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NOTESEdd b: Jh Duc
eometric Prperties he amm Func
Ptrick hem d Wlter Rudi
The fact that log x is convex on the positive real axis is one of te rucial properties of f. Combined with the nctional euation fx + 1 = xx and the
normalzation f 1=
1 it uniuely characterizes the gama nction his theorem, due to Bohr and Mollerup, is the basis of Artins eegant treatent of xin [2]
We extend tis convexity property into the compex pane, use this eension tootain infrmation about the argument of z on vertica lies, and dscriesome features of the conformal mappings induced by f' /f and b log f. In spiteof the apparently inexhaustible supply of formulas, identities, eansions, adintegrals concerned with f, we have found no mention of such geometric properties in the literature
For brevity, we use the notation
Gz = log fz.We let a denote the right halfplane consisting of all complex z having e z > a,and denote its closure by a
1. A x � then e G" x + iy > 0 or a rea y.(B) x < then e G" x + iy < 0 or a scienty ae y.
Note that e
G" x + iy=
/x x + iy.he theorem asserts there
fore that log x + iy a conv nction o x in 1 , bt in no aer haane.oo We begin with (B) since its proof gave us a hint that 1/2 might be thecuto between A d (B)
Define s = � s n 0 1] s + 1 = s for all real s. ne wy of writing Stirlings formua is [; p 151]
(s)log fz = z - ) log z z + log27 + 1 - 1
s + z
for all z " 0 that are not negative real numbers Dierentiate this tice, then perform an integration by parts The result is
1 1 csG"(z) = - + + 6 r z 2 Jo (s +) (2
where c s = t dt. Since has meanvalue 0 c is periodic, hence ounded
NOTS [ctoer
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In fact, 0 : �s : 1/8. It follows that
" . x3 + x2 + 2 x - 1y2 3 (X R e G x + zy : 2 + -), 2·
2 (x2 + y2) 4 0 [(s + X)2 + y2]
The last integral is O(y-3), as y / 0. W hen 2 x 1 = 0, the dominant term, or fixed x and large y, is thus (x �)y-2, and this is negative when x < 1/2; (B)follows
Next, we appl log to the identit [; p 24]
fzfl - z = .
'
s z
then dierentiate twice, and obtain
"2 G"z + G"l - z
s2z
Since sin � + i y cos(y) cosh(y), it follows that
or
"2
2 RG"U+iy = h2
> 0COS "y
3
Finall, 2 shows that G" is bounded in 08, for eve 8 > o Since boundedharmonic nctions in a halfplane are the Poisson integrals of their bounda values, A follo ws from 3
The following monotonicit propert was needed in [1], but onl when b - a isan integer, and in that case a more direct proof exists
Therem . (i) I �: a < b then
fb + i yarg
fa + iy)
i an increasing nction oy on ( - 0, 0).(ii) Te same conclusion hol i 0 < a < � and b > 1 -a.
Poof: Let u and v be the real and imagina parts of G log f, respectivelen v !g f, the CauchRiemann equations give Ux vy' and hence vxy
uxx > 0 in 01/2, b heorem 1. his means that v /a + iy < v /b + iy, or
arg fa + iy) < - arg fb + iy).y y
is proves <.
1996] NTE 6 7 9
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To deduce (iO, note thatf(b+iy) f(b+iy)f(a+iy) f(1-a+)
1
f(l-a+iy)f(a+iy)
f(b+iy)
f(1 -a+iy)
f(a-iy)f(a -iy)
sin (a -iy). (4)
Since now 1 -a > 1/2, (iO follows from (i) and the fact that the argument of thelast factor in 4 is
arctan [cot a) tanh7Y] , which is an increasig nction of y when 0 < a < 1/2.
The proofs of the next two theorems will use the following suiciency criterionfor univalence
Suppose: (a) an open hapane that does not contain the on, (b)
hoomorphic in a conv region 0, and (c) 'z E or eve z E Then univaent in
hen =0 this is proved on p 47 of [3] It is clear that the criterion isrotation-invriant
3. (i) f'If is univaent n 0, but in no arger hapane.(iO Rf'/fx+ y � f'/fx in 0
(iii) Im('If is bounded in or eve { > O.
(iv) f'/fz<
/2 in /2'Po Since f If' = G, it follows from Theorem 1 and the sucien criterionthat f'If is univalent in 01/2 This proves only part of , but at least it points inthe right direction
Differentiation of f'/fx+iy = u+ivx+iy (where u and v are now the real and imaginay parts of f'/f wth respect to y gives
iG"(x+y) = u+ iv(x+iy).Thus V = Re G, hence (3) becomes
2v +iy = 2 h2( ) cos 7 y
If we integrate this we obtain7v +iy = tanh Y),
and this implies (iv)Differentiation of the logarithm of
_1_=zeZn (1+ ) ef( z) n
[4; p 150], where ' is Eulers constant, leads to
u( x+ iy) = - '- x +E (2 x+y2 nn+x )
(n+X)2+y2
680
0 yv(x+iy) = L 2 2o (n+x) + y
NOT
( 5 )
(6)
[ctober
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and 1
G"(z) =E 2'o (n z)
(7)
The right side of 5 is an increasing ction of y2, if x > O. For fixed x it
therefore attains its minimum when y = O. This proves (ii), and (iii) follows from 6, because
y yv (x + iy) < + E�-:x2 + y2
1n2+ y 2
1 y 1 < ( 2 dt = 2
2' 2 x 0 t y x
For the proof of (i), let + = {y> O}
and = {y <O}
be the upper and lower half-planes, respectivel
If z E n + then n Z2 E + for all n � 0 hene n Z-2 E , sothat G"z E -
Likewise, G"z E + if z E n It follows that f'/ is univalent in each of the quadrants n + and -.
By 6, f'/ maps n + into + and n into ; since f'/ isstrictly increasing on the positive real as, we conclude that f' / is univalent inall of .
Finally, f' / has a pole at z = 0 hence maps each neighborhood of 0 onto a
neighborhood of Since f' /n(x) / as x / it follows that f' / is not univalent in if { < OIn the next theorem, X is the unique positive number at which f'x = O
Since = 2, 1 < X < 2
. log univaent in xo but in no arger hapane.
Po! ecall that (log n' = f'J If x > x, it follows from part (iO of Theorem 3that
elog ) '(xiy) � (f'/)(x) > (f'/)(x) =
Thus log is univalent in xo Clearly, X has no neighborhood in which log is univalent, since (log 'x =
REFERENCS
1. P. Ahem, M Flores, and W. Rudin, A invariant ean-value property, 1 Functional Anal ,1993, 38097
2. E. Arin,
The Gamma FunctionHolt, Rinehar and Winston, 1964
3 P L. Duren, Univalent Functions Springer-Verlag, 193 4. E. C Titcharsh, e eo of Functions 2nd ed., Oxord Univ. Press, 19395 R Reert, Wieandt's theore about the f-unction, Amer Math Monthly 03, 1996, 214220
Depament o MathematicsUniversi o WconsinMadon
Madon Wisconsin [email protected]
1996] NOTS 681