Set of completed picture

The theorem of the completed image is a mathematical theorem from the branch of functional analysis . It makes a statement about when the image of a continuous linear operator is completed .

motivation

If there is a continuous linear operator between normalized spaces , then the dual operator is explained by . ${\displaystyle A:E\rightarrow F}$ ${\displaystyle A':F'\rightarrow E'}$ ${\displaystyle A'(\psi ):=\psi \circ A}$

For a subspace suppose that is the subspace in the dual space , which consists of all continuous linear functionals that vanish on. For a subspace one defines a subspace in analogously by the formula . (In the literature you can also find the designation for this and thus accept an ambiguity of the designation.) ${\displaystyle U\subset E}$ ${\displaystyle U^{\circ }:=\{\phi \in E'\mid \forall x\in U:\phi (x)=0\}}$ ${\displaystyle U}$ ${\displaystyle V\subset E\,'}$ ${\displaystyle E}$ ${\displaystyle ^{\circ }V:=\{x\in E\mid \forall \phi \in V:\phi (x)=0\}}$ ${\displaystyle V^{\circ }}$

With the help of the separation theorem (or Hahn-Banach's theorem ) one shows and , where “ker” and “im” stand for the core and image of an operator. Such a relationship is familiar from linear algebra . Correspondingly, one would expect an analogous formula , which, however, cannot apply in general, because it is always closed, whereas the image of a continuous linear operator is generally not. Is z. B. the Banach space of all zero sequences, then is a continuous linear operator with a dense (i.e. not closed) image. Such a phenomenon can be found in linear algebra; H. in finite-dimensional spaces, do not occur. In order to arrive at the formula expected from linear algebra, one must therefore assume that the image space is closed. This proves to be sufficient and equivalent to the corresponding statement about the dual operator: ${\displaystyle \operatorname {ker} (A\,')=\operatorname {im(A)} ^{\circ }}$ ${\displaystyle \operatorname {ker} (A)=\,^{\circ }\!\operatorname {im} (A')}$ ${\displaystyle \operatorname {im} (A)=\,^{\circ }\!\operatorname {ker} (A')}$ ${\displaystyle ^{\circ }\operatorname {ker} (A\,')}$ ${\displaystyle c_{0}}$ ${\displaystyle A:c_{0}\rightarrow c_{0},\,(a_{n})_{n}\mapsto ({\tfrac {1}{n}}\cdot a_{n})_{n}}$

Set of completed picture

Let be and Banach spaces and be a continuous linear operator. Then the following statements are equivalent: ${\displaystyle E}$ ${\displaystyle F}$ ${\displaystyle A:E\rightarrow F}$

• ${\displaystyle \operatorname {im} (A)^{}\,}$ is closed.
• ${\displaystyle \operatorname {im} (A)=\,^{\circ }\operatorname {ker} (A\,')}$.
• ${\displaystyle \operatorname {im} (A')}$ is closed.
• ${\displaystyle \operatorname {im} (A')=\operatorname {ker} (A)^{\circ }}$.

This sentence does not apply to general standardized spaces. So has z. B. a closed image (because is surjective !), But the dual operator, which with the usual identifications for sequence spaces is equal to the inclusion mapping , has no closed image. ${\displaystyle A=\mathrm {id} _{\ell ^{1}}:\ell ^{1}\rightarrow (\ell ^{1},\|\cdot \|_{\infty })}$ ${\displaystyle A}$ ${\displaystyle \ell ^{1}\rightarrow \ell ^{\infty }}$

application

If and are continuous linear operators between Banach spaces, one can derive the sequence ${\displaystyle A:E\rightarrow F}$ ${\displaystyle B:F\rightarrow G}$

${\displaystyle 0\rightarrow E{\stackrel {A}{\rightarrow }}F{\stackrel {B}{\rightarrow }}G\rightarrow 0}$

form, where 0 stands for the zero vector space , and ask the question of the exactness . The specified sequence is exact if and only if the dual sequence

${\displaystyle 0\rightarrow G'{\stackrel {B'}{\rightarrow }}F'{\stackrel {A'}{\rightarrow }}E'\rightarrow 0}$

is exact. If the output sequence is exact, the images from and are terminated with . Therefore, according to the above sentence, the pictures of and are also closed, and it follows ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle \operatorname {im} (A)=\operatorname {ker} (B)}$ ${\displaystyle A\,'}$ ${\displaystyle B\,'}$

${\displaystyle \operatorname {ker} (B')=\operatorname {im} (B)^{\circ }=G^{\circ }=0}$

${\displaystyle \operatorname {im} (B')=\operatorname {ker} (B)^{\circ }=\operatorname {im} (A)^{\circ }=\operatorname {ker} (A')}$

${\displaystyle \operatorname {im} (A')=\operatorname {ker} (A)^{\circ }=0^{\circ }=E'}$.

That means exactness of the dual sequence. Likewise, the exactness of the output sequence follows from the exactness of the dual sequence.

literature

• R. Meise, D. Vogt: Introduction to functional analysis . Vieweg, 1992, ISBN 3-528-07262-8 .