# Moore-Penrose inverse and applications to linear equations

Y. Estaremi and S. Shamsi  Y. Estaremi and S. Shamsi<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematics and Computer Sciences, Golestan
University, Gorgan, Iran,  Department of Mathematics, Noor University, p. o.
box: 19395-3697, Tehran, Iran.

###### Abstract.

In the present paper we investigate Moore-Penrose inverse and characteristic
matrix of unbounded WCT operators on the Hilbert space $L^{2}(\mu)$. Also, we
obtain some applications of the Moore-Penrose inverse of unbounded operators
on the Hilbert space $\mathcal{H}$ to variational regularization problem.
Moreover,some examples are provided to illustrate the applications of our
results in linear equations and specially Fredholm integral equations.

###### Key words and phrases:

Conditional expectation, Moore-Penrose inverse, Variational Regularization.

###### 2010 Mathematics Subject Classification:

47B20,47B38

##  1\.  Introduction and Preliminaries

The theory of conditional expectation operators is a connection between
measure theory and operator theory. Indeed for any $\sigma$-subalgebra of a
$\sigma$-algebra in a measure space we can define a unique bounded operator on
the space of measurable function spaces, named conditional expectation with
respect to the chosen $\sigma$-subalgebra. Also, the class of composition of
conditional expectation and multiplication operators contains a large class of
linear operators(bounded and unbounded) like integral operator and weighted
composition operators. The theory of weighted conditional type(WCT) operators
have been studied by, for example Moy, [14], De Pagter and Grobler [8], Rao,
[16], Douglas, [3], Lambert, [12] and Herron, [9]. Specially, bounded WCT
operators have been studied by many authors on $L^{p}$-spaces. Also, we
initiated to study unbounded WCT operators on $L^{2}(\mu)$-spaces in [5] and
on $L^{p}(\mu)$-spaces in [7, 13]. Most of the results in the literature are
almost completely pure. In this paper we have obtained that the Moore-Penrose
inverse of WCT operators ($T$) at each single point $f\in L^{2}(\mu)$, is the
minimal-norm solution of the linear equation $Tu=f$. Also, it is true for any
unbounded densely defined operator $T$ on the Hilbert space $\mathcal{H}$.

Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space. For any
$\sigma$-finite subalgebra $\mathcal{A}\subseteq\Sigma$, the $L^{2}$-space,
$L^{2}(X,\mathcal{A},\mu_{|_{\mathcal{A}}})$ is abbreviated by
$L^{2}(\mathcal{A})$, and its norm is denoted by $\|.\|_{2}$. All comparisons
between two functions or two sets are to be interpreted as holding up to a
$\mu$-null set. The support of a measurable function $f$ is defined as
$S(f)=\\{x\in X;f(x)\neq 0\\}$. We denote the vector space of all (equivalence
classes of) almost everywhere finite valued measurable functions on $X$ by
$L^{0}(\Sigma)$.

For a $\sigma$-finite subalgebra $\mathcal{A}\subseteq\Sigma$, the conditional
expectation operator associated with $\mathcal{A}$ is the mapping
$f\rightarrow E^{\mathcal{A}}f$, defined for all non-negative, measurable
function $f$ as well as for all $f\in L^{2}(\Sigma)$, where
$E^{\mathcal{A}}f$, by the Radon-Nikodym theorem, is the unique
$\mathcal{A}$-measurable function satisfying

$\int_{A}fd\mu=\int_{A}E^{\mathcal{A}}fd\mu,\ \ \ (A\in\mathcal{A}).$

As an operator on $L^{2}({\Sigma})$, $E^{\mathcal{A}}$ is idempotent and
$E^{\mathcal{A}}(L^{2}(\Sigma))=L^{2}(\mathcal{A})$. If there is no
possibility of confusion, we write $E(f)$ in place of $E^{\mathcal{A}}(f)$. A
detailed discussion of the properties of conditional expectation may be found
in [16]. Let $u\in L^{0}(\Sigma)$. Then $u$ is said to be conditionable with
respect to $E$ if $u\in\mathcal{D}(E):=\\{f\in L^{0}(\Sigma):E(|f|)\in
L^{0}(\mathcal{A})\\}$. Throughout this paper we take $u$ and $w$ in
$\mathcal{D}(E)$. We now define the class of operator under investigation.

######  Definition 1.1.

Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space and let $\mathcal{A}$
be a $\sigma$-subalgebra of $\Sigma$, such that $(X,\mathcal{A},\mathcal{A})$
is also $\sigma$-finite. Suppose that $E$ is the corresponding conditional
expectation operator on $L^{2}(\Sigma)$ relative to $\mathcal{A}$. If $w,u\in
L^{0}(\Sigma)$ such that $uf$ is conditionable and $wE(uf)\in L^{2}(\Sigma)$,
for all $f\in L^{2}(\Sigma)$, then the corresponding weighted conditional
type(WCT) operator is the linear transformation
$M_{w}EM_{u}:L^{2}(\Sigma)\rightarrow L^{2}(\Sigma)$ defined by $f\rightarrow
wE(uf)$.

In this paper we are concerned to unbounded WCT operators on the Hilbert space
$L^{2}(\Sigma)$ and we will discuss Moore-Penrose inverse of these operators.
Also, we give a spectral characterization of closed range WCT operators.
Moreover, we find the characteristic matrix of WCT operators. Finally, we
obtain that the value of any unbounded densely defined operator $T$ on the
Hilbert space $\mathcal{H}$ at any single point $f\in\mathcal{H}$, is the
minimal-norm solution of the linear equation $Tu=f$.

##  2\. Moore-Penrose inverse of unbounded WCT operators

Let $u,w\in\mathcal{D}(E)$ and $T=M_{w}EM_{u}$ be the WCT operator on the
Hilbert space $L^{2}(\mu)$. Here we need to recall conditional type Holder
inequality on $L^{2}(\mu)$, it asserts that for every $f,g\in L^{2}(\mu)$, the
following inequality holds:

$|E(fg)|^{2}\leq E(|f|^{2})E(|g|^{2}).$

For $f\in\mathcal{D}(M_{w}EM_{u})\subseteq L^{2}(\mu)$, we have
$Tf=M_{w}EM_{u}(f)=wE(uf)$. We know that $S(|w|)\subseteq S(E(|w|^{2}))$ and
also by conditional type Holder inequality we have

$|wE(fg)|^{2}\leq|w|^{2}E(|f|^{2})E(|g|^{2}).$

Since $S(|Tf|)\subseteq S(E(|Tf|^{2})$, then we have the following lemma.

######  Lemma 2.1.

Let $S=S(E(|w|^{2}))$ and $G=S(E(|u|^{2}))$ and $T=M_{w}EM_{u}$ be the WCT
operator on $L^{2}(\mu)$. Then $S(Tf)\subseteq S\cap G$.

Let $T=M_{w}EM_{u}$ be the WCT operator on $L^{2}(\mu)$, i.e.,

$T=M_{w}EM_{u}:\mathcal{D}(T)\subseteq
L^{2}(\mu)\rightarrow\mathcal{R}(T)\subseteq L^{2}(\mu).$

As we have studied unbounded WCT operators in [5]. The WCT operator $T$ is
densely defined on $L^{2}(\mu)$ if and only if $E(|w|^{2})E(|u|^{2})<\infty$,
a.e., $\mu$. This gives us that the WCT operator $M_{w}EM_{u}$ is a densely
defined operator on $L^{2}(\mu)$ if and only if the WCT operator $M_{u}EM_{w}$
is a densely defined operator on $L^{2}(\mu)$. Also, for densely defined
operator $T=M_{w}EM_{u}$ on $L^{2}(\mu)$, we have
$T^{*}=M_{\bar{u}}EM_{\bar{w}}$. So by the above observations we get that
$T=M_{w}EM_{u}$ is densely defined if and only if $T^{*}$ is densely defined.
This implies that if $T=M_{w}EM_{u}$ is a densely defined operator on
$L^{2}(\mu)$, then it is closed. And consequently for densely defined operator
$T=M_{w}EM_{u}$, we have $T^{**}=T$. Therefore, if $T=M_{w}EM_{u}$ and
$E(|w|^{2})E(|u|^{2})<\infty$, a.e., $\mu$, then $T$ and $T^{*}$ both are
closed. Now we provide some technical lemmas for later use.

######  Lemma 2.2.

Let $EM_{u}$ be a linear operator on $L^{2}(\mu)$ such that $EM_{u}(f)=E(uf)$
for $f\in\mathcal{D}(EM_{u})$. Then
$L^{2}(\mathcal{A})\subseteq\mathcal{D}(EM_{u})$ if and only if
$\int_{A}|E(u)|^{2}d\mu<\infty$, for all $A\in\mathcal{A}$, with
$0<\mu(A)<\infty$.

###### Proof.

$L^{2}(\mathcal{A})\subseteq\mathcal{D}(EM_{u})$ if and only if
$\int_{X}|E(u)f|^{2}d\mu=\int_{X}|E(uf)|^{2}d\mu<\infty$, for every $f\in
L^{2}(\mathcal{A})$. Since $L^{2}(\mathcal{A})$ is $\sigma$-finite, then
simple functions are dense in $L^{2}(\mathcal{A})$. Therefore we have
$\int_{X}|E(u)f|^{2}d\mu<\infty$, for all $f\in L^{2}(\mathcal{A})$ if and
only if $\int_{A}|E(u)|^{2}d\mu<\infty$, for all $A\in\mathcal{A}$ with
$\mu(A)<\infty$. ∎

######  Lemma 2.3.

If the measure space $(X,\Sigma,\mu)$ is $\sigma$-finite and $f$ is a
$\Sigma$-measurable function such that $f<\infty$, a.e., $\mu$, then there
exists an increasing sequence of $\Sigma$-measurable sets
$\\{B_{n}\\}_{n\in\mathbb{N}}$ such that $f.\chi_{B_{n}}<n$,
$\mu(B_{n})<\infty$ and $X=\cup^{\infty}_{n=1}B_{n}$.

###### Proof.

It is an easy exercise. ∎

###  $(\bigstar)$ Let $(X,\Sigma,\mu)$ be a measure space such that if
$X=\cup_{n\in\mathbb{N}}X_{n}$ and $\mu(X_{n})<\infty$, for all $n$, in which
$\\{X_{n}\\}$ is an increasing sequence, then for each $A\in\Sigma$ with
$\mu(A)<\infty$, there exists $n\in\mathbb{N}$ such that $A\subseteq X_{n}$.

######  Proposition 2.4.

Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space that satisfies the
condition $(\bigstar)$. If the WCT operator $M_{w}EM_{u}$ is densely defined
on $L^{2}(\mu)$, then $L^{2}(\mathcal{A})\subseteq\mathcal{D}(EM_{wu})$.

###### Proof.

Let $M_{w}EM_{u}$ be densely defined on $L^{2}(\mu)$. Then we have
$E(|u|^{2})E(|w|^{2})<\infty$, a.e., $\mu$. Hence by conditional type Holder
inequality we get that $|E(uw)|^{2}<\infty$, a.e., $\mu$. Now by Lemma 2.3 we
get that there is an increasing sequence
$\\{B_{n}\\}_{n\in\mathbb{N}}\subseteq\mathcal{A}$ such that
$|E(uw)|^{2}\chi_{B_{n}}<n$, $\mu(B_{n})<\infty$ and
$X=\cup^{\infty}_{n=1}B_{n}$. Let $A\in\mathcal{A}$ such that $\mu(A)<\infty$.
Then by $(\bigstar)$, there exists $n\in\mathbb{N}$ for which $A\subseteq
B_{n}$. This implies that

$\int_{A}|E(uw)|^{2}d\mu\leq\int_{B_{n}}|E(uw)|^{2}d\mu\leq
n\mu(B_{n})<\infty.$

Thus by Lemma 2.2 we get that
$L^{2}(\mathcal{A})\subseteq\mathcal{D}(EM_{uw})$. ∎

Now we have the following Remark.

######  Remark 2.5.

If the WCT operator $M_{w}EM_{u}$ is densely defined on $L^{2}(\mu)$ such that
the underlying measure space satisfies the condition $(\bigstar)$, then by
Corollary 2.11 of [13] and the Proposition 2.4 we get that

$\sigma(M_{w}EM_{u})\cup\\{0\\}=\text{ess.range}(E(uw))\cup\\{0\\}.$

For the Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, the set of all
closed linear operators $T$, with $\mathcal{D}(T)\subseteq\mathcal{H}_{1}$ and
$\mathcal{R}(T)\subseteq\mathcal{H}_{2}$, is denoted by
$\mathcal{C}(\mathcal{H}_{1},\mathcal{H}_{2})$. Also, a linear operator
$T\in\mathcal{C}(\mathcal{H}_{1},\mathcal{H}_{2})$ is called regular if it is
densely defined. As we stated in the beginning of this section every densely
defined WCT operator is closed, so every densely defined WCT operator is
regular. Here we recall the definition of the Moore-Penrose inverse and some
related concepts.

######  Definition 2.6.

Let $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ be Hilbert spaces and
$T\in\mathcal{C}(\mathcal{H}_{1},\mathcal{H}_{2})$ be a densely defined
operator. Then there exists a unique densely defined operator
$T^{\dagger}\in\mathcal{C}(\mathcal{H}_{1},\mathcal{H}_{2})$, for which
$\mathcal{D}(T^{\dagger})=\mathcal{R}(T)\oplus\mathcal{R}(T)^{\perp}$ and has
the following properties:
1) $TT^{\dagger}y=P_{\overline{\mathcal{R}(T)}}y,\ \ \ \ \ \ \ \
y\in\mathcal{D}(T^{\dagger})$;
2) $T^{\dagger}Tx=P_{\mathcal{N}(T)^{\perp}}x,\ \ \ \ \ \ \ \
x\in\mathcal{D}(T)$;
3) $\mathcal{N}(T^{\dagger})=\mathcal{R}(T)^{\perp}$.
The unique operator $T^{\dagger}$ is called the Moore-Penrose inverse of $T$.

In [10] the authors have obtained the Moore-Penrose inverse of the bounded WCT
operator $T=M_{w}EM_{u}:L^{2}(\mu)\rightarrow L^{2}(\mu)$, in the special case
of $w,u\geq 0$, as follow:

$T^{\dagger}f=\frac{\chi_{K}}{E(u^{2})E(w^{2})}M_{u}EM_{w}(f),\ \ \ \ \ \
f\in\mathcal{D}(T^{\dagger}),$

in which $K=S(E(w))\cap S(E(u))$. But in the case that $w$ and $u$ are complex
valued measurable functions for closed range bounded WCT operator
$T=M_{w}EM_{u}$ on $L^{2}(\mu)$ we have

$T^{\dagger}=M_{\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}}M_{\bar{u}}EM_{\bar{w}},$

in which $S=S(E(|u|^{2}))\cap S(E(|w|^{2}))$. Since $\chi_{S}Tf=Tf$ and
$\chi_{S}T^{*}g=T^{*}g$, for $f,g\in L^{2}(\mu)$, then easily one can check
that $T^{\dagger}$ satisfies the conditions of the definition of Moore-Penrose
inverse of closed range bounded operators, i.e.,

$TT^{\dagger}T=T,\ \ \ T^{\dagger}TT^{\dagger}=T^{\dagger},\ \ \
(T^{\dagger}T)^{*}=T^{\dagger}T,\ \ \ (TT^{\dagger})^{*}=TT^{\dagger}.$

Moreover, if $T=M_{w}EM_{u}$ is an unbounded densely defined operator on
$L^{2}(\mu)$, then it is closed. So by definition 2.6 there exists a unique
closed densely defined operator $T^{\dagger}$ that satisfies with conditions
of the Definition 2.6. Similar to the bounded case, for unbounded densely
defined WCT operator $T=M_{w}EM_{u}$, easily we get that

$T^{\dagger}f=\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}\bar{u}E(\bar{w}f),\ \ \ \
\ f\in\mathcal{D}(T^{\dagger}).$

Indeed, $T^{\dagger}=M_{\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}}T^{*}$ and so
$\mathcal{N}(T^{\dagger})=\mathcal{N}(T^{*})=\mathcal{R}(T)^{\perp}$. Also for
$f\in\mathcal{D}(T^{\dagger})$ we have

$\displaystyle T^{\dagger}T(T^{\dagger}f)$
$\displaystyle=\frac{\chi_{S}}{(E(|u|^{2}))^{2}(E(|w|^{2}))^{2}}M_{\bar{u}}EM_{\bar{w}}M_{w}EM_{u}M_{\bar{u}}EM_{\bar{w}}f$
$\displaystyle=\frac{E(|u|^{2})E(|w|^{2})\chi_{S}}{(E(|u|^{2}))^{2}(E(|w|^{2}))^{2}}M_{\bar{u}}EM_{\bar{w}}f$
$\displaystyle=\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}M_{\bar{u}}EM_{\bar{w}}f$
$\displaystyle=T^{\dagger}f.$

This implies that $T^{\dagger}TT^{\dagger}=T^{\dagger}$ and
$(T^{\dagger}T)^{2}=T^{\dagger}T$. Also, by definition, it is clear that
$(T^{\dagger}T)^{*}=T^{\dagger}T$. Therefore
$T^{\dagger}T=P_{\overline{\mathcal{R}(T^{\dagger})}}$. Similarly we get that
$TT^{\dagger}=P_{\overline{\mathcal{R}(T)}}$. Therefor we have the following
theorem.

######  Theorem 2.7.

Let $T=M_{w}EM_{u}$ be unbounded densely defined WCT operator on $L^{2}(\mu)$.
Then the Moore-Penrose inverse of $T$ is as follow:

$T^{\dagger}=M_{\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}}M_{\bar{u}}EM_{\bar{w}},\text{in
which}\ \ S=S(E(|u|^{2}))\cap S(E(|w|^{2})).$

In the next Theorem we give an equivalent condition for unbounded WCT
operators to have closed range.

######  Theorem 2.8.

If WCT operator $M_{w}EM_{u}$ is densely defined on $L^{2}(\mu)$ such that the
underlying measure space satisfies with ($\bigstar$) condition, then
$\mathcal{R}(M_{w}EM_{u})$ is closed if and only if $E(|u|^{2})E(|w|^{2})\geq
r$, $\mu$, a.e., for some $r>0$.

###### Proof.

By the Remark 2.5 we have
$\sigma(M_{w}EM_{u})\cup\\{0\\}=\text{ess.range}(E(|u|^{2})E(|w|^{2}))\cup\\{0\\}$.
Also, for $T=M_{w}EM_{u}$ we have $T^{*}T=M_{\bar{u}E(|w|^{2})}EM_{u}$.
Therefore by Theorem 3.3 of [2] we get that $\mathcal{R}(T)$ is closed if and
only if $\sigma(M_{\bar{u}E(|w|^{2})}EM_{u})\subseteq[r,\infty)\cup\\{0\\}$,
for some $r>0$. This implies that $\mathcal{R}(T)$ is closed if and only if
$\text{ess.range}(E(|u|^{2})E(|w|^{2})\subseteq[r,\infty)\cup\\{0\\}$, for
some $r>0$, if and only if $E(|u|^{2})E(|w|^{2})\geq r$, $\mu$, a.e., for some
$r>0$. ∎

Here we recall some results of [1] as the next proposition, in which
$\mathcal{C}(T)=\mathcal{D}(T)\cap\mathcal{N}(T)^{\perp}$ and the reduced
minimum modulus of $T$ is defined by
$\gamma(T)=\inf\\{\|Tx\|:x\in\mathcal{C}(T),\ \ \|x\|=1\\}$.

######  Proposition 2.9.

[1] Let $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ be Hilbert spaces and
$T:\mathcal{H}_{1}\rightarrow\mathcal{H}_{2}$ be a closed operator. Then the
following are equivalent.

  1. (1)

$\mathcal{R}(T)$ is closed.

  2. (2)

$\mathcal{R}(T^{*})$ is closed.

  3. (3)

$T_{0}=T|_{\mathcal{C}(T)}$ has a bounded inverse.

  4. (4)

$\gamma(T)>0$.

  5. (5)

$T^{\dagger}$ is bounded.

  6. (6)

$\gamma(T)=\gamma(T^{*})$.

  7. (7)

$\mathcal{R}(T^{*}T)$ is closed.

  8. (8)

$\mathcal{R}(TT^{*})$ is closed.

Since $T^{*}=M_{\bar{u}}EM_{\bar{w}}$, then by Theorem 2.8 and Proposition 2.9
we have the following corollary.

######  Corollary 2.10.

If the WCT operator $T=M_{w}EM_{u}$ is densely define on Hilbert space
$L^{2}(\mu)$, then $\mathcal{R}(T)$ is closed if and only if
$\mathcal{R}(T^{*})$ is closed if and only if $E(|u|^{2})E(|w|^{2})\geq r$,
$\mu$, a.e., for some $r>0$.

Let $T\in\mathcal{C}(\mathcal{H}_{1},\mathcal{H}_{2})$. By definition of
reduced minimum modulus of $T$, for each $x\in\mathcal{D}(T)$ with $\|x\|=1$,
we can write $\gamma(T)^{2}\leq\|Tx\|^{2}=\langle T^{*}Tx,x\rangle$ and so
$\|T\|^{2}=\|T^{*}T\|\geq\gamma(T)^{2}$. Then we have $\|T\|\geq\gamma(T)$. If
$T=M_{w}EM_{u}$ and $\mathcal{R}(T)$ is closed, then by Proposition 2.9 we get
that $T^{\dagger}$ is bounded. Thus

$\|\frac{1}{(E(|u|^{2}))^{\frac{1}{2}}{(E(|w|^{2}))^{\frac{1}{2}}}}\|_{\infty}=\|T^{\dagger}\|\geq\gamma(T^{\dagger}).$

By Proposition 2.12 of [2] we have

$\gamma(T)=\frac{1}{\|T^{\dagger}\|}=\|(E(|u|^{2}))^{\frac{1}{2}}{(E(|w|^{2}))^{\frac{1}{2}}}\|=\|T\|$

Now by the above results, we have a necessary and sufficient condition for
boundedness of the Moore-Penrose inverse of an unbounded WCT operator with
closed range.

######  Corollary 2.11.

Under the assumptions of Theorem 2.8, for $T=M_{w}EM_{u}$, we get that
$T^{\dagger}$ is bounded if and only if $E(|u|^{2})E(|w|^{2})\geq r$, $\mu$,
a.e., for some $r>0$. And in this case
$\|T^{\dagger}\|=\|\frac{1}{(E(|u|^{2}))^{\frac{1}{2}}(E(|w|^{2}))^{\frac{1}{2}}}\|_{\infty}$.

###### Proof.

By the Proposition 2.9 we now that the $\mathcal{R}(T)$ is closed if and only
if $T^{\dagger}$ is bounded. Since

$T^{\dagger}=M_{\frac{\chi_{S}}{E(|u|^{2})E(|w|^{2})}}M_{\bar{u}}EM_{\bar{w}},\text{in
which}\ \ S=S(E(|u|^{2}))\cap S(E(|w|^{2})),$

is a WCT operator, then by [6] we get that $T^{\dagger}$ is bounded if and
only if

$\frac{1}{(E(|u|^{2}))^{\frac{1}{2}}(E(|w|^{2}))^{\frac{1}{2}}}\in
L^{\infty}(\mathcal{A}),$

and in this case
$\|T^{\dagger}\|=\|\frac{1}{(E(|u|^{2}))^{\frac{1}{2}}(E(|w|^{2}))^{\frac{1}{2}}}\|_{\infty}$.
∎

Now we recall some properties of closed densely defined operators between two
Hilbert spaces.

######  Proposition 2.12.

Let $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ be Hilbert spaces and
$T:\mathcal{H}_{1}\rightarrow\mathcal{H}_{2}$ be a closed densely defined
operator. Then the followings hold:

  1. (1)

$\mathcal{N}(T)=\mathcal{R}(T^{*})^{\perp}$.

  2. (2)

$\mathcal{N}(T^{*})=\mathcal{R}(T)^{\perp}$.

  3. (3)

$\mathcal{N}(T^{*}T)=\mathcal{N}(T)$.

  4. (4)

$\overline{\mathcal{R}(T^{*}T)}=\overline{\mathcal{R}(T^{*})}$.

Let $T$ be a linear operator on the Hilbert space $\mathcal{H}$ and
$\mathcal{G}(T)=\\{(x,Tx):x\in\mathcal{D}(T)\\}$ be the graph of $T$. Let
$P=(P_{ij})$ be the projection from $\mathcal{H}\oplus\mathcal{H}$ onto
$\overline{\mathcal{G}(T)}$. The $2\times 2$ operator $P$ in $\mathcal{H}$ is
called the characteristic matrix of $T$. Since $P$ is a projection, then
$P^{2}=P^{*}=P$ and so

$P^{*}_{ij}=P_{ij},\ \ \ \ \ \ \ \ \ \Sigma_{k}P_{ik}P_{kj}=P_{ij}.$

Therefore $P_{11}$ and $P_{22}$ are selfadjoint. Moreover, $TP_{11}=P_{21}$
and $TP_{12}=P_{22}$. Now we recall the next Proposition for characteristic
matrix for regular operators.

######  Proposition 2.13.

[17] Let $T$ be a regular operator with characteristic matrix $P=(P_{ij})$.
Then we have the followings:

  * •

The entries $P_{ij}$ are given by

$P_{11}=(I+T^{*}T)^{-1}\ \ \ \ \ \ \ \ P_{12}=T^{*}(I+TT^{*})^{-1},$
$P_{21}=T(I+T^{*}T)^{-1}\ \ \ \ \ \ \ \ P_{22}=TT^{*}(I+TT^{*})^{-1}.$

  * •

$I-P_{22}=(I+TT^{*})^{-1}$.

Let $T$ be a closed densely defined operator on the Hilbert space
$\mathcal{H}$ and $\lambda>0$. Then the operator $\lambda I+T^{*}T$ is
strictly positive and bounded below. Indeed, for $\lambda>0$ and
$f\in\mathcal{D}(T^{*}T)$, we have

$\langle(\lambda
I+T^{*}T)f,f\rangle=\langle\lambda^{\frac{1}{2}}f,\lambda^{\frac{1}{2}}f\rangle+\langle
Tf,Tf\rangle\geq\langle\lambda^{\frac{1}{2}}f,\lambda^{\frac{1}{2}}f\rangle=\lambda\|f\|^{2}>0.$

This implies that the operator $\lambda I+T^{*}T$ is strictly positive,
bounded below, one to one and so invertible.

Here we recall some basic calculations in functional analysis to obtain some
properties of the positive operator $P_{\lambda}=(\lambda I+T^{*}T)^{-1}$. Let

$U,V:\mathcal{H}\times\mathcal{H}\rightarrow\mathcal{H}\times\mathcal{H},\text{such
that},\ \ \ \ V(f,g)=(g,-f),\ \ \ U(f,g)=(g,f).$

Then we have

$\mathcal{G}_{T}\oplus V\mathcal{G}_{T^{*}}=\mathcal{G}_{T^{*}}\oplus
V\mathcal{G}_{T}=\mathcal{H}\times\mathcal{H}.$

Let $h\in\mathcal{H}$. Then $(h,0)\in\mathcal{H}\times\mathcal{H}$ and so
$(h,0)=(x,Tx)+(T^{*}y,-y)$ for $x\in\mathcal{D}(T)$ and
$y\in\mathcal{D}(T^{*})$. Therefore $h=x+T^{*}y$ and $y=Tx$. So we have
$h=x+T^{*}Tx=(I+T^{*}T)x$ and consequently $x=(I+T^{*}T)^{-1}h=P_{11}h$ and
$y=Tx=TP_{11}h$. This means that the operators $P_{11}$ and $TP_{11}$ are
every where defined. Therefore, we get that $P_{\lambda}$ is also every where
defined.

In addition for $x\in\mathcal{H}$,

$\displaystyle\langle P_{\lambda}x,x\rangle$ $\displaystyle=\langle
P_{\lambda}x,(\lambda I+T^{*}T)P_{\lambda}x\rangle$
$\displaystyle=\lambda\langle P_{\lambda}x,P_{\lambda}x\rangle+\langle
P_{\lambda}x,T^{*}TP_{\lambda}x\rangle$ $\displaystyle=\lambda\langle
P_{\lambda}x,P_{\lambda}x\rangle+\langle TP_{\lambda}x,TP_{\lambda}x\rangle$
$\displaystyle\geq 0.$

Then $P_{\lambda}$ is positive. Moreover, for $x,y\in\mathcal{H}$,

$\displaystyle\langle P_{\lambda}x,y\rangle$ $\displaystyle=\langle
P_{\lambda}x,P_{\lambda}^{-1}P_{\lambda}y\rangle$ $\displaystyle=\langle
P_{\lambda}x,(\lambda I+T^{*}T)T_{\lambda}y\rangle$
$\displaystyle=\lambda\langle P_{\lambda}x,P_{\lambda}y\rangle+\langle
P_{\lambda}x,T^{*}Ty\rangle$ $\displaystyle=\lambda\langle
P_{\lambda}x,P_{\lambda}y\rangle+\langle T^{*}TP_{\lambda}x,y\rangle$
$\displaystyle=\langle(\lambda I+T^{*}T)T_{\lambda}x,P_{\lambda}y\rangle$
$\displaystyle=\langle x,P_{\lambda}y\rangle.$

This implies that $P_{\lambda}$ is symmetric. In the next theorem apply the
above observations to WCT operators.

######  Theorem 2.14.

Let $T=M_{w}EM_{u}$ be densely defined on $L^{2}(\mu)$. Then the
characteristic matrix of $T$ is:

  * •

$P_{11}=I-M_{h_{1}\bar{u}}EM_{u}$,

  * •

$P_{21}=M_{h}M_{w}EM_{u}$,

  * •

$P_{12}=M_{h}M_{\bar{u}}EM_{\bar{w}}$,

  * •

$P_{22}=M_{h_{2}}M_{w}EM_{\bar{w}}$,

in which

$h_{1}=\frac{E(|w|^{2})}{1+E(|u|^{2})E(|w|^{2})},\ \ \ \
h_{2}=\frac{E(|u|^{2})}{1+E(|u|^{2})E(|w|^{2})},\ \ \ \
h=\frac{1}{1+E(|u|^{2})E(|w|^{2})}.$

###### Proof.

In [4] we have found that for densely defined WCT operator $T=M_{w}EM_{u}$ on
$L^{2}(\mu)$ and $\lambda\notin\text{ess.range}(E(uw))$,

$(\lambda
I-M_{w}EM_{u})^{-1}=\frac{1}{\lambda}I+M_{\frac{w}{\lambda(\lambda-E(uw))}}EM_{u}.$

Also, we have proven that if $E(|w|^{2})E(|u|^{2})<\infty$, a.e., $\mu$, then
$T=M_{w}EM_{u}$ and $T^{*}$ are closed and densely defined on the Hilbert
space $L^{2}(\mu)$, and also $T^{**}=T$. This means that $T$ is regular.

By the fact that $T^{*}T=M_{E(|w|^{2})\bar{u}}EM_{u}$ and
$TT^{*}=M_{E(|u|^{2})w}EM_{\bar{w}}$ are positive operators and the above
formula from [4], and also by Proposition 2.13 we have the characteristic
matrix of WCT operator $T=M_{w}EM_{u}$ as follows:

$\displaystyle P_{11}$
$\displaystyle=(I+M_{\bar{u}E(|w|^{2})}E_{u})^{-1}=I-M_{h_{1}\bar{u}}EM_{u},$
$\displaystyle P_{21}$
$\displaystyle=M_{w}EM_{u}(I+M_{\bar{u}E(|w|^{2})}EM_{u})^{-1}=M_{h}M_{w}EM_{u},$
$\displaystyle P_{12}$
$\displaystyle=M_{\bar{u}}EM_{\bar{w}}(I+M_{wE(|u|^{2})}EM_{\bar{w}})^{-1}=M_{h}M_{\bar{u}}EM_{\bar{w}},$
$\displaystyle P_{22}$
$\displaystyle=M_{wE(|u|^{2})}EM_{\bar{w}}(I+M_{wE(|u|^{2})}EM_{\bar{w}})^{-1}=M_{h_{2}}M_{w}EM_{\bar{w}},$

in which

$h_{1}=\frac{E(|w|^{2})}{1+E(|u|^{2})E(|w|^{2})},\ \ \ \
h_{2}=\frac{E(|u|^{2})}{1+E(|u|^{2})E(|w|^{2})},\ \ \ \
h=\frac{1}{1+E(|u|^{2})E(|w|^{2})}.$

∎

In the sequel we get some similar results, indeed for every $\lambda>0$ we
have

$P_{\lambda}=(\lambda
I+M_{\bar{u}E(|w|^{2})}EM_{u})^{-1}=\frac{1}{\lambda}I-M_{\frac{E(|w|^{2})\bar{u}}{\lambda(E(|w|^{2})E(|u|^{2})+\lambda)}}EM_{u}$

$Q_{\lambda}=(\lambda
I+M_{wE(|u|^{2})}EM_{\bar{w}})^{-1}=\frac{1}{\lambda}I-M_{\frac{E(|u|^{2})w}{\lambda(E(|w|^{2})E(|u|^{2})+\lambda)}}EM_{\bar{w}}$

$TP_{\lambda}=\frac{1}{\lambda}M_{w}EM_{u}-M_{\frac{E(|w|^{2})E(|u|^{2})w}{\lambda(E(|w|^{2})E(|u|^{2}))+\lambda}}EM_{u}=M_{t_{\lambda}}T$

i.e.,

$TP_{\lambda}=Q_{\lambda}T=M_{t_{\lambda}}T=TM_{t_{\lambda}},$

in which $t_{\lambda}=\frac{1}{E(|u|^{2})E(|w|^{2})+\lambda}$.

By Theorem 2.7 we have

$(I-P_{11})^{\dagger}=M_{\chi_{S}\frac{\bar{u}}{(E(|u|^{2}))^{2}}}EM_{u}$

and

$(I-Q_{1})^{\dagger}=M_{\frac{\chi_{S}w}{E(|w|^{2})}}EM_{\bar{w}}.$

As we compute in [6],

$|T|=M_{g\bar{u}}EM_{u},\ \ \ \ \text{in which}\ \ \ \
g=(\frac{E(|w|^{2})}{E(|u|^{2})})^{\frac{1}{2}}\chi_{S},$

and so by Remark 2.5 we have

$\sigma(|T|)\backslash\\{0\\}=\text{ess.range}((E(|w|^{2}))^{\frac{1}{2}}(E(|u|^{2}))^{\frac{1}{2}})\backslash\\{0\\}.$

By the above observations we get the next theorem.

######  Theorem 2.15.

Let $T=M_{w}EM_{u}:L^{2}(\mu)\rightarrow L^{2}(\mu)$ and
$(E(|u|^{2}))^{\frac{1}{2}}(E(|w|^{2}))^{\frac{1}{2}}<\infty,\ \ \
\text{a.e.}$. For every $\lambda>0$, the operators $P_{\lambda}$ and
$TP_{\lambda}$ are every where defined, bounded and contraction. Moreover,
$P_{\lambda}$ and $Q_{\lambda}$ are positive and symmetric.

##  3\. Variational Regularization

In this section we assume that $Tu=f$ is a solvable linear equation in a
Hilbert space $\mathcal{H}$, $T$ is a linear, closed, densely defined operator
in $\mathcal{H}$, which is not boundedly invertible. In [15] the authors
obtained some results for unbounded densely defined operators on the Hilbert
space $\mathcal{H}$ and then they applied the results to the variational
regularization problem $F(u):=\|Tu-f\|^{2}+\lambda\|u\|^{2}=min$, in which
$f\in\mathcal{H}$. In this section we determine and prove that the unique norm
minimal-norm solution of the equation $Tu=f$ is $T^{\dagger}f$. First we
discuss it for the WCT operators on the Hilbert space $L^{2}(\mu)$.

As we have obtained, if $T=M_{w}EM_{u}$ is densely defined, then
$M_{t_{\lambda}}T^{*}$ is a densely defined, closed WCT operator. Now we have
a nice relation between Moore-Penrose inverse of the WCT operator $T$ and a
minimal norm solution of the equation $Tu=f$.

######  Theorem 3.1.

Let $T=M_{w}EM_{u}$ be a densely defined WCT operator on the Hilbert space
$L^{2}(\mu)$. For every $f\in L^{2}(\mu)$, the linear functional
$F(u)=\|Tu-f\|^{2}+\lambda\|u\|^{2}$ has a unique global minimizer
$u_{\lambda}=M_{t_{\lambda}}T^{*}(f)$, where $\lambda>0$ and
$P_{\lambda}T^{*}(f)=Q_{\lambda}T^{*}(f)=M_{t_{\lambda}}T^{*}(f)$, in which
$t_{\lambda}=\frac{1}{\lambda+E(|u|^{2})E(|w|^{2})}$. If $f\in\mathcal{R}(T)$,
then

$\lim_{\lambda\rightarrow 0}\|u_{\lambda}-y\|=0,$

where $u_{\lambda}$ is the unique global minimizer of $F(u)$ and $y$ is the
minimal-norm solution to the equation $Tu=f$. In this case the minimal-norm
solution is exactly the value of the Moore-Penrose inverse of $T$ at $f$,
i.e., $T^{\dagger}(f)$.

###### Proof.

By using Theorem 2 from [15] and Theorem 2.7 we get the proof. ∎

######  Example 3.2.

Let $X=[-1,1]$, $d\mu=\frac{1}{2}dx$ and $\mathcal{A}$ be the sigma algebra
generated by symmetric intervals in $[0,1]$i.e., $\mathcal{A}=<\\{(-a,a):0\leq
a\leq 1\\}>$.

Then we have

$E^{\mathcal{A}}(f)(t)=\frac{f(t)+f(-t)}{2},\ \ t\in X,$

where $E^{\mathcal{A}}(f)$ is defined. If $u(t)=e^{t}$ and
$w(t)=\sin(t)+\cos(t)$, then

$E^{\mathcal{A}}(\mid u\mid^{2})(t)=\cosh(2t),\ \ \ \ \ \ \ \ \
E^{\mathcal{A}}(\mid w\mid^{2})(t)=1.$

Also,

$Tf(t)=M_{w}EM_{u}(f)(t)=(\sin(t)+\cos(t))(\frac{e^{t}f(t)+e^{-1}f(-t)}{2}).$

It’s clear that $T=M_{w}EM_{u}$ is densely defined. Since

$E^{\mathcal{A}}(\mid u\mid^{2})(t)E^{\mathcal{A}}(\mid
w\mid^{2})(t)=E^{\mathcal{A}}(\mid u\mid^{2})(t),$

then it’s bounded away from zero on $[-1,1]$. Hence by Theorem 2.8 we get that
$T$ has closed range. So $T^{\dagger}$ exists and

$T^{\dagger}=M_{\frac{e^{t}}{\cosh(2t)}}EM_{\sin(t)+\cos(t)}.$

Hence by Theorem 3.1 we get that the minimal norm solution of the equation:

$(\sin(t)+\cos(t))(e^{t}f(t)+e^{-t}f(-t))=2g$

is

$f(t)=\frac{(\sin(t)+\cos(t))e^{t}g(t)+(\cos(t)-\sin(t))e^{t}g(-t)}{2\cosh(2t)}.$

Also, by Theorem 2.14 we have the characteristic matrix of $T$ as follow:

$P=\left(\begin{array}[]{cc}I-M_{\frac{e^{t}}{1+\cosh(2t)}}EM_{e^{t}}&M_{\frac{e^{t}}{1+\cosh(2t)}}EM_{\sin(t)+\cos(t)}\\\
M_{\frac{\sin(t)+\cos(t)}{1+\cosh(2t)}}EM_{e^{t}}&M_{\frac{\cosh(2t)(\sin(t)+\cos(t))}{1+\cosh(2\alpha
t)}}EM_{\sin(t)+\cos(t)}\\\ \end{array}\right).$

######  Example 3.3.

Let $X=[0,1]\times[0,1]$, $d\mu=dtdt^{\prime}$, $\Sigma$ the Lebesgue subsets
of $\Omega$ and let $\mathcal{A}=\\{A\times[0,1]:A\ \mbox{is a Lebesgue set
in}\ [0,1]\\}$. Then, for each $f$ in $L^{2}(\Sigma)$,
$(Ef)(t,t^{\prime})=\int_{0}^{1}f(t,s)ds$, which is independent of the second
coordinate. Hence for every finite valued measurable functions $u,w$ on $X$,
the linear operator

$T(f)(t,t^{\prime})=w(t,t^{\prime})\int^{1}_{0}u(t,s)f(t,s)ds$

is densely defined on $L^{2}(X)$. Also, by Theorem 3.1 we get that the
solution of each equation of the form:

$w(t,t^{\prime})\int^{1}_{0}u(t,s)f(t,s)ds=g(t,t^{\prime})$

is

$f(t,t^{\prime})=\frac{u(t,t^{\prime})}{\int^{1}_{0}\mid
w\mid^{2}(t,s)ds+\int^{1}_{0}\mid
u\mid^{2}(t,s)ds}\int^{1}_{0}w(t,s)g(t,s)ds.$

This example is valid in the general case as follows:
Let $(X_{1},\Sigma_{1},\mu_{1})$ and $(X_{2},\Sigma_{2},\mu_{2})$ be two
$\sigma$-finite measure spaces and $X=X_{1}\times X_{2}$,
$\Sigma=\Sigma_{1}\times\Sigma_{2}$ and $\mu=\mu_{1}\times\mu_{2}$. Put
$\mathcal{A}=\\{A\times X_{2}:A\in\Sigma_{1}\\}$. Then $\mathcal{A}$ is a
sub-$\sigma$-algebra of $\Sigma$. Then for all $f$ in domain $E^{\mathcal{A}}$
we have

$E^{\mathcal{A}}(f)(t_{1})=E^{\mathcal{A}}(f)(t_{1},t_{2})=\int_{X_{2}}f(t_{1},s)d\mu_{2}(s)\
\ \ \mu-a.e.$

on $X$.

Also, if $(X,\Sigma,\mu)$ is a finite measure space and $u:X\times
X\rightarrow\mathbb{C}$ and $w:X\rightarrow\mathbb{C}$ are finite valued
measurable functions, then the operator $T:L^{2}(\Sigma)\rightarrow
L^{2}(\Sigma)$ defined by

$Tf(t)=w(t)\int_{X}u(t,s)f(s)d\mu(s),\ \ \ \ \ f\in L^{2}(\Sigma),$

is called weighted kernel operator on $L^{2}(\Sigma)$). We show that $T$ is a
WCT operator.[8] For each $f\in L^{2}(\Sigma)$, let
$f^{\prime}(t_{1},t_{2})=f(t_{2})$, for $t_{1},t_{2}\in X$. If we set
$\mathcal{A}=\\{A\times X:A\in\Sigma\\}$. Then $\mathcal{A}$ is a
sub-$\sigma$-algebra of $\Sigma\times\Sigma$ and for every $f\in
L^{2}(\Sigma)$, $f^{\prime}$ is $\Sigma\times\Sigma$-measurable. Then for all
$f$ in domain $E^{\mathcal{A}}$ we have

$E^{\mathcal{A}}(f^{\prime})(t_{1})=E^{\mathcal{A}}(f^{\prime})(t_{1},t_{2})=\int_{X}f^{\prime}(t_{1},s)d\mu(s)=\int_{X}f(s)d\mu(s)\
\ \ \mu-a.e.$

on $X$.

It is clear that
$u(t_{1},t_{2})f^{\prime}(t_{1},t_{2})=u(t_{1},t_{2})f(t_{2})$ is a
$\Sigma\otimes\Sigma$-measurable function, when $f\in L^{2}(\Sigma)$.
Therefore we have

$\displaystyle w(t)E^{\mathcal{A}}(uf)(t)$
$\displaystyle=E^{\mathcal{A}}(uf^{\prime})(t,s)$
$\displaystyle=w(t)\int_{X}u(t,t^{\prime})f^{\prime}(,t^{\prime})d\mu(t^{\prime})$
$\displaystyle=w(t)\int_{X}u(t,t^{\prime})f(t^{\prime})d\mu(t^{\prime})$
$\displaystyle=Tf(t).$

Hence $T=M_{w}EM_{u}$, i.e, $T$ is a WCT operator. And since $w$ and $u$ are
finite valued, then the WCT operator $T$ is densely defined. Therefore, by
Theorem 3.1 we can find the solution of Fredholm integral equations, i.e., the
solution of the integral equation of the form

$w(t)\int^{b}_{a}u(t,t^{\prime})f(t^{\prime})d\mu(t^{\prime})=g(t),\ \ \ \ \ \
\ \ a,b,t\in\mathbb{R}$

is

$f(t)=\frac{u(t,t^{\prime})}{\int^{b}_{a}\mid w\mid^{2}(s)ds+\int^{b}_{a}\mid
u\mid^{2}(t,s)ds}\int^{b}_{a}w(s)g(s)ds.$

This means all assertions of this paper are valid for a class of integral type
operators.

We recall that if $T$ is a linear densely defined closed operator on the
Hilbert space $\mathcal{H}$, then the domain of $T^{*}$ is also dense in
$\mathcal{H}$ and so $T^{**}=(T^{*})^{*}$ exists and $T^{**}=T$. So in the
next proposition we find a limit for a unique global minimizer of a linear
functional

######  Proposition 3.4.

Let $\mathcal{H}$ be a Hilbert space $T$ be a linear, unbounded and densely
defined operator in $\mathcal{H}$. For any $x\in\mathcal{H}$ if we set
$R_{\lambda}(x):=u_{\lambda}=T^{*}Q_{\lambda}(x)$, then

$\lim_{\lambda\rightarrow 0}R_{\lambda}(x)=\lim_{\lambda\rightarrow
0}u_{\lambda}=T^{\dagger}(x).$

###### Proof.

It is easy to see that the operator $T^{*}Q_{\lambda}$ is a bounded operator.
And direct computations show that

$TR_{\lambda}=I-\lambda Q_{\lambda}$

and so

$R^{*}_{\lambda}T^{*}=I-\lambda Q_{\lambda}.$

Hence we get that

$\lim_{\lambda\rightarrow 0}R^{*}_{\lambda}T^{*}=I.$

This implies that

$T^{*}\lim_{\lambda\rightarrow 0}R^{*}_{\lambda}T^{*}=T^{*}$

and therefore

$TST=T,$

in which $S=\lim_{\lambda\rightarrow 0}R_{\lambda}$. Similarly we get that

$(ST)^{*}=ST$

and also we have $ST=P_{\overline{\mathcal{R}(S)}}$ and
$TS=P_{\overline{\mathcal{R}(T)}}$. All these properties confirm that
$S=T^{\dagger}$ and so $\lim_{\lambda\rightarrow
0}u_{\lambda}=T^{\dagger}(x)$.

∎

Now in the net theorem under the assumptions of the Proposition 3.4 we get
that the value of the Moore-Penrose inverse of an unbounded operator $T$ at
the point $f$ is a minimal norm solution of the equation $Tu=f$.

######  Theorem 3.5.

Let $f\in\mathcal{H}$ and $F(u)=\|Tu-f\|^{2}+\lambda\|u\|^{2}$. Then the
linear functional $F$ has a unique global minimizer
$u_{\lambda}=T^{*}Q_{\lambda}$, where $\lambda>0$,
$Q_{\lambda}=(TT^{*}+\lambda I)^{-1}$, $P_{\lambda}=(T^{*}T+\lambda I)^{-1}$
and

$T^{*}Q_{\lambda}f=P_{\lambda}T^{*}f,$

where $P_{\lambda}$ is the closure of $P_{\lambda}T^{*}$ that is defined on
$\mathcal{D}(T^{*})$. If $f\in\mathcal{R}(T)$, then

$\lim_{\lambda\rightarrow 0}\|u_{\lambda}-T^{\dagger}f\|=0,$

i.e., $T^{\dagger}f$ is the minimal norm solution to the equation $Tu=f$.

###### Proof.

It is a direct consequence of Theorem 2 of [15]. ∎

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