Positive and negative parts of an unbounded self-adjoint operator

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This question is about the proof of lemma 14 in 1, which deals with the decomposition of an unbounded self-adjoint operator $A$ in positive and negative parts: $A = A^+ - A^-$. I have trouble understanding the proof given at the bottom of page 27 that $A^- geq 0$.

Indeed, the author writes that $(A^+ + A^-)Ex = |A|Ex$, where $x in mathcalD(A)$ and $E$ is the orthogonal projector onto $mathcalN(A^+)$. This seems to assume that $Ex in mathcalD(A)$ but I don't see where this would come from. Any idea?

Bernau, S. (1968). The square root of a positive self-adjoint operator. Journal of the Australian Mathematical Society, 8(01), p.17.

EDIT: ~~In fact, I wonder whether one could not start to prove that $|A| - A geq 0$. Since $A^-$ is the closure of this operator, the result would follow, right?~~

functional-analysis operator-theory adjoint-operators

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edited Aug 2 at 19:20

asked Jul 31 at 10:33

IchKenneDeinenNamen

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This question is about the proof of lemma 14 in 1, which deals with the decomposition of an unbounded self-adjoint operator $A$ in positive and negative parts: $A = A^+ - A^-$. I have trouble understanding the proof given at the bottom of page 27 that $A^- geq 0$.

Indeed, the author writes that $(A^+ + A^-)Ex = |A|Ex$, where $x in mathcalD(A)$ and $E$ is the orthogonal projector onto $mathcalN(A^+)$. This seems to assume that $Ex in mathcalD(A)$ but I don't see where this would come from. Any idea?

Bernau, S. (1968). The square root of a positive self-adjoint operator. Journal of the Australian Mathematical Society, 8(01), p.17.

EDIT: ~~In fact, I wonder whether one could not start to prove that $|A| - A geq 0$. Since $A^-$ is the closure of this operator, the result would follow, right?~~

functional-analysis operator-theory adjoint-operators

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edited Aug 2 at 19:20

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

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up vote
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1

This question is about the proof of lemma 14 in 1, which deals with the decomposition of an unbounded self-adjoint operator $A$ in positive and negative parts: $A = A^+ - A^-$. I have trouble understanding the proof given at the bottom of page 27 that $A^- geq 0$.

Indeed, the author writes that $(A^+ + A^-)Ex = |A|Ex$, where $x in mathcalD(A)$ and $E$ is the orthogonal projector onto $mathcalN(A^+)$. This seems to assume that $Ex in mathcalD(A)$ but I don't see where this would come from. Any idea?

Bernau, S. (1968). The square root of a positive self-adjoint operator. Journal of the Australian Mathematical Society, 8(01), p.17.

EDIT: ~~In fact, I wonder whether one could not start to prove that $|A| - A geq 0$. Since $A^-$ is the closure of this operator, the result would follow, right?~~

functional-analysis operator-theory adjoint-operators

share|cite|improve this question

edited Aug 2 at 19:20

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

This question is about the proof of lemma 14 in 1, which deals with the decomposition of an unbounded self-adjoint operator $A$ in positive and negative parts: $A = A^+ - A^-$. I have trouble understanding the proof given at the bottom of page 27 that $A^- geq 0$.

Indeed, the author writes that $(A^+ + A^-)Ex = |A|Ex$, where $x in mathcalD(A)$ and $E$ is the orthogonal projector onto $mathcalN(A^+)$. This seems to assume that $Ex in mathcalD(A)$ but I don't see where this would come from. Any idea?

Bernau, S. (1968). The square root of a positive self-adjoint operator. Journal of the Australian Mathematical Society, 8(01), p.17.

EDIT: ~~In fact, I wonder whether one could not start to prove that $|A| - A geq 0$. Since $A^-$ is the closure of this operator, the result would follow, right?~~

functional-analysis operator-theory adjoint-operators

share|cite|improve this question

edited Aug 2 at 19:20

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 19:20

edited Aug 2 at 19:20

edited Aug 2 at 19:20

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

asked Jul 31 at 10:33

IchKenneDeinenNamen

64

64

add a comment | 

add a comment | 

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So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $mathcalD(A^+) cap mathcalD(A^-) = mathcalD(A)$.

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answered Aug 2 at 19:26

IchKenneDeinenNamen

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accepted

So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $mathcalD(A^+) cap mathcalD(A^-) = mathcalD(A)$.

share|cite|improve this answer

answered Aug 2 at 19:26

IchKenneDeinenNamen

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up vote
0
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accepted

So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $mathcalD(A^+) cap mathcalD(A^-) = mathcalD(A)$.

share|cite|improve this answer

answered Aug 2 at 19:26

IchKenneDeinenNamen

64

add a comment | 

up vote
0
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accepted

up vote
0
down vote



accepted

So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $mathcalD(A^+) cap mathcalD(A^-) = mathcalD(A)$.

share|cite|improve this answer

answered Aug 2 at 19:26

IchKenneDeinenNamen

64

So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $mathcalD(A^+) cap mathcalD(A^-) = mathcalD(A)$.

share|cite|improve this answer

answered Aug 2 at 19:26

IchKenneDeinenNamen

64

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 19:26

IchKenneDeinenNamen

64

answered Aug 2 at 19:26

IchKenneDeinenNamen

64

answered Aug 2 at 19:26

IchKenneDeinenNamen

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64

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