- Bieberbach conjecture
In
complex analysis , the Bieberbach conjecture or de Branges's theorem, asked by harvs|txt|first=Ludwig |last=Bieberbach|authorlink=Ludwig Bieberbach|year=1916 and proved by harvs|txt|authorlink=Louis de Branges de Bourcia|first=Louis |last=de Branges|year=1985, states anecessary condition on aholomorphic function to map the open unit disk of thecomplex plane injective ly to the complex plane.The statement concerns the Taylor coefficients "an" of such a function, normalized as is always possible so that "a"0 = 0 and "a"1 = 1. That is, we consider a
holomorphic function of the form:
which is defined and injective on the open unit disk (such functions are also called univalent or
schlicht functions ). The theorem then states that:
chlicht functions
The normalizations
:"a"0 = 0 and "a"1 = 1
mean that
:"f"(0) = 0 and "f" '(0) = 1;
this can always be assured by a
linear fractional transformation : starting with an arbitrary injective holomorphic function "g" defined on the open unit disk and setting:
Such functions "g" are of interest because they appear in the
Riemann mapping theorem .A family of schlicht functions are the rotated Köbe functions
:
with α a complex number of
absolute value 1. If "f" is a schlicht function and |"a""n"| = "n" for some "n" ≥ 2, then "f" is a rotated Köbe function.The condition of de Branges' theorem is not sufficient to show the function is schlicht, as the function:shows: it is holomorphic on the unit disc and satisfies |"a""n"|≤"n" for all "n", but it is not injective since "f"(-1/2+"z") = "f"(-1/2-"z").
History
harvtxt|Bieberbach|1916 proved "a"2≤2, and stated the conjecture "a""n"≤"n".
harvtxt|Löwner|1923 proved "a"3≤3, using the
Löwner equation . His work was used by most later attempts, and also used in the theory ofSchramm-Loewner evolution .harvtxt|Littlewood|1925|loc=theorem 20 proved that "a""n" ≤ "en" for all "n", showing that the Bieberbach conjecture is true up to a factor of "e" = 2.718... Several authors later reduced the constant "e". If "f"("z") = "z" +... is a schlicht function then φ("z") = "f"("z"2)1/2 is an odd schlicht function. harvs|txt|authorlink=Raymond Paley|last=Paley|author2-link=John Edensor Littlewood|last2=Littlewood|year=1932 showed that "b""k"≤14 for all "k". They conjectured that 14 can be replaced by 1 as a natural generalization of the Bieberbach conjecture. The Littlewood-Paley conjecture easily implies the Bieberbach conjecture using the Cauchy inequality, but it was soon disproved by harvtxt|Fekete|Szegö|1933, who showed there is an odd schlicht function with "b"5 = 1/2 + exp(−2/3) = 1.013..., and that this is the maximum possible value of "b"5. (Milin later showed that 14 can be replace by 1.14., and Hayman showed that the numbers "b""k" have a limit less than 1 if φ is not a Koebe function, so Littewood and Paley's conjecture is true for all but a finite number of coefficients of any function.) A weaker form of Littlwood and Paley's conjecture was found by harvtxt|Robertson|1936.The Robertson conjecture states that if :is an odd schlicht function in the unit disk with "b"1=1 then for all positive integers "n", :Robertson observed that his conjecture is still strong enough to imply the Bieberbach conjecture, and proved it for "n"=3. This conjecture introduced the key idea of bounding various quadratic functions of the coefficients rather than the coefficients themselves, which is equivalent to bounding norms of elements in certain Hilbert spaces of Schlicht function.
There were several proofs of the Bieberbach conjecture for certain higher valued of "n", in particular harvtxt|Garabedian|Schiffer|1955 proved "a"4≤4, harvtxt|Ozawa|1969 and harvtxt|Pederson|1968 proved "a"6≤6, and harvtxt|Pederson|Schiffer|1972 proved "a"5≤5.
harvtxt|Hayman|1955 proved that the limit of "a""n"/"n" exists, and has absolute value less than 1 unless "f" is a Koebe function. In particular this showed that for any "f" there are at most a finite number exceptions to the Bieberbach conjecture.
The Milin conjecture states that for each simple function on the unit disk, and for all positive integers "n", : where the logarithmic coefficients γ"n" of "f" are given by : harvtxt|Milin|1977 showed using the
Lebedev-Milin inequality that the Milin conjecture (later proved by de Branges) implies the Robertson conjecture and therefore the Bieberbach conjecture.Finally harvtxt|De Branges|1985 proved "a""n"≤"n" for all "n".
de Branges's proof
The proof use a type of
Hilbert space s ofentire function s. The study of these spaces grew into a sub-field of complex analysis and the spaces come to be calledde Branges space s and the functionsde Branges function s. De Branges proved the stronger Milin conjecture harv|Milin|1971 on logarithmic coefficients. This was already known to imply the Robertson conjecture harv|Robertson|1936 about odd univalent functions, which in turn was known to imply the Bieberbach conjecture about simple functions harv|Bieberbach|1916. His proof uses theLoewner equation , theAskey-Gasper inequality aboutJacobi polynomial s, and theLebedev-Milin inequality on exponentiated power series.de Branges reduced the conjecture to some inequalities for Jacobi polynomials, and verified the first few by hand. Walter Gautschi verified more of these inequalities by computer for de Branges (proving the Bieberbach conjecture for the first 30 or so coefficients) and then asked Richard Askey if he knew of any similar inequalities. Askey pointed out that he and Gasper had proved the necessary inequalities a few year before, which allowed de Branges to complete his proof. The first version was very long and had some minor mistakes which caused some skepticism about it, but these were corrected with the help of members of the
Steklov Mathematical Institute when de Branges visited in 1984.De Branges proved the following result, which for ν=0 implies the Milin conjecture (and therefore the Bieberbach conjecture). Suppose that ν > − 3/2 and σ"n" are real numbers for positive integers "n" with limit 0 and such that: is non-negative, non-increasing, and has limit 0. Then for all Riemann mapping functions "F"("z") = "z" + ... univalent in the unit disk with:the maximinum value of :is achieved by the Koebe function "z"/(1−"z")2.
References
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