title: Characterization of surjective opeartor
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permalink: characterization-of-surjective-opeartor
date: 2018-11-18 08:20:08
categories:

• 数学
• 学习笔记
  tags:
• 泛函分析

This article provides a useful characterization of surjective operator, which is called the method of a priori estimates.
The main result concerning surjective operators is the following. For the proofs we refer the reader to [1, Theorem 2.20, page 47].

Theorem 1 Let and be two Banach spaces. Let %20%5Csubset%20E%20%5Crightarrow%20F#card=math&code=A%3A%20D%28A%29%20%5Csubset%20E%20%5Crightarrow%20F) be an unbounded linear operator that is densely defined and closed. The following properties are equivalent:

1. is surjective,
2. there is a constant such that , for all in #card=math&code=D%28A%5E%5Cstar%29),
3. %3D%7B0%7D#card=math&code=N%28A%5E%5Cstar%29%3D%7B0%7D) and #card=math&code=R%28A%5E%5Cstar%29) is closed.

If restricting our attention to Hilbert spaces, there is a further conclusion, which was proved by Hormander [2, Lemma 4.1.1, page 78].

Theorem 2 Let and be two Hilbert spaces. Let %20%5Csubset%20E%20%5Crightarrow%20F#card=math&code=A%3A%20D%28A%29%20%5Csubset%20E%20%5Crightarrow%20F) be an unbounded linear operator that is densely defined and closed. Let be a closed subspace of containing the range #card=math&code=R%28A%29) of . Then #card=math&code=K%3DR%28A%29) if and only if for some constant

.%20%5Ctag%7B1%7D%20%5Clabel%7B1%7D%0A#card=math&code=%5C%7Cf%5C%7C_F%20%5Cleq%20C%20%5C%7CA%5E%5Cstar%20f%5C%7C_E%2C%20%5Cquad%20f%5Cin%20K%5Ccap%20D%28A%5E%5Cstar%29.%20%5Ctag%7B1%7D%20%5Clabel%7B1%7D%0A)

The following lemma gives a sufficient condition for , which can be found in [1, Theorem 3.24, page 72].

Lemma 3 Let and be two reflexive Banach spaces. Let %20%5Csubset%20E%20%5Crightarrow%20F#card=math&code=A%3A%20D%28A%29%20%5Csubset%20E%20%5Crightarrow%20F) be an unbounded linear operator that is densely defined and closed. Then #card=math&code=D%28A%5E%5Cstar%29) is dense in . Thus is well defined and it may also be viewed as an unbounded operator from into . Then we have .

Proof of Theorem 2 First assume that is valid and let . Since , the equation is equivalent to the identity

_E%3D(g%2Cf)_F%2C%20%5Cquad%20f%20%5Cin%20D(A%5E%5Cstar)%0A#card=math&code=%28u%2CA%5E%5Cstar%20f%29_E%3D%28g%2Cf%29_F%2C%20%5Cquad%20f%20%5Cin%20D%28A%5E%5Cstar%29%0A)

It suffices to prove

_F%7C%5Cleq%20C%5C%7Cg%5C%7C_F%20%5C%7CA%5E%5Cstar%20f%5C%7C_E%2C%20%5Cquad%20f%5Cin%20D(A%5E%5Cstar).%20%5Ctag%7B2%7D%20%5Clabel%7B2%7D%0A#card=math&code=%7C%28g%2Cf%29_F%7C%5Cleq%20C%5C%7Cg%5C%7C_F%20%5C%7CA%5E%5Cstar%20f%5C%7C_E%2C%20%5Cquad%20f%5Cin%20D%28A%5E%5Cstar%29.%20%5Ctag%7B2%7D%20%5Clabel%7B2%7D%0A)

Indeed, let be the antilinear functional _F#card=math&code=A%5E%5Cstar%20f%5Crightarrow%28g%2Cf%29_F) from #card=math&code=R%28A%5E%2A%29). The map is well defined, since is injective by . An application of the Hahn-Banach theorem to shows that there exists a linear functional from , which is bounded by . According to Riesz representation theorem, there exists such that _E%3D(g%2Cf)_F#card=math&code=%28u%2CA%5E%5Cstar%20f%29_E%3D%28g%2Cf%29_F) for all #card=math&code=f%20%5Cin%20D%28A%5E%5Cstar%29). Thus, the equation has a solution with

To prove we first note that, if is orthogonal to , we have _F%3D0#card=math&code=%28g%2Cf%29_F%3D0), and %3D(f%2CAv)%3D0#card=math&code=%28A%5E%5Cstar%20f%2C%20v%29%3D%28f%2CAv%29%3D0) for all #card=math&code=v%5Cin%20D%28A%29), since %5Csubset%20K#card=math&code=R%28A%29%5Csubset%20K), which implies . Hence it is sufficient to prove when #card=math&code=f%20%5Cin%20K%20%5Ccap%20D%28A%5E%5Cstar%29) and then it follows immediately from .

Conversely, assuming that %3DK#card=math&code=R%28A%29%3DK), we must prove that the set

%2C%20%5C%7CA%5E%5Cstar%20f%5C%7C_E%20%5Cleq%201%5C%7D%0A#card=math&code=B%3D%5C%7Bf%3Bf%5Cin%20K%20%5Ccap%20D%28A%5E%5Cstar%29%2C%20%5C%7CA%5E%5Cstar%20f%5C%7C_E%20%5Cleq%201%5C%7D%0A)

is bounded. To do so it is sufficient to prove that is weakly bounded in , this means that %7C_F#card=math&code=%7C%28f%2Cg%29%7C_F) is bounded when for every fixed (see following lemma). But by hypothesis we can choose #card=math&code=u%20%5Cin%20D%28A%29) such that , and this implies

_F%7C%3D%7C(A%5E%5Cstar%20f%2Cu)%7C_E%20%5Cleq%20%5C%7Cu%5C%7C_E%2C%20%5Cquad%20f%5Cin%20B%0A#card=math&code=%7C%28f%2Cg%29_F%7C%3D%7C%28A%5E%5Cstar%20f%2Cu%29%7C_E%20%5Cleq%20%5C%7Cu%5C%7C_E%2C%20%5Cquad%20f%5Cin%20B%0A)

The theorem is proved.

Lemma 4 (weakly bounded implies strongly bounded) Let be a Banach space. The subset is weakly bounded, this means t %7C%20%5Cleq%20M_f#card=math&code=%7Cf%28x%29%7C%20%5Cleq%20M_f) when for every fixed , where is a constant independent of . Then is bounded.

Proof of Lemma 4 Consider continuous linear functional given by %3Df(x)#card=math&code=F_x%20%28f%29%3Df%28x%29) for all when . It is obviously that . It follows from weakly boundedness of that

%5C%7C%20%5Cleq%20Mf%20%3C%2B%5Cinfty%2C%20%5Cquad%20%20f%5Cin%20E%5E%5Cstar.%0A#card=math&code=%5Csup%7Bx%5Cin%20B%7D%5C%7CF_x%28f%29%5C%7C%20%5Cleq%20M_f%20%3C%2B%5Cinfty%2C%20%5Cquad%20%20f%5Cin%20E%5E%5Cstar.%0A)

By uniform boundedness principle, we have , which implies for all . The lemma is proved.

If is surjective in the theorem 2, we have further information concerning right inverse of which is the following corollary.

Corollary 5 Under the assumption of Theorem 2, if holds for some , and is surjective, there exists a linear continuous operator from to such that

%3Df%2C%20%5Cquad%20%5Cforall%20f%5Cin%20F%2C%5C%5C%0A%5C%7CG(f)%5C%7C_E%20%5Cleq%20%5Cfrac%7B1%7D%7Bc%7D%20%5C%7Cf%5C%7C_F.%0A#card=math&code=A%5Ccirc%20G%28f%29%3Df%2C%20%5Cquad%20%5Cforall%20f%5Cin%20F%2C%5C%5C%0A%5C%7CG%28f%29%5C%7C_E%20%5Cleq%20%5Cfrac%7B1%7D%7Bc%7D%20%5C%7Cf%5C%7C_F.%0A)

Proof of Corollary 5 It follows from that is injective. According to bounded inverse theorem, there exists a bounded linear operator %5E%7B-1%7D#card=math&code=%28A%5E%5Cstar%29%5E%7B-1%7D) from #card=math&code=R%28A%5E%5Cstar%29) to which satisfies %5E%7B-1%7D%5C%7C%3C%5Cfrac%7B1%7D%7Bc%7D#card=math&code=%5C%7C%28A%5E%5Cstar%29%5E%7B-1%7D%5C%7C%3C%5Cfrac%7B1%7D%7Bc%7D). Indeed, it follows from

%5E%7B-1%7Dy%5C%7C%20%3D%20%5C%7C(A%5E%5Cstar)%5E%7B-1%7DA%5E%5Cstar%20x%5C%7C%3D%5C%7Cx%5C%7C%5Cleq%20%5Cfrac%7B1%7D%7Bc%7D%5C%7CA%5E%5Cstar%20x%5C%7C%3D%5Cfrac%7B1%7D%7Bc%7D%5C%7Cy%5C%7C%2C%20%5Cforall%20y%5Cin%20R(A%5E%5Cstar).%0A#card=math&code=%5C%7C%28A%5E%5Cstar%29%5E%7B-1%7Dy%5C%7C%20%3D%20%5C%7C%28A%5E%5Cstar%29%5E%7B-1%7DA%5E%5Cstar%20x%5C%7C%3D%5C%7Cx%5C%7C%5Cleq%20%5Cfrac%7B1%7D%7Bc%7D%5C%7CA%5E%5Cstar%20x%5C%7C%3D%5Cfrac%7B1%7D%7Bc%7D%5C%7Cy%5C%7C%2C%20%5Cforall%20y%5Cin%20R%28A%5E%5Cstar%29.%0A)

Hence, we have %5E%7B-1%7DA%5E%5Cstar)%5E%5Cstar%3DA((A%5E%5Cstar)%5E%7B-1%7D)%5E%5Cstar#card=math&code=I%3D%28%28A%5E%5Cstar%29%5E%7B-1%7DA%5E%5Cstar%29%5E%5Cstar%3DA%28%28A%5E%5Cstar%29%5E%7B-1%7D%29%5E%5Cstar). Letting %5E%7B-1%7D)%5E%5Cstar#card=math&code=G%3D%28%28A%5E%5Cstar%29%5E%7B-1%7D%29%5E%5Cstar) proves the corollary.

References

1. Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differentioal Equations
2. Lars Hormander, An introduction to complex analysis in several variables