Does the Polygonal Confinement Theorem hold on the set of entire functions?
$begingroup$
The Polygonal Confinement Theorem can be found in Section 2 of this paper by Rosenthal. I am interested in a generalization of Lemma 3.1 in the paper, which states:
$textbf{Lemma 3.1:} $ If $v_1,ldots,v_m in mathbb{R}^n$, and $left|displaystyle sum_{i=1}^m v_iright|<epsilon$, $|v_j|<epsilon$ for all $j$, then there is a constant $C$ which does not depend on $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that for each $1leq jleq m$,
$$ left|sum_{i=1}^j v_iright| leq Cepsilon.$$
I am wondering whether the following analogue for entire functions holds. In the question below, $|f|_R$ denotes the supremum of $f$ on the disk $|z|leq R$.
$textbf{Question:}$
Suppose $R>0$ and $f_i$ are entire functions on $mathbb{C}$. Let $epsilon>0$. If $displaystyle left| sum_{i=1}^m f_i right|_R< epsilon$ and $displaystyle |f_j|_R < epsilon$ for each $j$, does there exist a constant $C$ independent of $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that whenever $1leq jleq m$,
$$ left| sum_{i=1}^j f_{sigma(i)} right|_R< Cepsilon?$$
Lemma 3.1, called the rearrangement theorem by Rosenthal and the Steinitz lemma by others, was the crucial element for proving the Levy-Steinitz theorem (found in the Rosenthal paper). The Levy-Stenitz theorem was generalized to metrizable nuclear topological vector spaces by Banaszczyk in this paper. Since the space of entire functions is a Frechet space, and hence a nuclear space, I am hoping that the analgoue of Lemma 3.1 holds. In fact, it looks like the Corollary in the Banaszczyk paper on page 196 may be what I'm looking for. But the generality and technicality of the paper is well beyond my expertise.
Any insight or references would be most appreciated.
complex-analysis topological-vector-spaces
$endgroup$
add a comment |
$begingroup$
The Polygonal Confinement Theorem can be found in Section 2 of this paper by Rosenthal. I am interested in a generalization of Lemma 3.1 in the paper, which states:
$textbf{Lemma 3.1:} $ If $v_1,ldots,v_m in mathbb{R}^n$, and $left|displaystyle sum_{i=1}^m v_iright|<epsilon$, $|v_j|<epsilon$ for all $j$, then there is a constant $C$ which does not depend on $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that for each $1leq jleq m$,
$$ left|sum_{i=1}^j v_iright| leq Cepsilon.$$
I am wondering whether the following analogue for entire functions holds. In the question below, $|f|_R$ denotes the supremum of $f$ on the disk $|z|leq R$.
$textbf{Question:}$
Suppose $R>0$ and $f_i$ are entire functions on $mathbb{C}$. Let $epsilon>0$. If $displaystyle left| sum_{i=1}^m f_i right|_R< epsilon$ and $displaystyle |f_j|_R < epsilon$ for each $j$, does there exist a constant $C$ independent of $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that whenever $1leq jleq m$,
$$ left| sum_{i=1}^j f_{sigma(i)} right|_R< Cepsilon?$$
Lemma 3.1, called the rearrangement theorem by Rosenthal and the Steinitz lemma by others, was the crucial element for proving the Levy-Steinitz theorem (found in the Rosenthal paper). The Levy-Stenitz theorem was generalized to metrizable nuclear topological vector spaces by Banaszczyk in this paper. Since the space of entire functions is a Frechet space, and hence a nuclear space, I am hoping that the analgoue of Lemma 3.1 holds. In fact, it looks like the Corollary in the Banaszczyk paper on page 196 may be what I'm looking for. But the generality and technicality of the paper is well beyond my expertise.
Any insight or references would be most appreciated.
complex-analysis topological-vector-spaces
$endgroup$
add a comment |
$begingroup$
The Polygonal Confinement Theorem can be found in Section 2 of this paper by Rosenthal. I am interested in a generalization of Lemma 3.1 in the paper, which states:
$textbf{Lemma 3.1:} $ If $v_1,ldots,v_m in mathbb{R}^n$, and $left|displaystyle sum_{i=1}^m v_iright|<epsilon$, $|v_j|<epsilon$ for all $j$, then there is a constant $C$ which does not depend on $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that for each $1leq jleq m$,
$$ left|sum_{i=1}^j v_iright| leq Cepsilon.$$
I am wondering whether the following analogue for entire functions holds. In the question below, $|f|_R$ denotes the supremum of $f$ on the disk $|z|leq R$.
$textbf{Question:}$
Suppose $R>0$ and $f_i$ are entire functions on $mathbb{C}$. Let $epsilon>0$. If $displaystyle left| sum_{i=1}^m f_i right|_R< epsilon$ and $displaystyle |f_j|_R < epsilon$ for each $j$, does there exist a constant $C$ independent of $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that whenever $1leq jleq m$,
$$ left| sum_{i=1}^j f_{sigma(i)} right|_R< Cepsilon?$$
Lemma 3.1, called the rearrangement theorem by Rosenthal and the Steinitz lemma by others, was the crucial element for proving the Levy-Steinitz theorem (found in the Rosenthal paper). The Levy-Stenitz theorem was generalized to metrizable nuclear topological vector spaces by Banaszczyk in this paper. Since the space of entire functions is a Frechet space, and hence a nuclear space, I am hoping that the analgoue of Lemma 3.1 holds. In fact, it looks like the Corollary in the Banaszczyk paper on page 196 may be what I'm looking for. But the generality and technicality of the paper is well beyond my expertise.
Any insight or references would be most appreciated.
complex-analysis topological-vector-spaces
$endgroup$
The Polygonal Confinement Theorem can be found in Section 2 of this paper by Rosenthal. I am interested in a generalization of Lemma 3.1 in the paper, which states:
$textbf{Lemma 3.1:} $ If $v_1,ldots,v_m in mathbb{R}^n$, and $left|displaystyle sum_{i=1}^m v_iright|<epsilon$, $|v_j|<epsilon$ for all $j$, then there is a constant $C$ which does not depend on $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that for each $1leq jleq m$,
$$ left|sum_{i=1}^j v_iright| leq Cepsilon.$$
I am wondering whether the following analogue for entire functions holds. In the question below, $|f|_R$ denotes the supremum of $f$ on the disk $|z|leq R$.
$textbf{Question:}$
Suppose $R>0$ and $f_i$ are entire functions on $mathbb{C}$. Let $epsilon>0$. If $displaystyle left| sum_{i=1}^m f_i right|_R< epsilon$ and $displaystyle |f_j|_R < epsilon$ for each $j$, does there exist a constant $C$ independent of $m$ and a permutation $sigma$ of ${1,ldots,m}$ such that whenever $1leq jleq m$,
$$ left| sum_{i=1}^j f_{sigma(i)} right|_R< Cepsilon?$$
Lemma 3.1, called the rearrangement theorem by Rosenthal and the Steinitz lemma by others, was the crucial element for proving the Levy-Steinitz theorem (found in the Rosenthal paper). The Levy-Stenitz theorem was generalized to metrizable nuclear topological vector spaces by Banaszczyk in this paper. Since the space of entire functions is a Frechet space, and hence a nuclear space, I am hoping that the analgoue of Lemma 3.1 holds. In fact, it looks like the Corollary in the Banaszczyk paper on page 196 may be what I'm looking for. But the generality and technicality of the paper is well beyond my expertise.
Any insight or references would be most appreciated.
complex-analysis topological-vector-spaces
complex-analysis topological-vector-spaces
asked Jan 5 at 4:05
user122916user122916
431315
431315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The analogue for entire functions does not hold. I will take $R=1$
throughout.
First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $|cdot|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.
There is no constant $C$ independent of $n$ such that PCT holds for $ell^infty(n)$
where $ell^infty(n)$ is $mathbb R^n$ with the $ell^infty$ norm. To see this consider $m$ even, $n=binom{m}{m/2},$ enumerate the order $m/2$ subsets of ${1,dots,m}$ as $S_1,dots,S_n,$ and define $f_i(j)=1$ if $iin S_j$ and $f_i(j)=-1$ otherwise ($1leq ileq m$ and $1leq jleq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$
We can transfer this counterexample to $X$ using an approximate embedding $F:ell^infty(n)to X,$ i.e.
$$tfrac12|f|_infty leq |F(f)|_Rleq 2|f|_infty.tag{*}$$
where $|f|_infty=sup{|f(z)|:|z|<1}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$
It remains to construct $F$ satisfying (*). Take $n$ continuous functions $phi_1,dots,phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|phi_i(z)-phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $phi'_i$ is in fact a polynomial and hence entire. For any $cinell^infty(n),$ the norm $|sum_{i=1}^n c_iphi'_i|_infty$ is within $tfrac12 max_i|c_i|$ of $|sum_{i=1}^n c_iphi_i|_infty=max_i|c_i|,$ which gives (*).
$endgroup$
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062389%2fdoes-the-polygonal-confinement-theorem-hold-on-the-set-of-entire-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The analogue for entire functions does not hold. I will take $R=1$
throughout.
First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $|cdot|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.
There is no constant $C$ independent of $n$ such that PCT holds for $ell^infty(n)$
where $ell^infty(n)$ is $mathbb R^n$ with the $ell^infty$ norm. To see this consider $m$ even, $n=binom{m}{m/2},$ enumerate the order $m/2$ subsets of ${1,dots,m}$ as $S_1,dots,S_n,$ and define $f_i(j)=1$ if $iin S_j$ and $f_i(j)=-1$ otherwise ($1leq ileq m$ and $1leq jleq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$
We can transfer this counterexample to $X$ using an approximate embedding $F:ell^infty(n)to X,$ i.e.
$$tfrac12|f|_infty leq |F(f)|_Rleq 2|f|_infty.tag{*}$$
where $|f|_infty=sup{|f(z)|:|z|<1}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$
It remains to construct $F$ satisfying (*). Take $n$ continuous functions $phi_1,dots,phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|phi_i(z)-phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $phi'_i$ is in fact a polynomial and hence entire. For any $cinell^infty(n),$ the norm $|sum_{i=1}^n c_iphi'_i|_infty$ is within $tfrac12 max_i|c_i|$ of $|sum_{i=1}^n c_iphi_i|_infty=max_i|c_i|,$ which gives (*).
$endgroup$
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
add a comment |
$begingroup$
The analogue for entire functions does not hold. I will take $R=1$
throughout.
First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $|cdot|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.
There is no constant $C$ independent of $n$ such that PCT holds for $ell^infty(n)$
where $ell^infty(n)$ is $mathbb R^n$ with the $ell^infty$ norm. To see this consider $m$ even, $n=binom{m}{m/2},$ enumerate the order $m/2$ subsets of ${1,dots,m}$ as $S_1,dots,S_n,$ and define $f_i(j)=1$ if $iin S_j$ and $f_i(j)=-1$ otherwise ($1leq ileq m$ and $1leq jleq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$
We can transfer this counterexample to $X$ using an approximate embedding $F:ell^infty(n)to X,$ i.e.
$$tfrac12|f|_infty leq |F(f)|_Rleq 2|f|_infty.tag{*}$$
where $|f|_infty=sup{|f(z)|:|z|<1}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$
It remains to construct $F$ satisfying (*). Take $n$ continuous functions $phi_1,dots,phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|phi_i(z)-phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $phi'_i$ is in fact a polynomial and hence entire. For any $cinell^infty(n),$ the norm $|sum_{i=1}^n c_iphi'_i|_infty$ is within $tfrac12 max_i|c_i|$ of $|sum_{i=1}^n c_iphi_i|_infty=max_i|c_i|,$ which gives (*).
$endgroup$
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
add a comment |
$begingroup$
The analogue for entire functions does not hold. I will take $R=1$
throughout.
First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $|cdot|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.
There is no constant $C$ independent of $n$ such that PCT holds for $ell^infty(n)$
where $ell^infty(n)$ is $mathbb R^n$ with the $ell^infty$ norm. To see this consider $m$ even, $n=binom{m}{m/2},$ enumerate the order $m/2$ subsets of ${1,dots,m}$ as $S_1,dots,S_n,$ and define $f_i(j)=1$ if $iin S_j$ and $f_i(j)=-1$ otherwise ($1leq ileq m$ and $1leq jleq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$
We can transfer this counterexample to $X$ using an approximate embedding $F:ell^infty(n)to X,$ i.e.
$$tfrac12|f|_infty leq |F(f)|_Rleq 2|f|_infty.tag{*}$$
where $|f|_infty=sup{|f(z)|:|z|<1}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$
It remains to construct $F$ satisfying (*). Take $n$ continuous functions $phi_1,dots,phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|phi_i(z)-phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $phi'_i$ is in fact a polynomial and hence entire. For any $cinell^infty(n),$ the norm $|sum_{i=1}^n c_iphi'_i|_infty$ is within $tfrac12 max_i|c_i|$ of $|sum_{i=1}^n c_iphi_i|_infty=max_i|c_i|,$ which gives (*).
$endgroup$
The analogue for entire functions does not hold. I will take $R=1$
throughout.
First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $|cdot|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.
There is no constant $C$ independent of $n$ such that PCT holds for $ell^infty(n)$
where $ell^infty(n)$ is $mathbb R^n$ with the $ell^infty$ norm. To see this consider $m$ even, $n=binom{m}{m/2},$ enumerate the order $m/2$ subsets of ${1,dots,m}$ as $S_1,dots,S_n,$ and define $f_i(j)=1$ if $iin S_j$ and $f_i(j)=-1$ otherwise ($1leq ileq m$ and $1leq jleq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$
We can transfer this counterexample to $X$ using an approximate embedding $F:ell^infty(n)to X,$ i.e.
$$tfrac12|f|_infty leq |F(f)|_Rleq 2|f|_infty.tag{*}$$
where $|f|_infty=sup{|f(z)|:|z|<1}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$
It remains to construct $F$ satisfying (*). Take $n$ continuous functions $phi_1,dots,phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|phi_i(z)-phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $phi'_i$ is in fact a polynomial and hence entire. For any $cinell^infty(n),$ the norm $|sum_{i=1}^n c_iphi'_i|_infty$ is within $tfrac12 max_i|c_i|$ of $|sum_{i=1}^n c_iphi_i|_infty=max_i|c_i|,$ which gives (*).
edited Jan 9 at 19:50
answered Jan 8 at 14:46
DapDap
16.5k739
16.5k739
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
add a comment |
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
I don't understand in the $ell^infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $tau(i)$ is the index such that $S_{tau(i)}$ is the complement of $S_i$, then $f_i + f_{tau(i)} = 0$. So if you add these up pairing indices $i$ and $tau(i)$, the partial sums have alternating norms $0$ and $1$.
$endgroup$
– Lukas Geyer
Jan 8 at 22:20
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
@LukasGeyer: $i$ doesn't index a subset - note $ileq m$ not $ileq n.$ Given distinct $i_1,dots,i_{m/2},$ there is some $j$ such that $S_j={i_1,dots,i_{m/2}}$, so $f_{i_1}(j)+dots+f_{i_{m/2}}(j)=m/2.$
$endgroup$
– Dap
Jan 9 at 7:45
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
$begingroup$
Thanks for the clarification, I completely misunderstood it the first time I read it.
$endgroup$
– Lukas Geyer
Jan 9 at 18:18
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062389%2fdoes-the-polygonal-confinement-theorem-hold-on-the-set-of-entire-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown