n^2. where PROD[z_j] IS bounded? HINT: Construct g, above, JUST using the Blashke product ... it has the requisite zeros ... check for convergence as the number of terms in the product increases.) ============================================================================== From: spamless@Nil.nil Subject: Re: Complex Analysis Date: 15 Sep 1999 04:01:39 -0400 Newsgroups: sci.math spamless@nil.nil wrote: > (QUESTION: What if the original problem indicated a function with zeros > at 2, 4, 9, 16, 25, ... n^2. where PROD[z_j] IS bounded? (that should say: where PROD[z_n] "converges" (and remember, going to zero, is "diverging to zero" ... so this means a non-zero limit) ============================================================================== From: spamless@nil.nil Subject: Re: Complex Analysis Date: 22 Sep 1999 00:26:13 -0400 Newsgroups: sci.math spamless@nil.nil wrote: > lena wrote: >> 1) h is an analytic and bounded function on the half-plane D = >> {z:Re(z)>0}. And h(k)=0 for k=1,2,3,4...... >> Show that h vanishes identically. > Well, I changed the name of the function to f (and used h later for > something else) ... but ... > f analytic and bounded in the half plane: Re(z)>0 having zeros at > z=1,2,3,... implies that f is identically zero. > ==== > Consider g(z)=f(1/(1+z)-1/2) for |z|<1 > (1/(1+z)-1/2 maps |z|<1 onto the right half plane ... it is nothing but > a Moebius transformation) > g is analytic in the (open) unit disk and bounded there. Or ... once you see the idea, even simpler. Let f(z) be analytic in the right half plane and bounded there (|f(z)|<=M for RE(z)>0) with zeros at z_n=n. Consider, for a fixed k, f(z)*PROD[(z+z_n)/(z-z_n):n=1,2,...,k] which is analytic in that half plane. Use the maximum modulus principle on the curve: z=i*y+epsilon for y=R to -R z=SQRT(R^2+epsilon^2)*exp(i*theta) for theta=-pi/2 to pi/2 (semi circle of radius R against the imaginary axis ... OK, as the imaginary axis is not inside the region of analyticity, move it over epsilon to get it in the right half plane for some small epsilon>0) Now let R-->infinity and epsilon-->0+ (large semi-circle, moving to the imaginary axis). As |z|-->infinity, |z-z_n|/|z+z_n|-->1 and on the imaginary axis, |z-z_n|/|z+z_n|=1 too! So for a fixed k, we get: |f(z)|*|PROD[(z+z_n)/(z-z_n):n=1 to k]<=M or: |f(z)| <= M*PROD[|1-z/z_n|/|1+z/z_n|: n=1 to k] This is true for each k. Take the limit as k-->infinity to get |f(z)|<=0 since the product diverges to zero for z_n=n for every z (RE(z)>0). ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Re:Complex Analysis Date: 15 Sep 1999 23:29:01 GMT Newsgroups: sci.math In article <37DED01C.832D7DF6@math.nwu.edu>, lena writes: > 1) h is an analytic and bounded function on the half-plane D = > {z:Re(z)>0}. And h(k)=0 for k=1,2,3,4...... > Show that h vanishes identically. Hint: transform to a disk. There is a theorem that says if h is a bounded analytic function in the open unit disk, not identically 0, and z_n the zeros of h in the disk (counted according to multiplicity), then sum_n (1 - |z_n|) is finite. > 2) Is there a univalent analytic funtion from D={z:0<|z|<1} to > E={z:0.5<|z|<1}? Univalent means one-to-one, and I assume you also want it to be onto (otherwise the answer is trivially yes). Hint: the inverse image of a curve in E would be a curve in D, and a line integral around the curve in E can be expressed as an integral around the curve in D. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2