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Does $f(ntheta) to 0$ for all $theta>0$ and $f$ Darboux imply $f(x) to 0$ as $x to infty$?
$begingroup$
Recall that a Darboux function $f:mathbb{R} to mathbb{R}$ is one which satisfies the conclusion of the intermediate value theorem (i.e., connected sets are mapped to connected sets). Being Darboux is a weaker condition than continuity. If a theorem about continuous functions only uses the intermediate value theorem, then chances are it also holds for the entire class of Darboux functions. I find it interesting to study which theorems about continuous functions also hold for Darboux functions.
We have the following theorem, which is fairly well known and hinges on the Baire Categoery Theorem.
If $f:mathbb{R} to mathbb{R}$ is continuous and $f(ntheta) xrightarrow[n in mathbb{N}, ntoinfty]{} 0$ for every $theta in (0, infty)$, then $f(x) xrightarrow[x in mathbb{R}, xtoinfty]{} 0$.
A counterexample if we drop continuity is $f(x) = mathbf{1}_{{ exp(n) : n in mathbb{N}}}$. However, this counterexample isn't Darboux, and I haven't been able to come up with any counterexample which is Darboux. Thus, this leads me to my question.
Can the continuity condition in the theorem stated above be relaxed to Darboux?
In searching for counterexamples of this sort, one approach is playing around with $sin frac{1}{x}$. An alternative approach is considering highly pathological functions with the property that every nonempty open set is mapped to $mathbb{R}$ (for instance, Conway Base-13, or Brian's example here) and modifying these in such a way that they satisfy the hypotheses of the problem.
real-analysis
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
add a comment |
$begingroup$
Recall that a Darboux function $f:mathbb{R} to mathbb{R}$ is one which satisfies the conclusion of the intermediate value theorem (i.e., connected sets are mapped to connected sets). Being Darboux is a weaker condition than continuity. If a theorem about continuous functions only uses the intermediate value theorem, then chances are it also holds for the entire class of Darboux functions. I find it interesting to study which theorems about continuous functions also hold for Darboux functions.
We have the following theorem, which is fairly well known and hinges on the Baire Categoery Theorem.
If $f:mathbb{R} to mathbb{R}$ is continuous and $f(ntheta) xrightarrow[n in mathbb{N}, ntoinfty]{} 0$ for every $theta in (0, infty)$, then $f(x) xrightarrow[x in mathbb{R}, xtoinfty]{} 0$.
A counterexample if we drop continuity is $f(x) = mathbf{1}_{{ exp(n) : n in mathbb{N}}}$. However, this counterexample isn't Darboux, and I haven't been able to come up with any counterexample which is Darboux. Thus, this leads me to my question.
Can the continuity condition in the theorem stated above be relaxed to Darboux?
In searching for counterexamples of this sort, one approach is playing around with $sin frac{1}{x}$. An alternative approach is considering highly pathological functions with the property that every nonempty open set is mapped to $mathbb{R}$ (for instance, Conway Base-13, or Brian's example here) and modifying these in such a way that they satisfy the hypotheses of the problem.
real-analysis
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52
add a comment |
$begingroup$
Recall that a Darboux function $f:mathbb{R} to mathbb{R}$ is one which satisfies the conclusion of the intermediate value theorem (i.e., connected sets are mapped to connected sets). Being Darboux is a weaker condition than continuity. If a theorem about continuous functions only uses the intermediate value theorem, then chances are it also holds for the entire class of Darboux functions. I find it interesting to study which theorems about continuous functions also hold for Darboux functions.
We have the following theorem, which is fairly well known and hinges on the Baire Categoery Theorem.
If $f:mathbb{R} to mathbb{R}$ is continuous and $f(ntheta) xrightarrow[n in mathbb{N}, ntoinfty]{} 0$ for every $theta in (0, infty)$, then $f(x) xrightarrow[x in mathbb{R}, xtoinfty]{} 0$.
A counterexample if we drop continuity is $f(x) = mathbf{1}_{{ exp(n) : n in mathbb{N}}}$. However, this counterexample isn't Darboux, and I haven't been able to come up with any counterexample which is Darboux. Thus, this leads me to my question.
Can the continuity condition in the theorem stated above be relaxed to Darboux?
In searching for counterexamples of this sort, one approach is playing around with $sin frac{1}{x}$. An alternative approach is considering highly pathological functions with the property that every nonempty open set is mapped to $mathbb{R}$ (for instance, Conway Base-13, or Brian's example here) and modifying these in such a way that they satisfy the hypotheses of the problem.
real-analysis
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
Recall that a Darboux function $f:mathbb{R} to mathbb{R}$ is one which satisfies the conclusion of the intermediate value theorem (i.e., connected sets are mapped to connected sets). Being Darboux is a weaker condition than continuity. If a theorem about continuous functions only uses the intermediate value theorem, then chances are it also holds for the entire class of Darboux functions. I find it interesting to study which theorems about continuous functions also hold for Darboux functions.
We have the following theorem, which is fairly well known and hinges on the Baire Categoery Theorem.
If $f:mathbb{R} to mathbb{R}$ is continuous and $f(ntheta) xrightarrow[n in mathbb{N}, ntoinfty]{} 0$ for every $theta in (0, infty)$, then $f(x) xrightarrow[x in mathbb{R}, xtoinfty]{} 0$.
A counterexample if we drop continuity is $f(x) = mathbf{1}_{{ exp(n) : n in mathbb{N}}}$. However, this counterexample isn't Darboux, and I haven't been able to come up with any counterexample which is Darboux. Thus, this leads me to my question.
Can the continuity condition in the theorem stated above be relaxed to Darboux?
In searching for counterexamples of this sort, one approach is playing around with $sin frac{1}{x}$. An alternative approach is considering highly pathological functions with the property that every nonempty open set is mapped to $mathbb{R}$ (for instance, Conway Base-13, or Brian's example here) and modifying these in such a way that they satisfy the hypotheses of the problem.
real-analysis
real-analysis
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
edited Jan 20 at 1:49
MathematicsStudent1122
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 20 at 0:05
MathematicsStudent1122MathematicsStudent1122
8,67622467
8,67622467
$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52
add a comment |
$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52
$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52
add a comment |
1 Answer
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Non-measurable example
By the axiom of choice there is a $mathbb Q$-linear basis of $mathbb R.$ This basis has the same cardinality as $mathbb R$ so can be indexed as $a_r$ for $rinmathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set ${a_0+qa_rmid qinmathbb Q}$ is dense for each $r.$ But for each $theta>0,$ we can only have $f(qtheta)neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $theta.$ In particular $f(ntheta)to 0$ as $ntoinfty$ with $ninmathbb N.$
Measurable example
For $ngeq 2$ let $b_n=n!(n-1)!dots 2!.$ Each real has a unique "mixed radix" expression as
$x=lfloor xrfloor + sum_{ngeq 2}frac{x_n}{b_n}$ where $x_n$ is the unique representative of $lfloor b_n xrfloor$ modulo $n!$ lying in ${0,1,dots,n!-1}.$ For non-negative $x$ define $f(x)=lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m$ if this limit exists and $x_nleq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.
Now consider a real $x>0$ with $f(x)neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_nleq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m=0.$
Write $x=x'/b_{n-1}+1/b_n+epsilon/b_{n+1}$ where $x'$ is an integer and $0leqepsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+qepsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $qepsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .
This shows that for each $theta>0,$ the sequence $f(ntheta)$ takes at most one non-zero value, and in particular $f(ntheta)to 0.$
Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.
answered Jan 22 at 10:39
DapDap
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$begingroup$
Non-measurable example
By the axiom of choice there is a $mathbb Q$-linear basis of $mathbb R.$ This basis has the same cardinality as $mathbb R$ so can be indexed as $a_r$ for $rinmathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set ${a_0+qa_rmid qinmathbb Q}$ is dense for each $r.$ But for each $theta>0,$ we can only have $f(qtheta)neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $theta.$ In particular $f(ntheta)to 0$ as $ntoinfty$ with $ninmathbb N.$
Measurable example
For $ngeq 2$ let $b_n=n!(n-1)!dots 2!.$ Each real has a unique "mixed radix" expression as
$x=lfloor xrfloor + sum_{ngeq 2}frac{x_n}{b_n}$ where $x_n$ is the unique representative of $lfloor b_n xrfloor$ modulo $n!$ lying in ${0,1,dots,n!-1}.$ For non-negative $x$ define $f(x)=lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m$ if this limit exists and $x_nleq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.
Now consider a real $x>0$ with $f(x)neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_nleq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m=0.$
Write $x=x'/b_{n-1}+1/b_n+epsilon/b_{n+1}$ where $x'$ is an integer and $0leqepsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+qepsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $qepsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .
This shows that for each $theta>0,$ the sequence $f(ntheta)$ takes at most one non-zero value, and in particular $f(ntheta)to 0.$
Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.
answered Jan 22 at 10:39
DapDap
17.7k841
$endgroup$
add a comment |
$begingroup$
Non-measurable example
By the axiom of choice there is a $mathbb Q$-linear basis of $mathbb R.$ This basis has the same cardinality as $mathbb R$ so can be indexed as $a_r$ for $rinmathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set ${a_0+qa_rmid qinmathbb Q}$ is dense for each $r.$ But for each $theta>0,$ we can only have $f(qtheta)neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $theta.$ In particular $f(ntheta)to 0$ as $ntoinfty$ with $ninmathbb N.$
Measurable example
For $ngeq 2$ let $b_n=n!(n-1)!dots 2!.$ Each real has a unique "mixed radix" expression as
$x=lfloor xrfloor + sum_{ngeq 2}frac{x_n}{b_n}$ where $x_n$ is the unique representative of $lfloor b_n xrfloor$ modulo $n!$ lying in ${0,1,dots,n!-1}.$ For non-negative $x$ define $f(x)=lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m$ if this limit exists and $x_nleq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.
Now consider a real $x>0$ with $f(x)neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_nleq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m=0.$
Write $x=x'/b_{n-1}+1/b_n+epsilon/b_{n+1}$ where $x'$ is an integer and $0leqepsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+qepsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $qepsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .
This shows that for each $theta>0,$ the sequence $f(ntheta)$ takes at most one non-zero value, and in particular $f(ntheta)to 0.$
Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.
answered Jan 22 at 10:39
DapDap
17.7k841
$endgroup$
add a comment |
$begingroup$
Non-measurable example
By the axiom of choice there is a $mathbb Q$-linear basis of $mathbb R.$ This basis has the same cardinality as $mathbb R$ so can be indexed as $a_r$ for $rinmathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set ${a_0+qa_rmid qinmathbb Q}$ is dense for each $r.$ But for each $theta>0,$ we can only have $f(qtheta)neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $theta.$ In particular $f(ntheta)to 0$ as $ntoinfty$ with $ninmathbb N.$
Measurable example
For $ngeq 2$ let $b_n=n!(n-1)!dots 2!.$ Each real has a unique "mixed radix" expression as
$x=lfloor xrfloor + sum_{ngeq 2}frac{x_n}{b_n}$ where $x_n$ is the unique representative of $lfloor b_n xrfloor$ modulo $n!$ lying in ${0,1,dots,n!-1}.$ For non-negative $x$ define $f(x)=lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m$ if this limit exists and $x_nleq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.
Now consider a real $x>0$ with $f(x)neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_nleq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m=0.$
Write $x=x'/b_{n-1}+1/b_n+epsilon/b_{n+1}$ where $x'$ is an integer and $0leqepsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+qepsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $qepsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .
This shows that for each $theta>0,$ the sequence $f(ntheta)$ takes at most one non-zero value, and in particular $f(ntheta)to 0.$
Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.
answered Jan 22 at 10:39
DapDap
17.7k841
$endgroup$
Non-measurable example
By the axiom of choice there is a $mathbb Q$-linear basis of $mathbb R.$ This basis has the same cardinality as $mathbb R$ so can be indexed as $a_r$ for $rinmathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set ${a_0+qa_rmid qinmathbb Q}$ is dense for each $r.$ But for each $theta>0,$ we can only have $f(qtheta)neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $theta.$ In particular $f(ntheta)to 0$ as $ntoinfty$ with $ninmathbb N.$
Measurable example
For $ngeq 2$ let $b_n=n!(n-1)!dots 2!.$ Each real has a unique "mixed radix" expression as
$x=lfloor xrfloor + sum_{ngeq 2}frac{x_n}{b_n}$ where $x_n$ is the unique representative of $lfloor b_n xrfloor$ modulo $n!$ lying in ${0,1,dots,n!-1}.$ For non-negative $x$ define $f(x)=lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m$ if this limit exists and $x_nleq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.
Now consider a real $x>0$ with $f(x)neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_nleq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $lim_{ntoinfty} tfrac{1}{n}sum_{m=2}^n x_m=0.$
Write $x=x'/b_{n-1}+1/b_n+epsilon/b_{n+1}$ where $x'$ is an integer and $0leqepsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+qepsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $qepsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .
This shows that for each $theta>0,$ the sequence $f(ntheta)$ takes at most one non-zero value, and in particular $f(ntheta)to 0.$
Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.
answered Jan 22 at 10:39
DapDap
17.7k841
edited Jan 22 at 11:08
answered Jan 22 at 10:39
DapDap
17.7k841
answered Jan 22 at 10:39
DapDap
17.7k841
answered Jan 22 at 10:39
DapDap
17.7k841
17.7k841
add a comment |
add a comment |
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$begingroup$
(+1) Nice question, it might be pretty subtle.
$endgroup$
– Jack D'Aurizio
Jan 20 at 0:08
$begingroup$
(+1) Nice question. I would be nice if it is true that preimages of open sets under the Darboux function has non-empty interior, but neither do I see reasons for this to hold nor can I find a counter-example...
$endgroup$
– Sangchul Lee
Jan 20 at 7:31
$begingroup$
A related question.
$endgroup$
– Alex Ravsky
Jan 22 at 16:52