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Vector-Valued Holomorphic Functions of Constant Norm
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Suppose throughout that $E$ is a complex normed vector space.
Question: For which $E$ does it hold that if $DsubsetBbb C$ is a domain, $f:Dto E$ is holomorphic and $||f(z)||$ is constant then $f$ is constant?
For reference below I'm going to call such spaces cnc spaces (for "constant norm implies constant").
(Why the several-complex-variables tag: If $E$ is not cnc then in some sense the boundary of the unit ball contains analytic disks. So I wonder if maybe at least in the finite-dimensional case this has something to do with strict pseudo-convexity of the unit ball. I know nothing about scv...)
Context: The positive answers to this question show that $Bbb C^2$ with the euclidean norm is a cnc space. In fact one of the answers shows that any Hilbert space is cnc. (Replace the pair of power series in that answer by a single power series with $E$-valued coefficients...)
I conjectured that any Banach space is cnc. But no:
Example: $Bbb C^2$ with the $ell_infty$ norm $||z||_infty=max(|z_1|,|z_2|)$ is not cnc. Consider the function $f(z)=(1,z)$ in the unit disk.
My work so far: My failed proof that every Banach space is cnc leads naturally to a condition that does imply cnc. Having no idea whether it's equivalent to one of the standard [adverb]-convex conditions, I'm going to invent another silly name and call it qc, for "quite convex":
Definition $E$ is qc if for every $Lambdain E^*$ with $Lambdane0$ there exists exactly one $xin E$ with $||x||=1$ and $Lambda x=||Lambda||$.
So for example any Hilbert space is qc, as is $L^p(mu)$ for $sigma$-finite $mu$ and $1<p<infty$.
Easy Theorem. If $E$ is qc then $E$ is cnc.
Proof: Say $D$ is connected, $f:Dto E$ is holomorphic and $||f(z)||=1$ for all $z$. Fix $pin D$. Hahn-Banach shows there exists $Lambdain E^*$ with $||Lambda||=1$ and $Lambda f(p)=1$. Now since $|Lambda f(z)|le 1$ for every $z$, MMT shows that $Lambdacirc f$ is constant.
So for every $zin D$ we have $Lambda f(z)=Lambda f(p)=||f(z)||=||f(p)||=||Lambda||=1$; the definition of qc was contrived precisely so that this should imply $f(z)=f(p)$.
It doesn't seem obviously impossible that assuming $E$ is not qc one could construct an example analogous to the $f(z)=(1,z)$ above to show that $E$ is not cnc... Edit: No, the answer below shows that $L^1(mu)$ is cnc, while it certainly is not qt.
banach-spaces several-complex-variables
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
add a comment |Â
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1
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Suppose throughout that $E$ is a complex normed vector space.
Question: For which $E$ does it hold that if $DsubsetBbb C$ is a domain, $f:Dto E$ is holomorphic and $||f(z)||$ is constant then $f$ is constant?
For reference below I'm going to call such spaces cnc spaces (for "constant norm implies constant").
(Why the several-complex-variables tag: If $E$ is not cnc then in some sense the boundary of the unit ball contains analytic disks. So I wonder if maybe at least in the finite-dimensional case this has something to do with strict pseudo-convexity of the unit ball. I know nothing about scv...)
Context: The positive answers to this question show that $Bbb C^2$ with the euclidean norm is a cnc space. In fact one of the answers shows that any Hilbert space is cnc. (Replace the pair of power series in that answer by a single power series with $E$-valued coefficients...)
I conjectured that any Banach space is cnc. But no:
Example: $Bbb C^2$ with the $ell_infty$ norm $||z||_infty=max(|z_1|,|z_2|)$ is not cnc. Consider the function $f(z)=(1,z)$ in the unit disk.
My work so far: My failed proof that every Banach space is cnc leads naturally to a condition that does imply cnc. Having no idea whether it's equivalent to one of the standard [adverb]-convex conditions, I'm going to invent another silly name and call it qc, for "quite convex":
Definition $E$ is qc if for every $Lambdain E^*$ with $Lambdane0$ there exists exactly one $xin E$ with $||x||=1$ and $Lambda x=||Lambda||$.
So for example any Hilbert space is qc, as is $L^p(mu)$ for $sigma$-finite $mu$ and $1<p<infty$.
Easy Theorem. If $E$ is qc then $E$ is cnc.
Proof: Say $D$ is connected, $f:Dto E$ is holomorphic and $||f(z)||=1$ for all $z$. Fix $pin D$. Hahn-Banach shows there exists $Lambdain E^*$ with $||Lambda||=1$ and $Lambda f(p)=1$. Now since $|Lambda f(z)|le 1$ for every $z$, MMT shows that $Lambdacirc f$ is constant.
So for every $zin D$ we have $Lambda f(z)=Lambda f(p)=||f(z)||=||f(p)||=||Lambda||=1$; the definition of qc was contrived precisely so that this should imply $f(z)=f(p)$.
It doesn't seem obviously impossible that assuming $E$ is not qc one could construct an example analogous to the $f(z)=(1,z)$ above to show that $E$ is not cnc... Edit: No, the answer below shows that $L^1(mu)$ is cnc, while it certainly is not qt.
banach-spaces several-complex-variables
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
1
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50
add a comment |Â
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up vote
1
down vote
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Suppose throughout that $E$ is a complex normed vector space.
Question: For which $E$ does it hold that if $DsubsetBbb C$ is a domain, $f:Dto E$ is holomorphic and $||f(z)||$ is constant then $f$ is constant?
For reference below I'm going to call such spaces cnc spaces (for "constant norm implies constant").
(Why the several-complex-variables tag: If $E$ is not cnc then in some sense the boundary of the unit ball contains analytic disks. So I wonder if maybe at least in the finite-dimensional case this has something to do with strict pseudo-convexity of the unit ball. I know nothing about scv...)
Context: The positive answers to this question show that $Bbb C^2$ with the euclidean norm is a cnc space. In fact one of the answers shows that any Hilbert space is cnc. (Replace the pair of power series in that answer by a single power series with $E$-valued coefficients...)
I conjectured that any Banach space is cnc. But no:
Example: $Bbb C^2$ with the $ell_infty$ norm $||z||_infty=max(|z_1|,|z_2|)$ is not cnc. Consider the function $f(z)=(1,z)$ in the unit disk.
My work so far: My failed proof that every Banach space is cnc leads naturally to a condition that does imply cnc. Having no idea whether it's equivalent to one of the standard [adverb]-convex conditions, I'm going to invent another silly name and call it qc, for "quite convex":
Definition $E$ is qc if for every $Lambdain E^*$ with $Lambdane0$ there exists exactly one $xin E$ with $||x||=1$ and $Lambda x=||Lambda||$.
So for example any Hilbert space is qc, as is $L^p(mu)$ for $sigma$-finite $mu$ and $1<p<infty$.
Easy Theorem. If $E$ is qc then $E$ is cnc.
Proof: Say $D$ is connected, $f:Dto E$ is holomorphic and $||f(z)||=1$ for all $z$. Fix $pin D$. Hahn-Banach shows there exists $Lambdain E^*$ with $||Lambda||=1$ and $Lambda f(p)=1$. Now since $|Lambda f(z)|le 1$ for every $z$, MMT shows that $Lambdacirc f$ is constant.
So for every $zin D$ we have $Lambda f(z)=Lambda f(p)=||f(z)||=||f(p)||=||Lambda||=1$; the definition of qc was contrived precisely so that this should imply $f(z)=f(p)$.
It doesn't seem obviously impossible that assuming $E$ is not qc one could construct an example analogous to the $f(z)=(1,z)$ above to show that $E$ is not cnc... Edit: No, the answer below shows that $L^1(mu)$ is cnc, while it certainly is not qt.
banach-spaces several-complex-variables
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
Suppose throughout that $E$ is a complex normed vector space.
Question: For which $E$ does it hold that if $DsubsetBbb C$ is a domain, $f:Dto E$ is holomorphic and $||f(z)||$ is constant then $f$ is constant?
For reference below I'm going to call such spaces cnc spaces (for "constant norm implies constant").
(Why the several-complex-variables tag: If $E$ is not cnc then in some sense the boundary of the unit ball contains analytic disks. So I wonder if maybe at least in the finite-dimensional case this has something to do with strict pseudo-convexity of the unit ball. I know nothing about scv...)
Context: The positive answers to this question show that $Bbb C^2$ with the euclidean norm is a cnc space. In fact one of the answers shows that any Hilbert space is cnc. (Replace the pair of power series in that answer by a single power series with $E$-valued coefficients...)
I conjectured that any Banach space is cnc. But no:
Example: $Bbb C^2$ with the $ell_infty$ norm $||z||_infty=max(|z_1|,|z_2|)$ is not cnc. Consider the function $f(z)=(1,z)$ in the unit disk.
My work so far: My failed proof that every Banach space is cnc leads naturally to a condition that does imply cnc. Having no idea whether it's equivalent to one of the standard [adverb]-convex conditions, I'm going to invent another silly name and call it qc, for "quite convex":
Definition $E$ is qc if for every $Lambdain E^*$ with $Lambdane0$ there exists exactly one $xin E$ with $||x||=1$ and $Lambda x=||Lambda||$.
So for example any Hilbert space is qc, as is $L^p(mu)$ for $sigma$-finite $mu$ and $1<p<infty$.
Easy Theorem. If $E$ is qc then $E$ is cnc.
Proof: Say $D$ is connected, $f:Dto E$ is holomorphic and $||f(z)||=1$ for all $z$. Fix $pin D$. Hahn-Banach shows there exists $Lambdain E^*$ with $||Lambda||=1$ and $Lambda f(p)=1$. Now since $|Lambda f(z)|le 1$ for every $z$, MMT shows that $Lambdacirc f$ is constant.
So for every $zin D$ we have $Lambda f(z)=Lambda f(p)=||f(z)||=||f(p)||=||Lambda||=1$; the definition of qc was contrived precisely so that this should imply $f(z)=f(p)$.
It doesn't seem obviously impossible that assuming $E$ is not qc one could construct an example analogous to the $f(z)=(1,z)$ above to show that $E$ is not cnc... Edit: No, the answer below shows that $L^1(mu)$ is cnc, while it certainly is not qt.
banach-spaces several-complex-variables
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
edited Jul 24 at 18:10
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
asked Jul 24 at 14:40
David C. Ullrich
54.1k33481
54.1k33481
1
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50
add a comment |Â
1
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50
1
1
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50
add a comment |Â
1 Answer
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This was studied in
The strong maximum modulus theorem for analytic functions into a Banach space by Edward Thorp and Robert Whitley (Proc. Amer. Math. Soc. 18 (1967), 640-646). I quote the relevant parts.
Definition 2.1. A point $e$ of a convex subset $K$ of a complex Banach space
$X$ is a complex extreme point of $K$ if $e+zy: subset K$
for [some] $y$ in $X$ implies that $y = 0$.
Theorem 3.1. Let $X$ be a complex Banach space such that each point on the surface of the unit sphere is a complex extreme point of the unit sphere. Then the strong form of the maximum modulus theorem holds, i.e. for $D$ a domain and $f: Dto X$ an analytic function, either $|f(z)|$ has no maximum on $D$ or $f(z)$ is constant on $D$.
Conversely, if the surface of the unit sphere of $X$ contains a point which
is not a complex extreme point of the sphere, then there is a nonconstant
analytic function $f$ mapping the open unit disc into $X$ yet satisfying
$|f(z) | = 1$ for all $z$ in the disc.
Note that in their language, "unit sphere" is $x : $, while the set $x : $ is "the surface of the unit sphere".
If $X$ is strictly convex in the real sense (the unit sphere contains no nontrivial line segments), then the condition of Theorem 3.1 is satisfied. However, Thorp and Whitley show that some non-strictly convex spaces satisfy this condition too, most notably $L^1$ [over any measure space].
Theorem 4.2. Every point on the surface of the unit sphere of
$L^1$ is a complex extreme point.
For later developments, see papers by Patrick Dowling such as Extensions of the Maximum Principle for Vector-Valued Analytic and Harmonic Functions (Journal of Mathematical Analysis and Applications Volume 190, Issue 2, pages 599-604).
answered Jul 24 at 17:41
user357151
13.8k31140
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This was studied in
The strong maximum modulus theorem for analytic functions into a Banach space by Edward Thorp and Robert Whitley (Proc. Amer. Math. Soc. 18 (1967), 640-646). I quote the relevant parts.
Definition 2.1. A point $e$ of a convex subset $K$ of a complex Banach space
$X$ is a complex extreme point of $K$ if $e+zy: subset K$
for [some] $y$ in $X$ implies that $y = 0$.
Theorem 3.1. Let $X$ be a complex Banach space such that each point on the surface of the unit sphere is a complex extreme point of the unit sphere. Then the strong form of the maximum modulus theorem holds, i.e. for $D$ a domain and $f: Dto X$ an analytic function, either $|f(z)|$ has no maximum on $D$ or $f(z)$ is constant on $D$.
Conversely, if the surface of the unit sphere of $X$ contains a point which
is not a complex extreme point of the sphere, then there is a nonconstant
analytic function $f$ mapping the open unit disc into $X$ yet satisfying
$|f(z) | = 1$ for all $z$ in the disc.
Note that in their language, "unit sphere" is $x : $, while the set $x : $ is "the surface of the unit sphere".
If $X$ is strictly convex in the real sense (the unit sphere contains no nontrivial line segments), then the condition of Theorem 3.1 is satisfied. However, Thorp and Whitley show that some non-strictly convex spaces satisfy this condition too, most notably $L^1$ [over any measure space].
Theorem 4.2. Every point on the surface of the unit sphere of
$L^1$ is a complex extreme point.
For later developments, see papers by Patrick Dowling such as Extensions of the Maximum Principle for Vector-Valued Analytic and Harmonic Functions (Journal of Mathematical Analysis and Applications Volume 190, Issue 2, pages 599-604).
answered Jul 24 at 17:41
user357151
13.8k31140
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
add a comment |Â
up vote
2
down vote
accepted
This was studied in
The strong maximum modulus theorem for analytic functions into a Banach space by Edward Thorp and Robert Whitley (Proc. Amer. Math. Soc. 18 (1967), 640-646). I quote the relevant parts.
Definition 2.1. A point $e$ of a convex subset $K$ of a complex Banach space
$X$ is a complex extreme point of $K$ if $e+zy: subset K$
for [some] $y$ in $X$ implies that $y = 0$.
Theorem 3.1. Let $X$ be a complex Banach space such that each point on the surface of the unit sphere is a complex extreme point of the unit sphere. Then the strong form of the maximum modulus theorem holds, i.e. for $D$ a domain and $f: Dto X$ an analytic function, either $|f(z)|$ has no maximum on $D$ or $f(z)$ is constant on $D$.
Conversely, if the surface of the unit sphere of $X$ contains a point which
is not a complex extreme point of the sphere, then there is a nonconstant
analytic function $f$ mapping the open unit disc into $X$ yet satisfying
$|f(z) | = 1$ for all $z$ in the disc.
Note that in their language, "unit sphere" is $x : $, while the set $x : $ is "the surface of the unit sphere".
If $X$ is strictly convex in the real sense (the unit sphere contains no nontrivial line segments), then the condition of Theorem 3.1 is satisfied. However, Thorp and Whitley show that some non-strictly convex spaces satisfy this condition too, most notably $L^1$ [over any measure space].
Theorem 4.2. Every point on the surface of the unit sphere of
$L^1$ is a complex extreme point.
For later developments, see papers by Patrick Dowling such as Extensions of the Maximum Principle for Vector-Valued Analytic and Harmonic Functions (Journal of Mathematical Analysis and Applications Volume 190, Issue 2, pages 599-604).
answered Jul 24 at 17:41
user357151
13.8k31140
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This was studied in
The strong maximum modulus theorem for analytic functions into a Banach space by Edward Thorp and Robert Whitley (Proc. Amer. Math. Soc. 18 (1967), 640-646). I quote the relevant parts.
Definition 2.1. A point $e$ of a convex subset $K$ of a complex Banach space
$X$ is a complex extreme point of $K$ if $e+zy: subset K$
for [some] $y$ in $X$ implies that $y = 0$.
Theorem 3.1. Let $X$ be a complex Banach space such that each point on the surface of the unit sphere is a complex extreme point of the unit sphere. Then the strong form of the maximum modulus theorem holds, i.e. for $D$ a domain and $f: Dto X$ an analytic function, either $|f(z)|$ has no maximum on $D$ or $f(z)$ is constant on $D$.
Conversely, if the surface of the unit sphere of $X$ contains a point which
is not a complex extreme point of the sphere, then there is a nonconstant
analytic function $f$ mapping the open unit disc into $X$ yet satisfying
$|f(z) | = 1$ for all $z$ in the disc.
Note that in their language, "unit sphere" is $x : $, while the set $x : $ is "the surface of the unit sphere".
If $X$ is strictly convex in the real sense (the unit sphere contains no nontrivial line segments), then the condition of Theorem 3.1 is satisfied. However, Thorp and Whitley show that some non-strictly convex spaces satisfy this condition too, most notably $L^1$ [over any measure space].
Theorem 4.2. Every point on the surface of the unit sphere of
$L^1$ is a complex extreme point.
For later developments, see papers by Patrick Dowling such as Extensions of the Maximum Principle for Vector-Valued Analytic and Harmonic Functions (Journal of Mathematical Analysis and Applications Volume 190, Issue 2, pages 599-604).
answered Jul 24 at 17:41
user357151
13.8k31140
This was studied in
The strong maximum modulus theorem for analytic functions into a Banach space by Edward Thorp and Robert Whitley (Proc. Amer. Math. Soc. 18 (1967), 640-646). I quote the relevant parts.
Definition 2.1. A point $e$ of a convex subset $K$ of a complex Banach space
$X$ is a complex extreme point of $K$ if $e+zy: subset K$
for [some] $y$ in $X$ implies that $y = 0$.
Theorem 3.1. Let $X$ be a complex Banach space such that each point on the surface of the unit sphere is a complex extreme point of the unit sphere. Then the strong form of the maximum modulus theorem holds, i.e. for $D$ a domain and $f: Dto X$ an analytic function, either $|f(z)|$ has no maximum on $D$ or $f(z)$ is constant on $D$.
Conversely, if the surface of the unit sphere of $X$ contains a point which
is not a complex extreme point of the sphere, then there is a nonconstant
analytic function $f$ mapping the open unit disc into $X$ yet satisfying
$|f(z) | = 1$ for all $z$ in the disc.
Note that in their language, "unit sphere" is $x : $, while the set $x : $ is "the surface of the unit sphere".
If $X$ is strictly convex in the real sense (the unit sphere contains no nontrivial line segments), then the condition of Theorem 3.1 is satisfied. However, Thorp and Whitley show that some non-strictly convex spaces satisfy this condition too, most notably $L^1$ [over any measure space].
Theorem 4.2. Every point on the surface of the unit sphere of
$L^1$ is a complex extreme point.
For later developments, see papers by Patrick Dowling such as Extensions of the Maximum Principle for Vector-Valued Analytic and Harmonic Functions (Journal of Mathematical Analysis and Applications Volume 190, Issue 2, pages 599-604).
answered Jul 24 at 17:41
user357151
13.8k31140
answered Jul 24 at 17:41
user357151
13.8k31140
answered Jul 24 at 17:41
user357151
13.8k31140
answered Jul 24 at 17:41
user357151
13.8k31140
13.8k31140
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
add a comment |Â
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Cool. So sure enough, $E$ is non-cnc if and only if (in our terminology) the unit sphere contains an affine analytic disk. And no, cnc is not equivalent to qt. Not that I'd formulated it that way, but the question of whether "every point on the sphere is a complex extreme point" implies strict convexity had occurred to me. Cool.
– David C. Ullrich
Jul 24 at 18:08
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
Ah. The fact that every point of the unit sphere of $L^1$ is a complex extreme point is clear, because if $rne0$ then $frac12piint_0^2pi|1+re^it|dt>frac12piint_0^2piRe(1+re^it),dt=1$, hence if $gne0$ then $frac12piint_0^2pi||f+e^itg||_1>||f||_1$.
– David C. Ullrich
Jul 24 at 18:32
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1
Whilst I don't have an answer for your question, your condition "qc" should lie somewhere between strict convexity and uniform convexity of $E$. Strict convexity of $E$ is equivalent to the statement of qc with "exists exactly one" replaced with "exists at most one" and so is strictly weaker than qc. The fact that uniform convexity implies qc is proved in many functional analysis texts. I have never seen this stated as an equivalence so I expect that qc is weaker than uniform convexity, though I don't have an explicit example separating the two in mind.
– Rhys Steele
Jul 24 at 15:50