Let A : D ( A ) ⊂ X → Y be a linear, closed, densely defined unbounded operator, where X and Y are Hilbert spaces. Consider the equation

A u = f (1)

Problem-solving solution of Equation (1) is called ill-posed [1] if A is not boundedly invertible. This may happen if the null space N ( A ) = { u : A u = 0 } is not trivial, i.e. A is not injective, or if A is injective but A − 1 is unbounded, i.e. the range of A, R ( A ) is not closed [2].

If ‖ A ‖ < ∞ , problem-solving stable solution of Equation (1) has been extensively studied in the literature in detail ( [2] [3] [4] [5] [6] and references therein).

If f δ , the noisy data, are given

‖ f δ − f ‖ ≤ δ (2)

is a stable approximation to the unique minimal norn solution to Equation (1) was constructed by several methods (variational regularization, quasi solution, iterative regularization, ... [2] [3] [4] [5] [6] and references therein).

If A is a linear, closed, densely defined unbounded operator, problem (1) has been some recent research [2] [7] [8] [9] [10] [11], however, there are still many open problems such as parameter choice rules of regularization method with the linear closed, densely defined unbounded operator A : D ( A ) ⊂ X → Y .

Our aim is to study problem-solving stable approximation solution of Equation (1) when operator A is a linear, closed, and densely defined from space Hilbert X into space Hilbert Y. We shall present the regularization method for solving the problem (1), we shall present a priori and a posteriori parameter choice rules of regularization; at the same time give an application to the weak derivative operator equation.

The paper structure consists of 3 sections: Section 1 the introduction briefly summarizes the recent research results and come up with the problem that needs to be studied; Section 2 presents some main results; Section 3 presents an application of this method.

Lemma 1. [2] Let A : D ( A ) ⊂ X → Y be a linear, closed, densely defined operator, where X and Y are Hilbert spaces, then

1) the operators T = A ∗ A and Q = A A ∗ are densely defined, self-adjoint;

2) A ∗ is closed, densely defined and A ∗ ∗ = A ;

3) the operators A ˜ : = ( I X + A ∗ A ) − 1 : X → Y , A A ˜ : X → Y are both defined on all of X and are bounded, σ ( A ˜ ) ⊆ [ 0,1 ] . Also, A ˜ is self-adjoint;

4) the operator A ^ : = ( I Y + A A ∗ ) − 1 : Y → X is bounded and self-adjoint and A ∗ A ^ : Y → X is bounded.

Lemma 2. Let A : D ( A ) ⊂ X → Y be a linear, closed, densely defined operator, where X and Y are Hilbert spaces. If f = A y , y ⊥ N ( A ) then y is unique.

Proof. Suppose y 1 , and y 2 satisfy f = A y 1 , y 1 ⊥ N ( A ) , and f = A y 2 , y 2 ⊥ N ( A ) then A ( y 1 − y 2 ) = 0 . Thus y 1 − y 2 ∈ N ( A ) . There exits u ∈ N ( A ) such that y 1 − y 2 = u imply 〈 y 1 − y 2 , u 〉 = 〈 y 1 , u 〉 − 〈 y 2 , u 〉 = 0 = 〈 u , u 〉 . Thus u = 0 , it follows that y 1 = y 2 .

Theorem 1. For any f ∈ Y , the problem

F ( u ) = ‖ A u − f ‖ 2 + α ‖ u ‖ 2 → min ,   α = const > 0, (3)

has a unique solution u α = A ∗ ( A A ∗ + α I Y ) − 1 f , where I Y is the identity operator on Y.

Proof. Consider the equation

( A A ∗ + α I Y ) w α = f ,   α = const > 0 (4)

which is uniquely solvable w α = ( A A ∗ + α I Y ) − 1 f (Lemma 1). Let u α = A ∗ w α then

A u α = A A ∗ ( A A ∗ + α I Y ) − 1 f = ( A A ∗ + α I Y ) ( A A ∗ + α I Y ) − 1 f − α I Y ( A A ∗ + α I Y ) − 1 f = f − α w α ,

or

A u α − f = − α w α .

We have

F ( u + v ) = ‖ A u − f ‖ 2 + α ‖ u ‖ 2 + ‖ A v ‖ 2 + α ‖ v ‖ 2 + 2 Re [ ( A u − f , A v ) + α ( u , v ) ] , (5)

for any v ∈ D ( A ) . If u = u α , then

( A u α − f , A v ) + α ( u α , v ) = − α ( w α , A v ) + α ( u α , v ) = − α ( A ∗ w α , v ) + α ( u α , v ) = 0. (6)

Thus Equation (6) implies

F ( u α + v ) = F ( u α ) + ‖ A v ‖ 2 + α ‖ v ‖ 2 ≥ F ( u α ) (7)

and F ( u α + v ) = F ( u α ) if and only if v = 0 , so u α is the unique minimier of F ( u ) .

Theorem 1 is proved.

Theorem 2. If f = A y , y ⊥ N ( A ) then

lim α → 0 ‖ u α − y ‖ = 0 ,   u α = A ∗ ( A A ∗ + α I ) − 1 f . (8)

Proof. It follows from Lemma 2, y is unique. Write Equation (4) as A ( A ∗ w α − y ) = − α w α . Apply A ∗ , which is possible because w α ∈ D ( A ∗ ) , we obtain

A ∗ A ( u α − y ) = − α u α . (9)

Multiply Equation (9) by u α − y , we obtain

( A ∗ A ( u α − y ) , u α − y ) = − α ( u α , u α − y )

or

‖ A ( u α − y ) ‖ 2 = − α ( ‖ u α ‖ 2 − ( u α , y ) ) . (10)

Since α > 0 this implies

‖ u α ‖ 2 ≤ ( u α , y ) ,

so

‖ u α ‖ ≤ ‖ y ‖ ,   ∀ α > 0.

Therefore one may assume (taking a subsequence) that u α weakly converges to an element z, u n : = u α n ⇀ z , as α n → 0 .

It follows from Equation (10) that

lim n → ∞ ‖ A ( u n − y ) ‖ = 0,   i . e .   lim n → ∞ ‖ A u n − f ‖ = 0.

We shall prove that z = y .

Let γ run through the set such that { A ∗ A γ } is dense in N ⊥ , where N : = N ( A ) . Note that N ( T ) = N ( A ) , where T = A ∗ A . Because of the formulas X = R ( T ) ¯ ⊕ N ( T ) the { γ } = D ( T ) is dense in X, and the set { T γ } is dense in N ⊥ .

Multiply the equation T ( u α − y ) = − α u α by γ and pass to the limit α → 0 . We obtain

( z − y , T γ ) = 0.

We have assume y ⊥ N . If z ⊥ N , then z − y ⊥ N and z − y ⊥ N ⊥ , so z − y = 0 .

One may always assume that z ⊥ N because T u α = T u α ˜ , where u α ˜ is the orthogonal projection of u α onto N ⊥ .

Thus, we have u n : = u α n ⇀ z , ‖ u n ‖ ≤ ‖ y ‖ . Thus implies lim n → ∞ ‖ u n − y ‖ = 0 .

For convenience for the reader we prove this claim. Since u n : = u α n ⇀ z , one gets ‖ y ‖ ≤ lim _ n → ∞ ‖ u n ‖ . The inequality ‖ u n ‖ ≤ ‖ y ‖ implies lim ¯ n → ∞ ‖ u n ‖ ≤ ‖ y ‖ . Therefore lim n → ∞ ‖ u n ‖ = ‖ y ‖ . This and the weakly converge u n : = u α n ⇀ z imply strong convergence

‖ u n − y ‖ 2 = ‖ u n ‖ 2 + ‖ y ‖ 2 − 2 Re ( u n − y ) → 0,   as   n → ∞ .

Theorem 2 is proved.

Theorem 3. If ‖ f δ − f ‖ ≤ δ , f = A y , y ⊥ N ( A ) and

F δ ( u ) = ‖ A u − f δ ‖ 2 + α ‖ u ‖ 2 = min , (11)

then there exists a unique global minimier u α , δ to (11) and lim δ → 0 ‖ u δ − y ‖ = 0 , where u δ : = u α ( δ ) , δ and α ( δ ) is properly chosen, in particular lim δ → 0 α ( δ ) = 0 .

Proof. It follows from Lemma 2, y is unique. The existence and uniqueness of the minimizer u α , δ of F δ ( u ) follows from Theorem 1 and u α , δ = A ∗ ( Q + α I Y ) − 1 f δ . We have

‖ u α , δ − y ‖ ≤ ‖ u α , δ − u α ‖ + ‖ u α − f ‖ .

By Theorem 2, ‖ u α − f ‖ : = η ( α ) → 0 , as α → 0 .

Let us estimate

‖ u α , δ − u α ‖ = ‖ A ∗ ( Q + α I Y ) − 1 ( f δ − f ) ‖ ≤ δ ‖ A ∗ ( Q + α I Y ) − 1 ‖ .

By the polar decomposition theorem [12], one has A ∗ = U Q 1 / 2 , where U is a partial isometry, so ‖ U ‖ ≤ 1 . One has,

‖ A ∗ ( Q + α I Y ) − 1 ‖ = ‖ U Q 1 / 2 ( Q + α I Y ) − 1 ‖ ≤ ‖ Q 1 / 2 ( Q + α I Y ) − 1 ‖ = max λ ≥ 0 c λ 1 / 2 λ + α = 1 2 α ,

where the spectral representation for Q was used.

Thus

‖ u α , δ − y ‖ ≤ δ 2 α + η ( α ) . (12)

For a fixed small δ > 0 , choose α = α ( δ ) which minimizes the right side of Equation (12). Then lim δ → 0 α ( δ ) = 0 and lim δ → 0 ( δ 2 α ( δ ) + η ( α ( δ ) ) ) = 0.

Theorem 3 is proved.

Remark 1. We can also choose α ( δ ) = c δ k , with any k < 2 and c = const > 0 . The constant c can be arbitrary.

We can also choose α ( δ ) by a descrepancy principle. For example, consider the equation for finding α ( δ ) :

‖ A u α , δ − f δ ‖ = c δ ,   c = const > 1.

We assume that ‖ f δ ‖ > c δ .

That is the content of the following theorem.

Theorem 4. The equation

‖ A u α , δ − f δ ‖ = c δ ,   c = const > 1 ,   ‖ f δ ‖ > c δ , (13)

has a unique solution α = α ( δ ) > 0 , lim δ → 0 α ( δ ) = 0 , and if u δ : = u α ( δ ) , δ , then lim δ → 0 ‖ u δ − y ‖ = 0 .

Proof. Let us prove that Equation (13) has a unique root α ( δ ) > 0 , lim δ → 0 α ( δ ) = 0 . Indeed, using the spectial theorem [12], one gets

‖ A u α , δ − f δ ‖ 2 = ‖ [ A A ∗ ( Q + α I ) ] − 1 f δ ‖ 2 = ∫ 0 ∞ | s s + α − 1 | 2 d ( E s , f δ , f δ ) = α 2 ∫ 0 ∞ d ( E s , f δ , f δ ) ( s + α ) 2 : = g ( α , δ ) ,

where E s is the resolution of the identity of Q.

One has g ( ∞ , δ ) = ‖ f δ ‖ 2 > c 2 δ δ , and g ( + 0 , δ ) = ‖ P N ∗ f δ ‖ 2 , where P N ∗ is the orthoprojector onto the subspace N ∗ = N ( Q ) = N ( A ∗ ) = R ( A ) ⊥ .

Since f ∈ R ( A ) and ‖ f δ − f ‖ ≤ δ , it follows that ‖ P N ∗ f δ ‖ ≤ δ , so

g ( + 0, δ ) ≤ δ 2 . The function g ( α , δ ) for a fixed δ > 0 is a continuous strictly increasing function of α on [ 0, ∞ ) . Therefore there exists a unique α = α ( δ ) > 0 which solves Equation (13) if ‖ f δ ‖ > c δ and c > 1 . Clearly lim δ → 0 α ( δ ) = 0 , because lim δ → 0 c α ( δ ) = 0 and the relation lim δ → 0 α 2 ( δ ) ∫ 0 ∞ d ( E s , f δ , f δ ) ( s + α ( δ ) ) 2 = 0 implies lim δ → 0 α ( δ ) = 0 . The function α = α ( δ ) is a monotonically growing function of δ with α ( + 0 ) = 0 .

Let us prove that lim δ → 0 ‖ u δ − y ‖ = 0 , where u δ : = u α ( δ ) , δ , and α ( δ ) solves Equation (13). By the definition of u δ , we get

‖ A u α − f δ ‖ 2 + α ( δ ) ‖ u δ ‖ 2 ≤ ‖ A y − f δ ‖ 2 + α ( δ ) ‖ y ‖ 2 = δ 2 + α ( δ ) ‖ y ‖ 2 .

Since ‖ A u α − f δ ‖ 2 = c 2 δ 2 > δ 2 , it follows that ‖ u δ ‖ ≤ ‖ y ‖ . Thus u δ ⇀ z , and, as in the proof of Theorem 2, we obtain z = y and lim δ → 0 ‖ u δ − y ‖ = 0 .

Theorem 4 is proved.

Remark 2. Theorems 1 - 4 are well known in the case of a bounded operator A.

If A is bounded, then a necessary condition for the minimum of the functional f ( u ) = ‖ A u − f ‖ 2 + α ‖ u ‖ 2 is the equation

A ∗ A u + α u = A ∗ f . (14)

Hence in this case conditions are required f ∈ D ( A ∗ ) .

If A is unbounded, then f does not necessarily belong to D ( A ∗ ) , so Equation (14) may have no sence. Therefore, some changes in the usual theory are necessary. The changes are given in this paper. We prove, among other things, that for any f ∈ Y , in particular for f ∉ D ( A ∗ ) , the element u α = A ∗ ( A A ∗ + α I Y ) − 1 f is well defined for any α = const > 0 , provided that A is a closed, linear, densely defined operator in Hilbert space (Theorem 1).

As a simple concrete example of this type of approximation, consider differentiation in H = L 2 [ 0 , 1 ] .

We define the operator A : D ( A ) ⊂ H → H as follows

A f = d f d x ,   f ∈ D ( A ) ,

with D ( A ) = { f ∈ H : f   is   absolutely   continuous   on   [ 0,1 ]   and   f ′ ( x ) ∈ H } .

Then D ( A ) is dense in H since it contains the complete orthonormal set { sin n π x } n = 1 ∞ .

Clearly, A is a linear operator.

We show that A is a closed operator in Hilbert space H. Indeed, for suppose { f n } ⊂ D ( A ) and f n → f and f ′ n → g , in each case the convergence being in the L 2 [ 0,1 ] norm. Since

f n ( x ) = f n ( 0 ) + ∫ 0 x     f ′ n ( t ) d t ,

we see that the sequence of constant functions { f n ( 0 ) } converges in L 2 [ 0,1 ] and hence the numerical sequence { f n ( 0 ) } converges to some real number C.

Now define h ∈ D ( A ) by h ( x ) = C + ∫ 0 x     g ( t ) d t . Then, for any x ∈ [ 0,1 ] , we have by of the Cauchy-Schwarz inequality

| f n ( x ) − h ( x ) | = | f n ( 0 ) − C + ∫ 0 x ( f ′ n ( t ) − g ( t ) ) d t | ≤ | f n ( 0 ) − C | + ∫ 0 x | f ′ n ( t ) − g ( t ) | d t ≤ | f n ( 0 ) − C | + ‖ f ′ n − g ‖

and hence f n → h uniformly. Therefore, f = h ∈ D ( A ) and A f = f ′ = h ′ = g , verifying that the operator A is closed, linear, densely defined in L 2 [ 0,1 ] .

Let

D ∗ = { g ∈ D ( A ) : g ( 0 ) = g ( 1 ) = 0 } .

Then for f ∈ D ( A ) and g ∈ D ∗ , we have

〈 A f , g 〉 = ∫ 0 1     f ′ ( t ) g ( t ) d t = f ( t ) g ( t ) | 0 1 − ∫ 0 1     f ( t ) g ′ ( t ) d t = 〈 f , − g ′ 〉

Therefore D ∗ ⊂ D ( A ∗ ) and A ∗ g = − g ′ , for g ∈ D ∗ .

On the other hand, if g ∈ D ( A ∗ ) , let g ∗ = A ∗ g . Then

〈 A f , g 〉 = 〈 f , g ∗ 〉

for all f ∈ D ( A ) . In particular, for f ≡ 1 , we find that ∫ 0 1     g ∗ ( t ) d t = 0 .

Now let

h ( t ) = − ∫ 0 t     g ∗ ( s ) d s .

Then h ∈ D ∗ and A ∗ h = g ∗ = A ∗ g and hence h − g ∈ N ( A ∗ ) . Therefore, 〈 A f , h − g 〉 = 0 , for all f ∈ D ( A ) . But R ( A ) contains all continuous function and hence g = h ∈ D ∗ .

We conclude that

D ( A ∗ ) = D ∗ ,   and   A ∗ g = − g ′ .

According to Theorem 1, for any f ∈ Y = L 2 [ 0 , 1 ] , the problem

F ( u ) = ‖ A u − f ‖ 2 + α ‖ u ‖ 2 → min ,   α = const > 0,

has a unique solution u α = A ∗ ( A A ∗ + α I ) − 1 f , where I is the identity operator on Y = L 2 [ 0,1 ] . f ∈ Y = L 2 [ 0 , 1 ] does not necessarily belong to D ( A ∗ ) .

It follows from Theorem 2, that if f = A y , y ⊥ N ( A ) then

lim α → 0 ‖ u α − y ‖ = 0 ,   u α = A ∗ ( A A ∗ + α I ) − 1 f .

It follows from Theorem 3, that if ‖ f δ − f ‖ ≤ δ , f = A y , and

F δ ( u ) = ‖ A u − f δ ‖ 2 + α ‖ u ‖ 2 = min , (15)

then there exists a unique global minimier u α , δ to Equation (15) and lim δ → 0 ‖ u δ − y ‖ = 0 , where u δ : = u α ( δ ) , δ and α ( δ ) is properly chosen, in particular lim δ → 0 α ( δ ) = 0 .

It follows from Theorem 4, that the equation

‖ A u α , δ − f δ ‖ = c δ ,   c = const > 1 ,   ‖ f δ ‖ > c δ ,

has a unique solution α = α ( δ ) > 0 , lim δ → 0 α ( δ ) = 0 , and if u δ : = u α ( δ ) , δ , then lim δ → 0 ‖ u δ − y ‖ = 0 .

The author declares no conflicts of interest regarding the publication of this paper.

Van Kinh, N. (2022) On the Regularization Method for Solving Ill-Posed Problems with Unbounded Operators. Open Journal of Optimization, 11, 7-14. https://doi.org/10.4236/ojop.2022.112002