On the Regularization Method for Solving Ill-Posed Problems with Unbounded Operators

Nguyen Van Kinh

Faculty of Applied Science, Ho Chi Minh University of Food Industry, Ho Chi Minh City, Vietnam.

DOI: 10.4236/ojop.2022.112002                

Abstract

Let be a linear, closed, and densely defined unbounded operator, where X and Y are Hilbert spaces. Assume that A is not boundedly invertible. Suppose the equation Au=f is solvable, and instead of knowing exactly f only know its approximation satisfies the condition: In this paper, we are interested a regularization method to solve the approximation solution of this equation. This approximation is a unique global minimizer of the functional , for any , defined as follows: . We also study the stability of this method when the regularization parameter is selected a priori and a posteriori. At the same time, we give an application of this method to the weak derivative operator equation in Hilbert space.

Keywords

Ill-Posed Problem, Regularization Method, Unbounded Linear Operator

1. Introduction

Let $A\mathrm{<mo>\; :\; </mo>}D\left(A\right)\subset X\to Y$ be a linear, closed, densely defined unbounded operator, where X and Y are Hilbert spaces. Consider the equation

$Au=f$ (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 $N\left(A\right)=\left\{u:Au=0\right\}$ 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, $R\left(A\right)$ is not closed [2].

If $\Vert A\Vert <\infty$, 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 $f_{\delta }$, the noisy data, are given

$\Vert f_{\delta }-f\Vert \le \delta$ (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 $A\mathrm{<mo>\; :\; </mo>}D\left(A\right)\subset X\to Y$.

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.

2. Some Main Results

Lemma 1. [2] Let $A\mathrm{<mo>\; :\; </mo>}D\left(A\right)\subset X\to Y$ be a linear, closed, densely defined operator, where X and Y are Hilbert spaces, then

1) the operators $T=A^{\ast }A$ and $Q=A{A}^{\ast }$ are densely defined, self-adjoint;

2) $A^{\ast }$ is closed, densely defined and $A^{\ast \ast }=A$ ;

3) the operators $\tilde{A}:={\left(I_X+A^{\ast }A\right)}^{-1}:X\to Y$, $A\tilde{A}:X\to Y$ are both defined on all of X and are bounded, $\sigma \left(\tilde{A}\right)\subseteq \left[\mathrm{0,1}\right]$. Also, $\tilde{A}$ is self-adjoint;

4) the operator $\hat{A}:={\left(I_Y+A{A}^{\ast }\right)}^{-1}:Y\to X$ is bounded and self-adjoint and $A^{\ast }\hat{A}\mathrm{<mo>\; :\; </mo>}Y\to X$ is bounded.

Lemma 2. Let $A\mathrm{<mo>\; :\; </mo>}D\left(A\right)\subset X\to Y$ be a linear, closed, densely defined operator, where X and Y are Hilbert spaces. If $f=Ay$, $y\perp N\left(A\right)$ then y is unique.

Proof. Suppose $y_1$, and $y_2$ satisfy $f=A{y}_1$, $y_1\perp N\left(A\right)$, and $f=A{y}_2$, $y_2\perp N\left(A\right)$ then $A\left(y_1-y_2\right)=0$. Thus $y_1-y_2\in N\left(A\right)$. There exits $u\in N\left(A\right)$ such that $y_1-y_2=u$ imply $\langle y_1-y_2,u\rangle =\langle y_1,u\rangle -\langle y_2,u\rangle =0=\langle u,u\rangle$. Thus $u=0$, it follows that $y_1=y_2$.

Theorem 1. For any $f\in Y$, the problem

$F\left(u\right)={\Vert Au-f\Vert }^2+\alpha {\Vert u\Vert }^2\to \min \mathrm{,}\alpha =\text{const}>\mathrm{0,}$ (3)

has a unique solution $u_{\alpha }=A^{\ast }{\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f$, where $I_Y$ is the identity operator on Y.

Proof. Consider the equation

$\left(A{A}^{\ast }+\alpha I_Y\right)w_{\alpha }=f\mathrm{,}\alpha =\text{const}>0$ (4)

which is uniquely solvable $w_{\alpha }={\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f$ (Lemma 1). Let $u_{\alpha }=A^{\ast }{w}_{\alpha }$ then

$\begin{array}{c}A{u}_{\alpha }=A{A}^{\ast }{\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f\\ =\left(A{A}^{\ast }+\alpha I_Y\right){\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f-\alpha I_Y{\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f\\ =f-\alpha w_{\alpha },\end{array}$

or

$A{u}_{\alpha }-f=-\alpha w_{\alpha }.$

We have

$F\left(u+v\right)={\Vert Au-f\Vert }^2+\alpha {\Vert u\Vert }^2+{\Vert Av\Vert }^2+\alpha {\Vert v\Vert }^2+2\mathrm{Re}\left[\left(Au-f,Av\right)+\alpha \left(u,v\right)\right],$ (5)

for any $v\in D\left(A\right)$. If $u=u_{\alpha }$, then

$\begin{array}{c}\left(A{u}_{\alpha }-f,Av\right)+\alpha \left(u_{\alpha },v\right)=-\alpha \left(w_{\alpha },Av\right)+\alpha \left(u_{\alpha },v\right)\\ =-\alpha \left(A^{\ast }{w}_{\alpha },v\right)+\alpha \left(u_{\alpha },v\right)=0.\end{array}$ (6)

Thus Equation (6) implies

$F\left(u_{\alpha }+v\right)=F\left(u_{\alpha }\right)+{\Vert Av\Vert }^2+\alpha {\Vert v\Vert }^2\ge F\left(u_{\alpha }\right)$ (7)

and $F\left(u_{\alpha }+v\right)=F\left(u_{\alpha }\right)$ if and only if $v=0$, so $u_{\alpha }$ is the unique minimier of $F\left(u\right)$.

Theorem 1 is proved.

Theorem 2. If $f=Ay$, $y\perp N\left(A\right)$ then

$\underset{\alpha \to 0}{\lim }\Vert u_{\alpha }-y\Vert =0,u_{\alpha }=A^{\ast }{\left(A{A}^{\ast }+\alpha I\right)}^{-1}f.$ (8)

Proof. It follows from Lemma 2, y is unique. Write Equation (4) as $A\left(A^{\ast }{w}_{\alpha }-y\right)=-\alpha w_{\alpha }$. Apply $A^{\ast }$, which is possible because $w_{\alpha }\in D\left(A^{\ast }\right)$, we obtain

$A^{\ast }A\left(u_{\alpha }-y\right)=-\alpha u_{\alpha }\mathrm{.}$ (9)

Multiply Equation (9) by $u_{\alpha }-y$, we obtain

$\left(A^{\ast }A\left(u_{\alpha }-y\right),u_{\alpha }-y\right)=-\alpha \left(u_{\alpha },u_{\alpha }-y\right)$

or

${\Vert A\left(u_{\alpha }-y\right)\Vert }^2=-\alpha \left({\Vert u_{\alpha }\Vert }^2-\left(u_{\alpha },y\right)\right).$ (10)

Since $\alpha >0$ this implies

${\Vert u_{\alpha }\Vert }^2\le \left(u_{\alpha }\mathrm{,}y\right)\mathrm{,}$

so

$\Vert u_{\alpha }\Vert \le \Vert y\Vert ,\forall \alpha >0.$

Therefore one may assume (taking a subsequence) that $u_{\alpha }$ weakly converges to an element z, $u_n\mathrm{<mo>\; :\; </mo>}=u_{{\alpha }_n}\rightharpoonup z$, as ${\alpha }_n\to 0$.

It follows from Equation (10) that

$\underset{n\to \infty }{\lim }\Vert A\left(u_n-y\right)\Vert =\mathrm{0,}i\mathrm{.}e\mathrm{.}\underset{n\to \infty }{\lim }\Vert A{u}_n-f\Vert =0.$

We shall prove that $z=y$.

Let $\gamma$ run through the set such that $\left\{A^{\ast }A\gamma \right\}$ is dense in $N^{\perp }$, where $N:=N\left(A\right)$. Note that $N\left(T\right)=N\left(A\right)$, where $T=A^{\ast }A$. Because of the formulas $X=\overline{R\left(T\right)}\oplus N\left(T\right)$ the $\left\{\gamma \right\}=D\left(T\right)$ is dense in X, and the set $\left\{T\gamma \right\}$ is dense in $N^{\perp }$.

Multiply the equation $T\left(u_{\alpha }-y\right)=-\alpha u_{\alpha }$ by $\gamma$ and pass to the limit $\alpha \to 0$. We obtain

$\left(z-y\mathrm{,}T\gamma \right)=0.$

We have assume $y\perp N$. If $z\perp N$, then $z-y\perp N$ and $z-y\perp N^{\perp }$, so $z-y=0$.

One may always assume that $z\perp N$ because $T{u}_{\alpha }=T\widetilde{u_{\alpha }}$, where $\widetilde{u_{\alpha }}$ is the orthogonal projection of $u_{\alpha }$ onto $N^{\perp }$.

Thus, we have $u_n\mathrm{<mo>\; :\; </mo>}=u_{{\alpha }_n}\rightharpoonup z$, $\Vert u_n\Vert \le \Vert y\Vert$. Thus implies $\underset{n\to \infty }{\lim }\Vert u_n-y\Vert =0$.

For convenience for the reader we prove this claim. Since $u_n\mathrm{<mo>\; :\; </mo>}=u_{{\alpha }_n}\rightharpoonup z$, one gets $\Vert y\Vert \le \underset{n\to \infty }{\underset{\_}{\lim }}\Vert u_n\Vert$. The inequality $\Vert u_n\Vert \le \Vert y\Vert$ implies $\underset{n\to \infty }{\bar{\lim }}\Vert u_n\Vert \le \Vert y\Vert$. Therefore $\underset{n\to \infty }{\lim }\Vert u_n\Vert =\Vert y\Vert$. This and the weakly converge $u_n\mathrm{<mo>\; :\; </mo>}=u_{{\alpha }_n}\rightharpoonup z$ imply strong convergence

${\Vert u_n-y\Vert }^2={\Vert u_n\Vert }^2+{\Vert y\Vert }^2-2\mathrm{Re}\left(u_n-y\right)\to \mathrm{0,}\text{as}n\to \infty \mathrm{.}$

Theorem 2 is proved.

Theorem 3. If $\Vert f_{\delta }-f\Vert \le \delta$, $f=Ay$, $y\perp N\left(A\right)$ and

$F_{\delta }\left(u\right)={\Vert Au-f_{\delta }\Vert }^2+\alpha {\Vert u\Vert }^2=\min ,$ (11)

then there exists a unique global minimier $u_{\alpha \mathrm{,}\delta }$ to (11) and $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$, where $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ and $\alpha \left(\delta \right)$ is properly chosen, in particular $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$.

Proof. It follows from Lemma 2, y is unique. The existence and uniqueness of the minimizer $u_{\alpha \mathrm{,}\delta }$ of $F_{\delta }\left(u\right)$ follows from Theorem 1 and $u_{\alpha \mathrm{,}\delta }=A^{\ast }{\left(Q+\alpha I_Y\right)}^{-1}{f}_{\delta }$. We have

$\Vert u_{\alpha \mathrm{,}\delta }-y\Vert \le \Vert u_{\alpha \mathrm{,}\delta }-u_{\alpha }\Vert +\Vert u_{\alpha }-f\Vert \mathrm{.}$

By Theorem 2, $\Vert u_{\alpha }-f\Vert \mathrm{<mo>\; :\; </mo>}=\eta \left(\alpha \right)\to 0$, as $\alpha \to 0$.

Let us estimate

$\Vert u_{\alpha ,\delta }-u_{\alpha }\Vert =\Vert A^{\ast }{\left(Q+\alpha I_Y\right)}^{-1}\left(f_{\delta }-f\right)\Vert \le \delta \Vert A^{\ast }{\left(Q+\alpha I_Y\right)}^{-1}\Vert .$

By the polar decomposition theorem [12], one has $A^{\ast }=U{Q}^{1/2}$, where U is a partial isometry, so $\Vert U\Vert \le 1$. One has,

$\begin{array}{c}\Vert A^{\ast }{\left(Q+\alpha I_Y\right)}^{-1}\Vert =\Vert U{Q}^{1/2}{\left(Q+\alpha I_Y\right)}^{-1}\Vert \le \Vert Q^{1/2}{\left(Q+\alpha I_Y\right)}^{-1}\Vert \\ =\underset{\lambda \ge 0c}{\max }\frac{{\lambda }^{1/2}}{\lambda +\alpha }=\frac{1}{2\sqrt{\alpha }},\end{array}$

where the spectral representation for Q was used.

Thus

$\Vert u_{\alpha \mathrm{,}\delta }-y\Vert \le \frac{\delta }{2\sqrt{\alpha }}+\eta \left(\alpha \right)\mathrm{.}$ (12)

For a fixed small $\delta >0$, choose $\alpha =\alpha \left(\delta \right)$ which minimizes the right side of Equation (12). Then $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$ and $\underset{\delta \to 0}{\lim }\left(\frac{\delta }{2\sqrt{\alpha \left(\delta \right)}}+\eta \left(\alpha \left(\delta \right)\right)\right)=0.$

Theorem 3 is proved.

Remark 1. We can also choose $\alpha \left(\delta \right)=c{\delta }^k$, with any $k<2$ and $c=\text{const}>0$. The constant c can be arbitrary.

We can also choose $\alpha \left(\delta \right)$ by a descrepancy principle. For example, consider the equation for finding $\alpha \left(\delta \right)$ :

$\Vert A{u}_{\alpha \mathrm{,}\delta }-f_{\delta }\Vert =c\delta \mathrm{,}c=\text{const}>1.$

We assume that $\Vert f_{\delta }\Vert >c\delta$.

That is the content of the following theorem.

Theorem 4. The equation

$\Vert A{u}_{\alpha ,\delta }-f_{\delta }\Vert =c\delta ,c=\text{const}>1,\Vert f_{\delta }\Vert >c\delta ,$ (13)

has a unique solution $\alpha =\alpha \left(\delta \right)>0$, $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$, and if $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$, then $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$.

Proof. Let us prove that Equation (13) has a unique root $\alpha \left(\delta \right)>0$, $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$. Indeed, using the spectial theorem [12], one gets

$\begin{array}{c}{\Vert A{u}_{\alpha ,\delta }-f_{\delta }\Vert }^2={\Vert {\left[A{A}^{\ast }\left(Q+\alpha I\right)\right]}^{-1}{f}_{\delta }\Vert }^2={\displaystyle {\int }_0^{\infty }}{\left|\frac{s}{s+\alpha }-1\right|}^{2}d\left(E_s,f_{\delta },f_{\delta }\right)\\ ={\alpha }^2{\displaystyle {\int }_0^{\infty }}\frac{d\left(E_s,f_{\delta },f_{\delta }\right)}{{\left(s+\alpha \right)}^2}:=g\left(\alpha ,\delta \right),\end{array}$

where $E_s$ is the resolution of the identity of Q.

One has $g\left(\infty ,\delta \right)={\Vert f_{\delta }\Vert }^2>c^2{\delta }^{\delta }$, and $g\left(+0,\delta \right)={\Vert P_{N^{\ast }}{f}_{\delta }\Vert }^2$, where $P_{N^{\ast }}$ is the orthoprojector onto the subspace $N^{\ast }=N\left(Q\right)=N\left(A^{\ast }\right)=R{\left(A\right)}^{\perp }$.

Since $f\in R\left(A\right)$ and $\Vert f_{\delta }-f\Vert \le \delta$, it follows that $\Vert P_{N^{\ast }}{f}_{\delta }\Vert \le \delta$, so

$g\left(+\mathrm{0,}\delta \right)\le {\delta }^2$. The function $g\left(\alpha \mathrm{,}\delta \right)$ for a fixed $\delta >0$ is a continuous strictly increasing function of $\alpha$ on $\left[\mathrm{0,}\infty \right)$. Therefore there exists a unique $\alpha =\alpha \left(\delta \right)>0$ which solves Equation (13) if $\Vert f_{\delta }\Vert >c\delta$ and $c>1$. Clearly $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$, because $\underset{\delta \to 0}{\lim }c\alpha \left(\delta \right)=0$ and the relation $\underset{\delta \to 0}{\lim }{\alpha }^2\left(\delta \right){\displaystyle {\int }_0^{\infty }}\frac{d\left(E_s,f_{\delta },f_{\delta }\right)}{{\left(s+\alpha \left(\delta \right)\right)}^2}=0$ implies $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$. The function $\alpha =\alpha \left(\delta \right)$ is a monotonically growing function of $\delta$ with $\alpha \left(+0\right)=0$.

Let us prove that $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$, where $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$, and $\alpha \left(\delta \right)$ solves Equation (13). By the definition of $u_{\delta }$, we get

${\Vert A{u}_{\alpha }-f_{\delta }\Vert }^2+\alpha \left(\delta \right){\Vert u_{\delta }\Vert }^2\le {\Vert Ay-f_{\delta }\Vert }^2+\alpha \left(\delta \right){\Vert y\Vert }^2={\delta }^2+\alpha \left(\delta \right){\Vert y\Vert }^2.$

Since ${\Vert A{u}_{\alpha }-f_{\delta }\Vert }^2=c^2{\delta }^2>{\delta }^2$, it follows that $\Vert u_{\delta }\Vert \le \Vert y\Vert$. Thus $u_{\delta }\rightharpoonup z$, and, as in the proof of Theorem 2, we obtain $z=y$ and $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$.

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 $f\left(u\right)={\Vert Au-f\Vert }^2+\alpha {\Vert u\Vert }^2$ is the equation

$A^{\ast }Au+\alpha u=A^{\ast }f.$ (14)

Hence in this case conditions are required $f\in D\left(A^{\ast }\right)$.

If A is unbounded, then f does not necessarily belong to $D\left(A^{\ast }\right)$, 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 $f\in Y$, in particular for $f\notin D\left(A^{\ast }\right)$, the element $u_{\alpha }=A^{\ast }{\left(A{A}^{\ast }+\alpha I_Y\right)}^{-1}f$ is well defined for any $\alpha =\text{const}>0$, provided that A is a closed, linear, densely defined operator in Hilbert space (Theorem 1).

3. Applications

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

We define the operator $A\mathrm{<mo>\; :\; </mo>}D\left(A\right)\subset H\to H$ as follows

$Af=\frac{\text{d}f}{\text{d}x}\mathrm{,}f\in D\left(A\right)\mathrm{,}$

with $D\left(A\right)=\left\{f\in H\mathrm{<mo>\; :\; </mo>}f\text{is}\text{absolutely}\text{continuous}\text{on}\left[\mathrm{0,1}\right]\text{and}{f}^{\prime }\left(x\right)\in H\right\}$.

Then $D\left(A\right)$ is dense in H since it contains the complete orthonormal set ${\left\{\sin n\pi x\right\}}_{n=1}^{\infty }$.

Clearly, A is a linear operator.

We show that A is a closed operator in Hilbert space H. Indeed, for suppose $\left\{f_n\right\}\subset D\left(A\right)$ and $f_n\to f$ and ${f^{\prime }}_n\to g$, in each case the convergence being in the $L^2\left[\mathrm{0,1}\right]$ norm. Since

$f_n\left(x\right)=f_n\left(0\right)+{\displaystyle {\int }_0^x}{f^{\prime }}_n\left(t\right)\text{d}t\mathrm{,}$

we see that the sequence of constant functions $\left\{f_n\left(0\right)\right\}$ converges in $L^2\left[\mathrm{0,1}\right]$ and hence the numerical sequence $\left\{f_n\left(0\right)\right\}$ converges to some real number C.

Now define $h\in D\left(A\right)$ by $h\left(x\right)=C+{\displaystyle {\int }_0^x}g\left(t\right)\text{d}t$. Then, for any $x\in \left[\mathrm{0,1}\right]$, we have by of the Cauchy-Schwarz inequality

$\begin{array}{c}\left|f_n\left(x\right)-h\left(x\right)\right|=\left|f_n\left(0\right)-C+{\displaystyle {\int }_0^x}\left({f^{\prime }}_n\left(t\right)-g\left(t\right)\right)\text{d}t\right|\\ \le \left|f_n\left(0\right)-C\right|+{\displaystyle {\int }_0^x}\left|{f^{\prime }}_n\left(t\right)-g\left(t\right)\right|\text{d}t\\ \le \left|f_n\left(0\right)-C\right|+\Vert {f^{\prime }}_n-g\Vert \end{array}$

and hence $f_n\to h$ uniformly. Therefore, $f=h\in D\left(A\right)$ and $Af=f^{\prime }=h^{\prime }=g$, verifying that the operator A is closed, linear, densely defined in $L^2\left[\mathrm{0,1}\right]$.

Let

$D^{\ast }=\left\{g\in D\left(A\right):g\left(0\right)=g\left(1\right)=0\right\}.$

Then for $f\in D\left(A\right)$ and $g\in D^{\ast }$, we have

$\langle Af,g\rangle ={\displaystyle {\int }_0^1}{f}^{\prime }\left(t\right)g\left(t\right)\text{d}t={f\left(t\right)g\left(t\right)|}_0^1-{\displaystyle {\int }_0^1}f\left(t\right)g^{\prime }\left(t\right)\text{d}t=\langle f,-g^{\prime }\rangle$

Therefore $D^{\ast }\subset D\left(A^{\ast }\right)$ and $A^{\ast }g=-g^{\prime }$, for $g\in D^{\ast }$.

On the other hand, if $g\in D\left(A^{\ast }\right)$, let $g^{\ast }=A^{\ast }g$. Then

$\langle Af,g\rangle =\langle f\mathrm{,}{g}^{\ast }\rangle$

for all $f\in D\left(A\right)$. In particular, for $f\equiv 1$, we find that ${\displaystyle {\int }_0^1}{g}^{\ast }\left(t\right)\text{d}t=0$.

Now let

$h\left(t\right)=-{\displaystyle {\int }_0^t}{g}^{\ast }\left(s\right)\text{d}s\mathrm{.}$

Then $h\in D^{\ast }$ and $A^{\ast }h=g^{\ast }=A^{\ast }g$ and hence $h-g\in N\left(A^{\ast }\right)$. Therefore, $\langle Af,h-g\rangle =0$, for all $f\in D\left(A\right)$. But $R\left(A\right)$ contains all continuous function and hence $g=h\in D^{\ast }$.

We conclude that

$D\left(A^{\ast }\right)=D^{\ast }\mathrm{,}\text{and}{A}^{\ast }g=-g^{\prime }.$

According to Theorem 1, for any $f\in Y=L^2\left[0,1\right]$, the problem

$F\left(u\right)={\Vert Au-f\Vert }^2+\alpha {\Vert u\Vert }^2\to \min \mathrm{,}\alpha =\text{const}>\mathrm{0,}$

has a unique solution $u_{\alpha }=A^{\ast }{\left(A{A}^{\ast }+\alpha I\right)}^{-1}f$, where I is the identity operator on $Y=L^2\left[\mathrm{0,1}\right]$. $f\in Y=L^2\left[0,1\right]$ does not necessarily belong to $D\left(A^{\ast }\right)$.

It follows from Theorem 2, that if $f=Ay$, $y\perp N\left(A\right)$ then

$\underset{\alpha \to 0}{\lim }\Vert u_{\alpha }-y\Vert =0,u_{\alpha }=A^{\ast }{\left(A{A}^{\ast }+\alpha I\right)}^{-1}f.$

It follows from Theorem 3, that if $\Vert f_{\delta }-f\Vert \le \delta$, $f=Ay$, and

$F_{\delta }\left(u\right)={\Vert Au-f_{\delta }\Vert }^2+\alpha {\Vert u\Vert }^2=\min ,$ (15)

then there exists a unique global minimier $u_{\alpha \mathrm{,}\delta }$ to Equation (15) and $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$, where $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ and $\alpha \left(\delta \right)$ is properly chosen, in particular $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$.

It follows from Theorem 4, that the equation

$\Vert A{u}_{\alpha ,\delta }-f_{\delta }\Vert =c\delta ,c=\text{const}>1,\Vert f_{\delta }\Vert >c\delta ,$

has a unique solution $\alpha =\alpha \left(\delta \right)>0$, $\underset{\delta \to 0}{\lim }\alpha \left(\delta \right)=0$, and if $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$, then $\underset{\delta \to 0}{\lim }\Vert u_{\delta }-y\Vert =0$.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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