Analytic Functions
[13/100%]Is the space of square-integrable analytic functions on a region a Hilbert space, or does it have to be completed before it becomes one?
A region is a non-empty open connected subset of the complex plane. Let be planar Lebesgue measure in , and let be the set of all complex-valued functions that are analytic throughout and square-integrable with respect to . The set is a vector space with respect to pointwise addition and scalar multiplication, and it is an inner-product space with respect to the inner product defined by
When is the open unit disc, , the corresponding function space is denoted simply by .
If for and , then the 's form an orthonormal basis for . If , with Taylor series , then for .
If is a real function in , then is a constant.
Let be the unit circle in the complex plane, , and let be Lebesgue measure (the extension of arc length) on the Borel sets of , normalized so that . If for (), then the functions form an orthonormal basis for . The space is, by definition, the subspace of spanned by the 's with ; equivalently is the orthogonal complement in of . The elements of are called the analytic elements of . All statements such as "" or " is real" are to be interpreted as holding almost everywhere.
Is the Fourier series of a product the same as the formal product of the Fourier series?
A necessary and sufficient condition that
is that the numbers
be bounded.
Give an example of a Hilbert space of functions such that the vector operations are pointwise, but not all the evaluation functionals are bounded.
A functional Hilbert space is a Hilbert space of complex-valued functions on a (non-empty) set ; the Hilbert space structure of is related to in two ways. It is required that (1) if and are in and if and are scalars, then for each in , i.e., the evaluation functionals on are linear, and (2) to each in there corresponds a positive constant , such that for all in , i.e., the evaluation functionals on are bounded.
If is an orthonormal basis for a functional Hilbert space , then the kernel function of is given by
What are the kernel functions of and of ?
If is a functional Hilbert space, over say, then the linear functional on is bounded for each in , and, consequently, there exists, for each in , an element of such that for all . The function on , defined by , is called the kernel function or the reproducing kernel of . The kernel functions of and of are known, respectively, as the Bergman kernel and the Szegő kernel. If , its image in (the set of those functions analytic in the unit disc whose series of Taylor coefficients is square-summable) may be spoken of as the extension of into the interior.
If is an element of a self-conjugate functional Hilbert space, does it follow that ?
If is an element of a functional Hilbert space , the complex conjugate may fail to belong to ; the spaces and yield examples. A functional Hilbert space is called self-conjugate if it is closed under the formation of complex conjugates; an example is the sequence space . A more sophisticated example is the set of all square-integrable complex harmonic functions in the unit disc (the functions of the form , where each of and is the real part of some analytic function).
Whenever the answer is yes for all in the space, then a routine polarization argument shows that for all and .
The extension mapping (from to ) is continuous not only in the Hilbert space sense, but also in the sense appropriate to analytic functions. That is: if in , then for , and, in fact, the convergence is uniform on each disc , .
If an element of is such that the corresponding analytic function in is bounded, then is bounded (i.e., ).
If , does it follow that is bounded?
Is the mapping multiplicative?
To each real function in there corresponds a unique real function in such that and such that . Equivalently, to each real in there corresponds a unique in such that is real and such that .
The relation between and is expressed by saying that they are conjugate functions; alternatively, is the Hilbert transform of .