_{1}

^{*}

This paper addresses the optimal recovery of functions from Hilbert spaces of functions on the unit disc. The estimation, or recovery, is performed from inaccurate information given by integration along radial paths. For a holomorphic function expressed as a series, three distinct situations are considered: where the information error in L
_{2} norm is bound by δ＞0 or for a finite number of terms the error in l
_{2}
^{N} norm is bound by δ＞0 or lastly the error in the j
^{th} coefficient is bound by δ
_{j}＞0. The results are applied to the Hardy-Sobolev and Bergman-Sobolev spaces.

Let W be a subset of a linear space X, let Z be a normed linear space, and T the linear operator that we are trying to recover on from given information. This information is provided by a linear operator where Y is a normed linear space. For any we know some that is near. That is, we know such that

for some. The value is our inaccurate information. Now we try to approximate the value of from using an algorithm or method,. Define a method to be any mapping, and regard as the approximation to from the information. Our goal is to minimize the difference of

and in, i.e. minimize

However, the size of varies since

can be chosen to be any satisfying (1). Furthermore varies depending on the chosen. So the error of any single method is defined as the worst case error

Now the optimal error is that of the method with the smallest error. Thus the error of optimal recovery is defined as

For the problems addressed in this paper, let be linear spaces with semi-inner norms and linear operators,. We want to recover for

(where if we let), if we know the values

satisfying for.

Define the extremal problem

This problem is dual to (2).

The following results of G. G. Magaril-Il’yaev and K. Yu. Osipenko [

Theorem 1: Assume that there exist, such that the solution of the extremal problem

is the same as in (3). Assume also that for each

there exists

which is a solution to

Then for all,

and the method

is optimal.

Theorem 1 gives a constructive approach to finding an optimal method from the information. It follows from results obtained in [1-7] (see also [

In order to apply Theorem 1 the values of extremal problems (4) and the dual problem (3) must agree. The following result, also due to G. G. Magaril-Il’yaev and K. Yu Osipenko [

Typically, when one encounters extremal problems, one approach is to construct the Lagrange function. For an extremal problem of the form of (4), the corresponding Lagrange function is

Furthermore, is called an extremal element if

for and thus admissible in (4)

and

Theorem 2: Let and be such that

for and 1)

2)

Then is an extremal element and

If we wish to combine Theorems 1 and 2 to determine an optimal error and method then we must show the posed problem is able to satisfy equating extremal problems (3) and (4). Through Theorem 2 we have such a means available.

Consider the class of functions defined on the unit disc given by

for, satisfying

and

Therefore, any is holomorphic in the unit disc by (6). We define the semi-norm in as

and

Let, , be a linear operator given by

That is, is the radial integral of. To see that, by (7) we have for all but finitely many,

for some. Thus if then

.

We assume to know given with a level of accuracy. That is, for a given, we know a such that

The problem of optimal recovery is to find an optimal recovery method of the function in the class from the information satisfying (9). The error of a given method is measured in the norm defined by

Any method is admitted as a recovery method. Let be sequences of non-negative real numbers such that

Define to be the convex hull. Define for by

Lemma 1: The piecewise linear function with points of break , with for given by is such that.

Proof. Assume that It means that

. Since and as there is a such that and. Then the interval between and belongs to. Consequently, and is not a point of break of.

Assume that. Since the interval between and belongs to. Geometrically, the line to will lie above the line. It means that contradicting that is a point of break of.

Note that as then for any fixed the slopes between points and also tends to 0 as

Consider the points in given by

and define the convex hull of the origin and this collection of points as:

Let

thus is a piecewise linear function. Let, be the points of break of with. By (7) the assumption for Lemma 1 is satisfied by and.

Theorem 3: Suppose that with . Let

Then the error of optimal recovery is

and

is an optimal method of recovery. If then

and is an optimal method.

Proof. Consider the dual extremal problem

which can be written as

where. Define the corresponding Lagrange function as

Let the line segment between successive points

and be given by.

That is. Thus are given by (12). Take any, then by definition of the function we have

Thus for all

and hence for any.

We proceed to the construction of a function admissable in (15) that also satisfies

Assume.

As if and only if and then if and only if or. Let be the indices that satisfy

and

.

We let for, and choose so that they satisfy the conditions

From these conditions let

and

Now if with or and

the function is admissible in (15) and

, that is minimizes

and condition 1) of Theorem 2 is satisfied. Furthermore, by construction, satisfies condition 2) of Theorem 2.

If, that is and, define as in (17). Then as

So let and we have

Thus the function is admissable in (15) and satisfies 1) and 2) of Theorem 2. It should be noted that in this case are simply and.

Now we proceed to the extremal problem

This problem may be rewritten as

which has solution

So for, by Theorems 1 and 2, (14) is an optimal method and the error of optimal recovery is given by (13). If then

and is an optimal method.

It should be noted that for fixed, that is for a fixed, the terms

will have the property, and as. So smooths approximate values of the coefficients of by the filter.

Our next problem of optimal recovery remains to recover from inaccurate information pertaining to the radial integral of f. However, the inaccurate information we are given are the values

such that

where is the coefficient of the radial integral,

Denote

We again consider the space of functions given by (5) and and defined by (10) and (11) respectively but now add the condition

The problem of optimal recovery on the class given by (8) is to determine the optimal error

and an optimal method obtaining this error.

Define as the largest index such that

which by (7) exists, and

Theorem 4: Suppose with. If

let and.

Then the optimal error is

and

is an optimal method. If then

and is an optimal method.

If and then with

and the error of optimal recovery is (22) and (23) is an optimal method. For

, and is an optimal method.

Proof. For the cases with we simply apply the same structure of proof as in Theorem 3. For the case there remains some work.

Our construction will depend on whether or not, that is whether or not with

.

First we notice. Assume not. Then if we also know since for all we assumed. Since we know. Then by definition of we know for,

and substituting we have

which contradicts the definition of. Therefore

and if then

.

In either case, or, the dual problem is of the form

The corresponding Lagrange function is then

where is the characteristic function of

.

Case 1):

If let correspond to the index satisfying

To determine let be the line through the point that is parallel to the line from the origin to. That is, let

So for any point of break we have and for any index, we obtain

If then

Thus for the chosen and and any we have.

To construct admissable in (24), let for and define by the system

and since this becomes

So let and.

Then for the function

is admissable in (24) with

Therefore and by construction we have and

so that

and conditions (a) and (b) of Theorem 2 are satisfied.

Case 2):

If then, and, as this is the only point in the set

with a -coordinate of. Furthermore, as is a point of break of we know for all. Since then by the definition of we know. As

then we obtain equality,.

Define by (25) so and. If we let be then

In addition is admissable in extremal problem (24)

as and.

To justify simply note that as

satisfies (26) and for all then

. So we have

. Since

then minimizes.

For both cases, we now consider extremal problem

This problem can be written as

which will have solution

So by Theorems 1 and 2 we have obtained the optimal error and an optimal method for all scenarios. In each case i and ii, and are given by (25). In each case, the error of optimal recovery is

which for case 2) simplifies to. Also for each case, a method of optimal recovery is given by where in case 2) this simplifies to since in case 2),.

One may be able to reduce the amount information needed without affecting the error of optimal recovery. Therefore, by reducing the number of terms in the optimal method we reduce the compututaions needed. The following ideas are in [

for where if define

. Now consider the same problem as stated in Theorem 4 using only information. For

, we have and so

. In this situation, with, it was shown that the error of optimal recovery only involves the two points

then the reduction in information from to will not change the error. That is

and if, an optimal method is

where.

In Theorems 3 and 4 the inaccuracy of the information given is a total inaccuracy. That is, the inaccuracy is an upper bound on the sum total of the inaccuracies in each term, be it a finite or infinite sum. For Theorems 3 and 4 however, there is no way to tell how the inaccuracy is distributed. In particular, with regards to Theorem 4, the situations in which the given information satisfies

or for some particular satisfying

are treated the same. For the next problem of optimal recovery we address this ambiguity. The problem of optimal recovery is to determine an optimal method and the optimal error of recovering, from the information satisfying

for some prescribed and.

To define use conditions (6) and (20) as previously but impose an additional restriction. We add the condition

Define where are the levels of accuracy. If define

So and furthermore. The case will be treated seperately.

Theorem 5: If let

then the error of optimal recovery is given by

and

is an optimal method.

If then and is an optimal method.

Proof. The dual problem in this situation is

with the corresponding Lagrange function

The method of proof will be to first determine

with and admissable in (31) and satisfying 1) and 2) of Theorem 2.

If, define and as follows:

To verify assume in which case

and hence

To show for the chosen and any,

, we consider the cases or.

For we know by assumption

and hence

For

Thus for any,. For the constructed, it can be shown that as desired. and thus minimizes the Lagrange function.

To show is admissable in (31) we can clearly see that for,. It remains to show for. Assume not, then

which occurs if and only if

which contradicts the definition of unless . If then and hence we no longer need the condition in order for to satisfy (31).

Furthermore

and so is admissable in (31).

By the construction of we also have the results

and for while

for. Thus satisfies 2) of Theorem 2 as

We now proceed to the extremal problem

Notice the upper bound on the sum is as for any. This extremal problem will have solution

Therefore the error of optimal recovery is given by

and

is an optimal method.

Now we proceed to the case. Choose and for. Then as for all

Thus for all. Let

and and notice and clearly

so is admissable in (31). Furthermore

and so. Also,

Therefore and

is an optimal method.

The optimal method may not use all of the information provided as may be less than. Thus increasing may not change and hence not change the error or the method. If, then

and we can reduce the amount of information needed for a given optimal error.

If we may be able to reduce the error of optimal recovery if we have more information available. Fix. The greater number of terms we have of then the better we may be able to approximate, that is the smaller the optimal error of recovery. Let

and for

for any. If we know the first terms with some errors, then further increasing the terms will not yield a decrease in the error of optimal recovery.

We now apply the general results to the Hardy-Sobolev and Bergman-Sobolev spaces of functions on the unit disc. Let denote the set of functions holomorphic on the unit disc. Define the Hardy space of functions

as the set of all, with where

The Hardy-Sobolev space of functions, , are those such that and

is the class consisting of those

with. The Bergman space of functions

is the space of all such that

That is, is the space of all holomorphic functions in. The Bergman-Sobolev space of functions, , consists of with

and as the class of all

with.

So each space can be considered as the space with

For each space of functions we have the collection of points. If

then for

Therefore for

In this case we consider the collection of points

It is easy to see that if then the piecewise linear function will have points of break

For the space, the points to consider are

Again let and thus the points of break of will be precisely

For the special case of, the function has only a single point of break at the origin as

so that for. Furthermore, does not satisfy (7) as

Thus, in the applications of the general results, this case will be treated separately.

For notational purposes, let, be the points of break of for the space.

Corollary 1. Let or. If with or then the error of optimal recovery is given by (13) and (14) is an optimal method. If and then and is optimal.

Proof. For the spaces or, if and only if and. Thus if and only if or. Thus apply Theorem 3 to obtain the result for all spaces except. The dual problem in the case leads to a simple Lagrange function. The dual problem is specifically

Therefore the Lagrange function is simply given by

Now if we let and then

for any. So now proceed as in Theorem 3. As any will minimize, choose as in (18). The extremal problem (19) is solved similarly, and as then for.

It should be noted that the optimal method described is stable with respect to the inaccurate information data.

We now apply Theorem 4 to the Hardy-Sobolev spaces and Bergman-Sobolev spaces in which is explicitly defined to be the smallest nonnegative integer satisfying

For the case, for all. Thus does not depend on. So

and hence for any we are in the case.

Corollary 2. Let or. Suppose with. If or then let

be given by (12) and the optimal error is given by

(13) and (23) is an optimal method. If and then and is an optimal method.

Otherwise suppose. If or then the optimal error is given by (13) and (23) is an optimal method with and

. If and then

and is an optimal method.

Proof. As previously stated, if the only break point of is and furthermore as then given by (21) does not exist so we treat this special case. In this case, the dual extremal problem is

and the corresponding Lagrange function is simply

If and then for any

. Now proceed as in the proof of Theorem 4 to obtain the result.

We now apply Theorem 5 to the spaces or for. In this situation will be a non-decreasing sequence for all. Also, for any we have and we are always in the case. For then for both the Hardy and Bergman spaces and so the condition will be satisfied if we know satisfying

Corollary 3. Let or with or and and given by (27). Let, be given by (28). Then the error of optimal recovery is given by (29) and (38) is an optimal method. If and then and is an optimal method.

Proof. For Theorem 5 we simply used conditions (6) and (20), both of which are satisfied by and for all.

As a direct consequence of Theorem 5, we consider the situation in which we have a uniform bound on the inaccuracy of each of the first terms of. That is we take for every. If we define similarly as

and the apriori information is given by the values such that

Again we will only need the values for an optimal method.

As previously noted, since the optimal method and error of optimal recovery only use up to the term then any information beyond may be disregarded if as additional information will not decrease the error of optimal recovery.