Perturbation Results for Multivalued Linear Operators 113
y ∈ E}), and the minimum modulus (cf. [1, II.2.2])
γ(I
E
)=
∞, if E is dense in X,
inf
x/∈E
x+E
x+E
, otherwise
=
∞, if
E = X,
1, if E = X.
Proposition 2.2. Let S ∈ M L(X) satisfy D(S)=X and S < 1.Then,the
operator I − S has a dense range.
Proof. Write E := S(0) (it is known, [1, I.2.4], that S(0) is a linear subspace of
X). Since γ(I
E
) > 0, I
E
is open (see [1, II.3.2(b)]). Also, R(I
E
)=X, S(0) = E =
I
E
(0), and S < 1 ≤ γ(I
E
). Hence, by [1, III.7.5], I
E
−S has a dense range. Since
I − S = I
E
−S, the result follows.
Obviously, if X is finite dimensional, then D((I − S)
−1
)=X.
Proposition 2.3. Let S ∈ ML(X) satisfy D(S)=X and S < 1,andletS(0) be
closed. Then, (I − S)
−1
∈ M L(X) is continuous if and only if R(I −S) is closed.
Proof. By Proposition 2.2, if R(I −S)isclosedthenI −S is surjective, therefore,
D((I − S)
−1
)=X which implies, by the closed graph theorem (see [1, III.4.2]),
that (I −S)
−1
is continuous. Conversely, if (I − S)
−1
is continuous then I −S is
open (see [1, II.3.1]) and, by [1, III.4.2(b)], R(I − S)isclosed.
Let us now prove the main proposition.
Proposition 2.4. Let S ∈ ML(X) satisfy D(S)=X and S < 1,andletS(0) be
closed. Then, (I − S)
−1
∈ ML(X) is everywhere defined and continuous.
Proof. In order to prove that the operator (I − S)
−1
is continuous, it is enough,
by Proposition 2.3, to show that R(I − S) is closed. By [1, III.4.4], R(I − S)is
closed if and only if the range of the adjoint R((I −S)
) is closed. By [1, III.1.5(b)],
(I − S)
= I −S
. Thus, we are going to show that R(I −S
)isclosed.
By [1, II.3.2(a)] and [1, III.1.13], the adjoint operator S
is continuous. Then,
by [1, III.4.2(a)], D(S
) is a closed subspace of X
.SinceD(S)=X then, by [1,
III.1.4(b)], S
is single-valued. By [1, III.1.13], S
≤S < 1. Denote S
=
1 − d,where0<d<1. Some trivial calculations show that the operator I − S
is injective. Indeed, let (I −S
)x
= 0. Then, 0 = (I − S
)x
≥x
−S
x
≥
x
−S
x
= dx
≥0, i.e., x
= 0. Let us now show that I −S
is open. By
[1, II.2.1], since I −S
is injective, γ(I −S
)= inf
x
=0,x
∈D(S
)
(I−S
)x
x
≥ d>0. This
proves, by [1, II.3.2(b)], that I −S
is open as a map from D(I −S
)=D(S
)onto
R(I − S
), i.e., the inverse (I − S
)
−1
is continuous, by [1, II.3.1] and, therefore,
D(S
)andR(I −S
) are isomorphic. This implies, in turn, that R(I −S
)isclosed
since D(S
) is closed. Therefore, as mentioned above, R(I − S) is closed and, by
Proposition 2.3, the operator (I − S)
−1
is continuous.
Combining Proposition 2.2 with the fact that R(I −S) is closed, we conclude
that R(I −S)=X.