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 $L^2\left[0,1\right]$.

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 $\hat{A}:={\left(I_Y+A{A}^{*}\right)}^{-1}:Y\to Y$ is bounded and self-adjoint and $A^{*}\hat{A}:Y\to X$ 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.

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 $\underset{\delta \to 0}{\text{lim}}\Vert u_{\delta }-x_0\Vert =0$.