Suppose is a holomorphic function in the unit disk such that in the disk. How large can its Taylor polynomial be in the disk?
We should not expect to be bounded by 1 as well. Indeed, the Möbius transformation has Taylor expansion , so in this case. This turns out to be the worst case: in general is bounded by 5/4 in the disk.
For the second-degree polynomial the sharp bound is , attained when ; the image of the unit circle under the extremal is shown below. Clearly, there is something nontrivial going on.
Edmund Landau established the sharp bound for in his paper Abschätzung der Koeffizientensumme einer Potenzreihe, published in Archiv der Mathematik und Physik (3) 21 in 1913. Confusingly, there are two papers with the same title in the same issue of the journal: one on pages 42-50, the other on pages 250-255, and they appear in different volumes of Landau’s Collected Works. The sharp bound is in the second paper.
By rotation, it suffices to bound , which is . As is often done, we rescale a bit so that it’s holomorphic in a slightly larger disk, enabling the use of the Cauchy integral formula on the unit circle . The Cauchy formula says . Hence
It is natural to use now, which leads to
Here we can use the geometric sum formula and try to estimate the integral of on the unit circle. This is what Landau does in the first of two papers; the result is which is the correct rate of growth (this is essentially the Dirichlet kernel estimate from the theory of Fourier series). But there is a way to do better and get the sharp bound.
First idea: the factor could be replaced by any polynomial as long as the coefficients of powers up to stay the same. Higher powers contribute nothing to the integral that evaluates , but they might reduce the integral of .
Second idea: we should choose to be the square of some polynomial, , because can be computed exactly: it is just the sum of squares of the coefficients of , by Parseval’s formula.
Since is the -th degree Taylor polynomial of , it is natural to choose to be the -th degree Taylor polynomial of . Indeed, if , then as desired (asymptotics as ). The binomial formula tells us that
The coefficient of here can be written out as or rewritten as which shows that in lowest terms, its denominator is a power of 2. To summarize, is bounded by the sum of squares of the coefficients of . Such sums are referred to as the Landau constants,
A number of asymptotic and non-asymptotic formulas have been derived for , for example Brutman (1982) shows that is between 1 and 1.0663.
To demonstrate the sharpness of the bound , we want and on the unit circle. Both are arranged by taking which is a Blaschke product of degree . Note that the term can also be written as . Hence which is simply when . Equality holds in all the estimates above, so they are sharp.
Here are the images of the unit circle under extremal Taylor polynomials and .
These polynomials attain large values only on a short subarc of the circle; most of the time they oscillate at levels less than 1. Indeed, the mean value of cannot exceed the mean of which is at most 1. Here is the plot of the roots of extremal : they are nearly uniform around the circle, except for a gap near 1.
But we are not done…
Wait a moment. Does define a holomorphic function in the unit disk? We are dividing by here. Fortunately, has no zeros in the unit disk, because its coefficients are positive and decreasing as the exponent increases. Indeed, if with , then has constant term and other coefficients , , … , . Summing the absolute values of the coefficients of nonconstant terms we get . So, when these coefficients are attached to with , the sum of nonconstant terms is strictly less than in absolute value. This proves in the unit disk. Landau credits Adolf Hurwitz with this proof.
In fact, the zeros of (Taylor polynomials of ) lie just outside of the unit disk.
The zeros of the Blaschke products formed from are the reciprocals of the zeros of , so they lie just inside the unit circle, much like the zeros of (though they are different).