On the Regularization Method to Stable Approximate Solution of Equations of the First Kind with Unbounded Operators

Nguyen Van Kinh

Faculty of Information Technology, Gia Dinh University, Ho Chi Minh City, Vietnam.

DOI: 10.4236/ojop.2025.141001                

Abstract

Let $A:D\left(A\right)\subset X\to Y$ be a linear, closed, densely defined unbounded operator, where $X$ and $Y$ are Hilbert spaces. Assume that $A$ is not boundedly invertible. If equation (1) $Au=f$ is solvable, and $\Vert f_{\delta }-f\Vert \le \delta$ then the following results are provided: Problem $F_{\alpha ,\delta }\left(u\right):={\Vert Au-f_{\delta }\Vert }^2+\alpha {\Vert u\Vert }^2$ has a unique global minimizer $u_{\alpha ,\delta }$ for any $f_{\delta }\in Y$ , and $u_{\alpha ,\delta }=A^{*}{\left(A{A}^{*}+\alpha I_Y\right)}^{-1}{f}_{\delta }$ . Then there is a function $\alpha \left(\delta \right)$ , $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ such that $\underset{\delta \to 0}{\text{lim}}\Vert u_{\alpha \left(\delta \right),\delta }-x_0\Vert =0$ , where $x_0$ is the unique minimal-norm solution to (1). In this paper we introduce the regularization method solving Equation (1) with $A$ being a linear, closed, densely defined unbounded operator. At the same time, an application is given to the weak derivative operator equation.

Keywords

Ill-Posed Problem, Regularization Method, Unbounded Linear Operator

1. Introduction

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

$Au=f$ (1)

Problem finding approximate solution of (1) is called ill-posed (instability) in the sense of Hadamard [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 $\Vert A\Vert <\infty$ , and $f_{\delta }$ , the noisy data, are given

$\Vert f_{\delta }-f\Vert \le \delta$ (2)

problem finding approximate solution of (1) has been extensively studied in the literature in detail [2]-[6] and references therein.

If $A$ is a linear, closed, densely defined operator or is a linear unbounded operator, problem finding approximate solution of (1) has been some recent research [7]-[13], however, there are still many open problems such as parameter choice rules of regularization...

Our aim is to study problem finding approximate of (1) when unbounded operator $A$ is a linear, closed, densely defined and the noisy data $f_{\delta }$ satisfies (2). We shall introduce the regularization method for solving the problem, and introduce a priori and a posteriori parameter choice rules of regularization; at the same time give an application to the weak derivative operator equation in Hilbert space of measurable functions, Lebesgue squares integrable $L^2\left[0,1\right]$ .

2. The Some Main Results

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

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

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

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

4) the operator $\widehat{A}:={\left(I_Y+A{A}^{*}\right)}^{-1}:Y\to Y$ is bounded and self-adjoint and $A^{*}\widehat{A}:Y\to X$ is bounded, with $I_Y$ be the identity mapping on $Y$ .

Lemma 2. Let $A:D\left(A\right)\subset X\to Y$ be a linear, closed, densely defined operator, where $X$ and $Y$ are Hilbert spaces, if Equation (1) is solvable then minimal-norm solution of (1) is unique, we denote by $x_0$ , furthermore $x_0\perp N\left(A\right)$ .

Proof. Because $A:D\left(A\right)\subset X\to Y$ is a linear operator, the set of solutions of (1) is a convex set in the Hilbert space $X$ , which we denote by $X_0$ . Furthermore, $A$ is a closed operator then $X_0$ is a closed set. Indeed, suppose $x_n\in X_0$ , $x_n\to x$ . Because, $x_n\in X_0$ , then $A{x}_n=f,\forall n$ . Therefore, $A{x}_n\to f$ , then $x\in D\left(A\right)$ and $Ax=f$ or $x\in X_0$ . So, $X_0$ is a closed set in Hilbert $X$ .

Because $X_0$ is a convex, closed set in Hilbert $X$ , so it contains only one element with minimal norm $x_0$ [4] and $x_0\perp N\left(A\right)$ [4].

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

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

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

Proof. Consider the equation

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

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

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

or

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

We have

$\begin{array}{c}{F}_{\alpha }\left(u+v\right)={\Vert A\left(u+v\right)-f\Vert }^2+\alpha {\Vert u+v\Vert }^2={\Vert \left(Au-f\right)+Av\Vert }^2+\alpha {\Vert u+v\Vert }^2\\ ={\Vert Au-f\Vert }^2+{\Vert Av\Vert }^2+2\mathrm{Re}\langle Au-f,Av\rangle +\alpha \left({\Vert u\Vert }^2+{\Vert v\Vert }^2+2\mathrm{Re}\langle u,v\rangle \right)\\ ={\Vert Au-f\Vert }^2+{\Vert Av\Vert }^2+\alpha \left({\Vert u\Vert }^2+{\Vert v\Vert }^2\right)+2\mathrm{Re}\left(\langle Au-f,Av\rangle +\alpha \langle u,v\rangle \right)\end{array}$ (5)

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

$\begin{array}{c}\langle A{u}_{\alpha }-f,Av\rangle +\alpha \langle u_{\alpha },v\rangle =-\alpha \langle w_{\alpha },Av\rangle +\alpha \langle u_{\alpha },v\rangle \\ =-\alpha A^{*}\langle w_{\alpha },v\rangle +\alpha \langle u_{\alpha },v\rangle \\ =-\alpha \langle u_{\alpha },v\rangle +\alpha \langle u_{\alpha },v\rangle =0.\end{array}$ (6)

Thus (5) and (6) imply

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

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

Theorem 1 is proved.

Theorem 2. If $x_0$ is the unique minimal-norm solution of (1) and $u_{\alpha }$ is the unique minimizer of $F_{\alpha }\left(u\right)$ then

$\underset{\alpha \to 0}{\text{lim}}\Vert u_{\alpha }-x_0\Vert =0,\text{with}{u}_{\alpha }=A^{*}{\left(A{A}^{*}+\alpha I_Y\right)}^{-1}f.$ (7)

Proof. Write (4) as $A\left(A^{*}{w}_{\alpha }-x_0\right)=-\alpha w_{\alpha }$ . Apply $A^{*}$ , which is possible because $w_{\alpha }\in D\left(A^{*}\right)$ , we obtain

$A^{*}A\left(u_{\alpha }-x_0\right)=-\alpha u_{\alpha }.$ (8)

Multiply (8) by $u_{\alpha }-x_0$ , we obtain

$\langle A^{*}A\left(u_{\alpha }-x_0\right),u_{\alpha }-x_0\rangle =-\alpha \langle u_{\alpha },u_{\alpha }-x_0\rangle$

or

${\Vert A\left(u_{\alpha }-x_0\right)\Vert }^2=-\alpha \left({\Vert u_{\alpha }\Vert }^2-u_{\alpha },x_0\right).$ (9)

Since $\alpha >0$ this implies

${\Vert u_{\alpha }\Vert }^2\le \langle u_{\alpha },x_0\rangle ,$

So

$\Vert u_{\alpha }\Vert \le \Vert x_0\Vert ,\forall \alpha >0.$

Therefore, one may assume (taking a subsequence) that the sequence $u_{\alpha }$ weakly converges to an element $z$ , denoted by $u_n:=u_{{\alpha }_n}\rightharpoonup z$ , as ${\alpha }_n\to 0$ .

It follows from (9) that

$\underset{n\to \infty }{\text{lim}}\Vert A\left(u_n-x_0\right)\Vert =0,i.e.\underset{n\to \infty }{\text{lim}}\Vert A{u}_n-f\Vert =0.$

We shall prove that $z=x_0$ .

Let $\gamma$ run through the set such that $\left\{A^{*}A\gamma \right\}$ is dense in $N^{\perp }$ , where $N:=N\left(A\right)$ . Note that $N\left(T\right)=N\left(A\right)$ , where $T=A^{*}A$ [2].

Because of the formulas $X=\overline{R\left(T\right)}\oplus N\left(T\right)$ [2], then $\left\{\gamma \right\}=D\left(T\right)$ is dense in $X$ , and the set $\left\{T\gamma \right\}$ is dense in $N^{\perp }$ .

Multiply the equation $T\left(u_{\alpha }-x_0\right)=-\alpha u_{\alpha }$ by $\gamma$ and pass to the limit $\alpha \to 0$ . We obtain

$\left(z-x_0,T\gamma \right)=0.$

We have $x_0\perp N$ . If $z\perp N,$ then $z-x_0\perp N$ and $z-x_0\perp N^{\perp }$ , so $z-x_0=0.$

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 $N^{\perp }$ .

Since the sequence $\left\{u_n\right\}$ converges weakly to $z$ , and $\Vert u_n\Vert \le \Vert x_0\Vert$ , so $\underset{n\to \infty }{\text{lim}}\Vert u_n-x_0\Vert =0$ [14].

For convenience for the reader, we prove this claim as follows:

Since $u_n:=u_{{\alpha }_n}\rightharpoonup z$ , one gets $\Vert x_0\Vert \le \underset{n\to \infty }{\underset{\_}{\text{lim}}}\Vert u_n\Vert$ . The inequality $\Vert u_n\Vert \le \Vert x_0\Vert$ implies $\underset{n\to \infty }{\overline{\text{lim}}}\Vert u_n\Vert \le \Vert x_0\Vert$. Therefore $\underset{n\to \infty }{\text{lim}}\Vert u_n\Vert =\Vert x_0\Vert$ . This and the weakly converge $u_n:=u_{{\alpha }_n}\rightharpoonup z$ imply strong convergence

${\Vert u_n-x_0\Vert }^2={\Vert u_n\Vert }^2+{\Vert x_0\Vert }^2-2\mathrm{Re}\langle u_n,x_0\rangle \to 0,\text{as}n\to \infty .$

Theorem 2 is proved.

Theorem 3. If $\Vert f_{\delta }-f\Vert \le \delta$ , $x_0$ is the unique minimal-norm solution of (1), and

$F_{\alpha ,\delta }\left(u\right)={\Vert Au-f_{\delta }\Vert }^2+\alpha {\Vert u\Vert }^2=\text{min},$ (10)

then there exists a unique global minimizer $u_{\alpha ,\delta }$ to (10) and $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ , where $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ and $\alpha \left(\delta \right)$ is properly chosen, in particular $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ .

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

$\Vert u_{\alpha ,\delta }-x_0\Vert \le \Vert u_{\alpha ,\delta }-u_{\alpha }\Vert +\Vert u_{\alpha }-x_0\Vert .$

By Theorem 2, $\Vert u_{\alpha }-x_0\Vert :=\eta \left(\alpha \right)\to 0$ , as $\alpha \to 0$ , with $u_{\alpha }=A^{*}{\left(A{A}^{*}+\alpha I_Y\right)}^{-1}f=A^{*}{\left(Q+\alpha I_Y\right)}^{-1}f$ .

Let us estimate

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

By the polar decomposition theorem [15] [16], one has $A^{*}=U{Q}^{1/2}$ , where $U$ is a partial isometry, so $\Vert U\Vert \le 1$ . One has,

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

where the spectral representation for $Q$ was used.

Thus

$\Vert u_{\alpha ,\delta }-x_0\Vert \le \Vert u_{\alpha ,\delta }-u_{\alpha }\Vert +\Vert u_{\alpha }-x_0\Vert \le \frac{\delta }{2\sqrt{\alpha }}+\eta \left(\alpha \right).$ (11)

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

Theorem 3 is proved.

Remark 1. From (11), we can also choose $\alpha \left(\delta \right)=c{\delta }^k$ , with any $0<k<2$ and $c=\text{const}>0$ . The constant $c$ can be arbitrary.

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

$\Vert A{u}_{\alpha ,\delta }-f_{\delta }\Vert =c\delta ,c=\text{const}>1.$ (12)

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 $\alpha =\alpha \left(\delta \right)>0$ , $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ , and if $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ , then $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ .

Proof. Let us prove that Equation (13) has a unique root $\alpha \left(\delta \right)>0$ , $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ . Indeed, using the spectral theorem [15] [16], one gets

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

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

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

Since $f\in R\left(A\right)$ and $\Vert f_{\delta }-f\Vert \le \delta$ , it follows that $\Vert P_{N^{*}}{f}_{\delta }\Vert \le \delta$ , so $g\left(0^{+},\delta \right)\le {\delta }^2$ . The function $g\left(\alpha ,\delta \right)$ for a fixed $\delta >0$ is a continuous strictly increasing function of $\alpha$ on $\left[0,\infty \right)$ . Therefore, there exists a unique $\alpha =\alpha \left(\delta \right)>0$ which solves Eq. (13) if $\Vert f_{\delta }\Vert >c\delta$ and $c>1$ . Clearly $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ , because $\underset{\delta \to 0}{\text{lim}}c\alpha \left(\delta \right)=0$ and the relation $\underset{\delta \to 0}{\text{lim}}{\alpha }^2\left(\delta \right){\displaystyle {\int }_0^{\infty }\frac{d\left(E_s,f_{\delta },f_{\delta }\right)}{{\left(s+\alpha \left(\delta \right)\right)}^2}}=0$ implies $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ . The function $\alpha =\alpha \left(\delta \right)$ is a monotonically growing function of $\delta$ with $\alpha \left(0^{+}\right)=0$ .

Let us prove that $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ , 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

${\Vert A{u}_{\delta }-f_{\delta }\Vert }^2+\alpha \left(\delta \right){\Vert u_{\delta }\Vert }^2\le {\Vert A{x}_0-f_{\delta }\Vert }^2+\alpha \left(\delta \right){\Vert x_0\Vert }^2={\delta }^2+\alpha \left(\delta \right){\Vert x_0\Vert }^2.$

Since ${\Vert A{u}_{\delta }-f_{\delta }\Vert }^2=c^2{\delta }^2>{\delta }^2$ , it follows that $\Vert u_{\delta }\Vert \le \Vert x_0\Vert$ . Thus the sequence $u_{\delta }$ weakly converges to $z$ , denoted by $u_{\delta }\rightharpoonup z$ , and similar argument as in proof of Theorem 2, we obtain $z=x_0$ and $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ .

Theorem 4 is proved.

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

If $A$ is bounded, then a necessary condition for the minimum of the functional $F_{\alpha }\left(u\right)={\Vert Au-f\Vert }^2+\alpha {\Vert u\Vert }^2$ is the equation

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

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

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

According to [8] [12], we can replace $x_0$ is unique minimal-norm solution of (1) by $y$ is solution of (1) satisfying condition $y\perp N\left(A\right)$ , then Theorems 1-4 are still true.

3. Applications

We will give a concrete example applying the regularization method presented in Section 2.

Let $X=L^2\left[0,1\right]$ to be a Hilbert space of measurable functions, Lebesgue squares integrable, with scalar product

$\langle x,y\rangle ={\displaystyle {\int }_0^{1}x\left(t\right)\overline{y\left(t\right)}\text{d}t}$

We define the operator $A:D\left(A\right)\subset X\to X$ be the weak derivative operator in $X=L^2\left[0,1\right]$ , denoted by

$Ax=\frac{\text{d}x}{\text{d}t},x\in D\left(A\right),$

with domain $D\left(A\right)=\left\{x\in X:x\left(t\right)\text{is}\text{absolutely}\text{continuous}\text{on}\left[0,1\right]\text{and}\frac{\text{d}x}{\text{d}t}\in X\right\}$ .

Then, A is a linear, closed operator with dense domain.

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

Clearly, $A$ is a linear operator.

We now prove that $A$ is a closed operator in Hilbert $X$ . Indeed, for suppose $\left\{x_n\right\}\subset D\left(A\right)$ and $x_n\to x$ and ${x^{\prime }}_n\to g$ , in each case the convergence being in the $L^2\left[0,1\right]$ norm. Since

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

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

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

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

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

Let

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

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

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

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

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

$\langle Ax,g\rangle =\langle x,g^{*}\rangle$

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

Now let

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

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

We conclude that

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

If the equation

$Ax=f$ (15)

is solvable, then (15) has the unique minimal-normal solution $x_0$ . (Lemma 2)

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

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

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

It follows from Theorem 2, then

$\underset{\alpha \to 0}{\text{lim}}\Vert u_{\alpha }-x_0\Vert =0,u_{\alpha }=A^{*}{\left(A{A}^{*}+\alpha I\right)}^{-1}f.$

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

$F_{\alpha ,\delta }\left(u\right)={\Vert Au-f_{\delta }\Vert }^2+\alpha {\Vert u\Vert }^2=\text{min},$ (16)

then there exists a unique global minimizer $u_{\alpha ,\delta }$ to (16) and $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ , where $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ and $\alpha \left(\delta \right)$ is properly chosen, in particular $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ .

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 $\alpha =\alpha \left(\delta \right)>0$ , $\underset{\delta \to 0}{\text{lim}}\alpha \left(\delta \right)=0$ , and if $u_{\delta }:=u_{\alpha \left(\delta \right),\delta }$ , then $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$ .

Conflicts of Interest

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

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