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What is known about the reciprocal $1/f$ of a holomorphic Banach-valued function $f$?
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Let $U subseteq mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U to A$ holomorphic and invertible in a punctured neighborhood of $0 in U$, so that $0$ is an isolated singularity of $1/f$.
Must $1/f$ be meromorphic at $0$? I.e. has the Laurent expansion at $0$ a finite singular part?
In general, I think the answer is no, because if $f(0)$ is nonzero and not a zero divisor, then looking at the Laurent expansion shows that if $1/f$ is meromorphic at $0$, it must be holomorphic, which gives a contradiction if we suppose in addition that $f(0)$ is not invertible.
I'm interested in the case of the algebra of operators on a complex Banach algebra, and $f(0)$ of the form $1+$ a compact operator, so that this counterexample cannot occur (by the Fredholm alternative).
Are there counterexamples with $f(0) = 1 + K$ with $K$ compact ?
I managed to prove that if $K$ is compact, normal and trace class, and $f : U to mathbb C$ is holomorphic, then even when $(1+f(s)K)^-1$ is not meromorphic it does send meromorphic functions to meromorphic functions. The proof analyzes the Laurent-expansion and uses the spectral theorem to eventually reduce to the finite-dimensional case.
functional-analysis banach-algebras compact-operators holomorphic-functions meromorphic-functions
asked Jul 14 at 22:20
barto
13k32479
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up vote
2
down vote
favorite
Let $U subseteq mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U to A$ holomorphic and invertible in a punctured neighborhood of $0 in U$, so that $0$ is an isolated singularity of $1/f$.
Must $1/f$ be meromorphic at $0$? I.e. has the Laurent expansion at $0$ a finite singular part?
In general, I think the answer is no, because if $f(0)$ is nonzero and not a zero divisor, then looking at the Laurent expansion shows that if $1/f$ is meromorphic at $0$, it must be holomorphic, which gives a contradiction if we suppose in addition that $f(0)$ is not invertible.
I'm interested in the case of the algebra of operators on a complex Banach algebra, and $f(0)$ of the form $1+$ a compact operator, so that this counterexample cannot occur (by the Fredholm alternative).
Are there counterexamples with $f(0) = 1 + K$ with $K$ compact ?
I managed to prove that if $K$ is compact, normal and trace class, and $f : U to mathbb C$ is holomorphic, then even when $(1+f(s)K)^-1$ is not meromorphic it does send meromorphic functions to meromorphic functions. The proof analyzes the Laurent-expansion and uses the spectral theorem to eventually reduce to the finite-dimensional case.
functional-analysis banach-algebras compact-operators holomorphic-functions meromorphic-functions
asked Jul 14 at 22:20
barto
13k32479
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $U subseteq mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U to A$ holomorphic and invertible in a punctured neighborhood of $0 in U$, so that $0$ is an isolated singularity of $1/f$.
Must $1/f$ be meromorphic at $0$? I.e. has the Laurent expansion at $0$ a finite singular part?
In general, I think the answer is no, because if $f(0)$ is nonzero and not a zero divisor, then looking at the Laurent expansion shows that if $1/f$ is meromorphic at $0$, it must be holomorphic, which gives a contradiction if we suppose in addition that $f(0)$ is not invertible.
I'm interested in the case of the algebra of operators on a complex Banach algebra, and $f(0)$ of the form $1+$ a compact operator, so that this counterexample cannot occur (by the Fredholm alternative).
Are there counterexamples with $f(0) = 1 + K$ with $K$ compact ?
I managed to prove that if $K$ is compact, normal and trace class, and $f : U to mathbb C$ is holomorphic, then even when $(1+f(s)K)^-1$ is not meromorphic it does send meromorphic functions to meromorphic functions. The proof analyzes the Laurent-expansion and uses the spectral theorem to eventually reduce to the finite-dimensional case.
functional-analysis banach-algebras compact-operators holomorphic-functions meromorphic-functions
asked Jul 14 at 22:20
barto
13k32479
Let $U subseteq mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U to A$ holomorphic and invertible in a punctured neighborhood of $0 in U$, so that $0$ is an isolated singularity of $1/f$.
Must $1/f$ be meromorphic at $0$? I.e. has the Laurent expansion at $0$ a finite singular part?
In general, I think the answer is no, because if $f(0)$ is nonzero and not a zero divisor, then looking at the Laurent expansion shows that if $1/f$ is meromorphic at $0$, it must be holomorphic, which gives a contradiction if we suppose in addition that $f(0)$ is not invertible.
I'm interested in the case of the algebra of operators on a complex Banach algebra, and $f(0)$ of the form $1+$ a compact operator, so that this counterexample cannot occur (by the Fredholm alternative).
Are there counterexamples with $f(0) = 1 + K$ with $K$ compact ?
I managed to prove that if $K$ is compact, normal and trace class, and $f : U to mathbb C$ is holomorphic, then even when $(1+f(s)K)^-1$ is not meromorphic it does send meromorphic functions to meromorphic functions. The proof analyzes the Laurent-expansion and uses the spectral theorem to eventually reduce to the finite-dimensional case.
functional-analysis banach-algebras compact-operators holomorphic-functions meromorphic-functions
asked Jul 14 at 22:20
barto
13k32479
edited Jul 15 at 8:45
asked Jul 14 at 22:20
barto
13k32479
asked Jul 14 at 22:20
barto
13k32479
asked Jul 14 at 22:20
barto
13k32479
13k32479
add a comment |Â
add a comment |Â
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