Mixing
Equivalence norm between two fundamental descompotition (krein spaces).
$begingroup$
Im working on a corolary in this article: http://www.people.virginia.edu/~jlr5m/Papers/p46.pdf corolary 2, page 5.
I understand the argument shown, more precisely how to prove that
$$
f'_+=Xf_++(I-Y)f_-,
$$
$$
f'_-=(I-X)f_++Yf_-
$$
But im stuck getting a costant $m$ to show that $m|| f ||_{|mathfrak{H}|} le || f ||_{|mathfrak{H}|'}$, i know those relations will make it but i dont really know how to approach. Any help is welcome.
linear-algebra functional-analysis inner-product-space spectral-theory
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
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add a comment |
$begingroup$
Im working on a corolary in this article: http://www.people.virginia.edu/~jlr5m/Papers/p46.pdf corolary 2, page 5.
I understand the argument shown, more precisely how to prove that
$$
f'_+=Xf_++(I-Y)f_-,
$$
$$
f'_-=(I-X)f_++Yf_-
$$
But im stuck getting a costant $m$ to show that $m|| f ||_{|mathfrak{H}|} le || f ||_{|mathfrak{H}|'}$, i know those relations will make it but i dont really know how to approach. Any help is welcome.
linear-algebra functional-analysis inner-product-space spectral-theory
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
$endgroup$
add a comment |
$begingroup$
Im working on a corolary in this article: http://www.people.virginia.edu/~jlr5m/Papers/p46.pdf corolary 2, page 5.
I understand the argument shown, more precisely how to prove that
$$
f'_+=Xf_++(I-Y)f_-,
$$
$$
f'_-=(I-X)f_++Yf_-
$$
But im stuck getting a costant $m$ to show that $m|| f ||_{|mathfrak{H}|} le || f ||_{|mathfrak{H}|'}$, i know those relations will make it but i dont really know how to approach. Any help is welcome.
linear-algebra functional-analysis inner-product-space spectral-theory
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
$endgroup$
Im working on a corolary in this article: http://www.people.virginia.edu/~jlr5m/Papers/p46.pdf corolary 2, page 5.
I understand the argument shown, more precisely how to prove that
$$
f'_+=Xf_++(I-Y)f_-,
$$
$$
f'_-=(I-X)f_++Yf_-
$$
But im stuck getting a costant $m$ to show that $m|| f ||_{|mathfrak{H}|} le || f ||_{|mathfrak{H}|'}$, i know those relations will make it but i dont really know how to approach. Any help is welcome.
linear-algebra functional-analysis inner-product-space spectral-theory
linear-algebra functional-analysis inner-product-space spectral-theory
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
asked May 11 '18 at 3:16
ipreferpiipreferpi
277
277
add a comment |
add a comment |
1 Answer
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The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then
begin{align}
|f|_{|mathfrak H|'}^2&=|f_+'|_{|mathfrak H|'}^2+|f_-'|_{|mathfrak H|'}^2\ \
&=|Xf_++(I-Y)f_-|_{|mathfrak H|'}^2+|(I-X)f_++Yf_-|_{|mathfrak H|'}^2\ \
&leq (|X|,|f_+|_{|mathfrak H|}+|I-Y|,|f_-|_{|mathfrak H|})^2
+(|I-X|,|f_+|_{|mathfrak H|}+|Y|,|f_-|_{|mathfrak H|})^2\ \
&leq M(|f_+|_{|mathfrak H|}^2+|f_-|_{|mathfrak H|}^2)\ \
&=M,|f|_{|mathfrak H|}^2,
end{align}
where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
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$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then
begin{align}
|f|_{|mathfrak H|'}^2&=|f_+'|_{|mathfrak H|'}^2+|f_-'|_{|mathfrak H|'}^2\ \
&=|Xf_++(I-Y)f_-|_{|mathfrak H|'}^2+|(I-X)f_++Yf_-|_{|mathfrak H|'}^2\ \
&leq (|X|,|f_+|_{|mathfrak H|}+|I-Y|,|f_-|_{|mathfrak H|})^2
+(|I-X|,|f_+|_{|mathfrak H|}+|Y|,|f_-|_{|mathfrak H|})^2\ \
&leq M(|f_+|_{|mathfrak H|}^2+|f_-|_{|mathfrak H|}^2)\ \
&=M,|f|_{|mathfrak H|}^2,
end{align}
where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
add a comment |
$begingroup$
The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then
begin{align}
|f|_{|mathfrak H|'}^2&=|f_+'|_{|mathfrak H|'}^2+|f_-'|_{|mathfrak H|'}^2\ \
&=|Xf_++(I-Y)f_-|_{|mathfrak H|'}^2+|(I-X)f_++Yf_-|_{|mathfrak H|'}^2\ \
&leq (|X|,|f_+|_{|mathfrak H|}+|I-Y|,|f_-|_{|mathfrak H|})^2
+(|I-X|,|f_+|_{|mathfrak H|}+|Y|,|f_-|_{|mathfrak H|})^2\ \
&leq M(|f_+|_{|mathfrak H|}^2+|f_-|_{|mathfrak H|}^2)\ \
&=M,|f|_{|mathfrak H|}^2,
end{align}
where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
add a comment |
$begingroup$
The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then
begin{align}
|f|_{|mathfrak H|'}^2&=|f_+'|_{|mathfrak H|'}^2+|f_-'|_{|mathfrak H|'}^2\ \
&=|Xf_++(I-Y)f_-|_{|mathfrak H|'}^2+|(I-X)f_++Yf_-|_{|mathfrak H|'}^2\ \
&leq (|X|,|f_+|_{|mathfrak H|}+|I-Y|,|f_-|_{|mathfrak H|})^2
+(|I-X|,|f_+|_{|mathfrak H|}+|Y|,|f_-|_{|mathfrak H|})^2\ \
&leq M(|f_+|_{|mathfrak H|}^2+|f_-|_{|mathfrak H|}^2)\ \
&=M,|f|_{|mathfrak H|}^2,
end{align}
where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then
begin{align}
|f|_{|mathfrak H|'}^2&=|f_+'|_{|mathfrak H|'}^2+|f_-'|_{|mathfrak H|'}^2\ \
&=|Xf_++(I-Y)f_-|_{|mathfrak H|'}^2+|(I-X)f_++Yf_-|_{|mathfrak H|'}^2\ \
&leq (|X|,|f_+|_{|mathfrak H|}+|I-Y|,|f_-|_{|mathfrak H|})^2
+(|I-X|,|f_+|_{|mathfrak H|}+|Y|,|f_-|_{|mathfrak H|})^2\ \
&leq M(|f_+|_{|mathfrak H|}^2+|f_-|_{|mathfrak H|}^2)\ \
&=M,|f|_{|mathfrak H|}^2,
end{align}
where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
edited Jan 13 at 4:49
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
answered May 14 '18 at 4:45
Martin ArgeramiMartin Argerami
127k1182182
127k1182182
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
add a comment |
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
$begingroup$
I understand. Thank you very much.
$endgroup$
– ipreferpi
May 14 '18 at 18:41
add a comment |
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