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Jeffrey Hellrung Thursday, June 08, 2006 Math 246A, Homework 09
- From a qualifying exam:
Let f and g be continuous integrable functions on the real line related by the Fourier transform,
f (x) =
R
g(k)e−^2 πixkdk,
g(k) =
R
f (x)e^2 πixkdx.
Prove that it is impossible that both f and g are compactly supported. Hint: Consider holomorphic extensions. Solution
- From a qualifying exam:
Find an explicit conformal mapping from the region {|z| < 1 }[0, 1) onto the upper half plane {ℑ(z) > 0 }. Hint: It is a composition of simpler maps such as power functions z 7 → zα^ and linear fractional transformations. Solution Suppose supp f is compact. Then g extends to a holomorphic function:
g(z) =
R
f (x)e^2 πixz^ dx.
To see this, consider g(z + h) − g(z) =
R
f (x)e^2 πixz^
e^2 πixh^ − 1
dx.
Now ∣ ∣e^2 πixh^ − 1
∣^2 πixh
(2πixh) +
(2πixh)^2 + · · ·
∣ ≤^2 π|x||h|e
2 π|x||h| (^) ≤ 2 π|x||h|e 2 π|x|
for |h| ≤ 1, so (^) ∣ ∣ ∣ ∣
h
(g(z + h) − g(z))
∣ ≤^2 π
R
|f (x)||x|e^2 π|x|(|z|+1)dx,
which is finite since the integrand |f (x)||x|e^2 π|x|(|z|+1)^ is a compactly supported continuous function. By the Dominated Convergence Theorem, then,
g′(z) = lim h→ 0
h
(g(z + h) − g(z))
exists, and hence g is holomorphic. It follows that the restriction of g to R is not compact unless g is constant (the zeros of a holomorphic function are isolated), in which case g ≡ 0 and hence f ≡ 0.
- From a qualifying exam:
Find an explicit conformal mapping from the region {|z| < 1 }[0, 1) onto the upper half plane {ℑ(z) > 0 }. Hint: It is a composition of simpler maps such as power functions z 7 → zα^ and linear fractional transformations. Solution We use the following composition of conformal mappings.
z = e(log^ z)/^2 maps {|z| < 1 }[0, 1) conformally to {|z| < 1 , ℑ(z) > 0 }, where we choose the branch cut of log to be {z > 0 } and such that log(−1) = πi.
- z 7 → (1 + z)/(1 − z) maps {|z| < 1 , ℑ(z) > 0 } conformally to {ℜ(z) > 0 , ℑ(z) > 0 }.
- z 7 → z^2 maps {ℜ(z) > 0 , ℑ(z) > 0 } conformally to {ℑ(z) > 0 }.
- From a qualifying exam:
Let D be the domain in the complex plane C that is the intersection of the two open disks centered at ±1 whose boundary circles passes through ±i. Find a conformal map f of D onto the open disk ∆ = {|w| < 1 } such that f (i) = 1 and f (−i) = −1. You may express f as a composition of other specific maps. What are the images of arcs of circles passing through ±i under your map? Justify your answer. Solution We use the following composition of conformal mappings.
- z 7 → e−^3 πi/^4 (z + i)/(z − i) maps D conformally to {ℜ(z) > 0 , ℑ(z) > 0 }.
- z 7 → z^2 maps {ℜ(z) > 0 , ℑ(z) > 0 } conformally to {ℑ(z) > 0 }.
- z 7 → i(z − i)/(z + i) maps {ℑ(z) > 0 } conformally to ∆.
The image of i and −i under the above composition of maps is i and −i, respectively, so we use a final multiplication by −i to complete the construction of f. The images of arcs of a circles whose endpoints are ±i under the above bulleted conformal maps give, consecutively, rays from 0, rays from 0, and arcs of circles whose endpoints are ±i. Indeed, each of the maps above may be inverted, and we see that this provides a self-bijection.
- From a qualifying exam:
Let f be any conformal mapping from the strip S = {z ∈ C | − 1 < ℑ(z) < 1 } onto the unit disc D = {z | |z| < 1 } such that, uniformly in y ∈ (− 1 , 1),
lim x→∞ f (x + iy) = 1,
lim x→−∞ f (x + iy) = − 1.
Find the images in D of the set of horizontal lines in S and the set of vertical lines in S. Hint: In each case the set of images does not depend on the choice of f. Solution
