Looking for a function such that The Next CEO of Stack Overflowname this functionConvex...

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Looking for a function such that



The Next CEO of Stack Overflowname this functionConvex functions - two questionsStudy of a function and other factsProve that there is no $(x,n,m)$ such that $f^n(x)=f^{n+km}(x)$ if $f$ is injectivereal analysis -functional equationProve that $f : [a,b] rightarrow mathbb{R}$ is a bijection from $[a, b]$ to $[f(a), f(b)]$A sufficient condition on a real smooth functionFinding $f(x)$ in a Fourier IntegralProving a limit of this growing functionLemma used in Proof of L'Hôpital's Rule for Indeterminate Types of $infty/infty$












1












$begingroup$


Let $Minmathbb{R}^{ntimes n}$ and $yinmathbb{R}^n$. We define $l:,mathbb{R}^ntomathbb{R}$ by $xmapsto ||Mx-y||^2_2$. Now I´m looking for a a function $f:,mathbb{R}_{geq 0}tomathbb{R}$, which satisfies $f(||x||_2)leq l(x)$. In addition $f$ should be monotone increasing with $lim_{alphatoinfty}frac{f(alpha)}{alpha}=infty$.
This looks very simple and maybe it is, but unfornately I couldn't find such a function.



Edit: We assume $Mneq0$. If it helps we can also assume that $M$ is a diagonal binary matrix.



Thank you in advance,



Chris






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    I updated my answer to reflect the edit.
    $endgroup$
    – user159517
    Mar 17 at 14:01
















1












$begingroup$


Let $Minmathbb{R}^{ntimes n}$ and $yinmathbb{R}^n$. We define $l:,mathbb{R}^ntomathbb{R}$ by $xmapsto ||Mx-y||^2_2$. Now I´m looking for a a function $f:,mathbb{R}_{geq 0}tomathbb{R}$, which satisfies $f(||x||_2)leq l(x)$. In addition $f$ should be monotone increasing with $lim_{alphatoinfty}frac{f(alpha)}{alpha}=infty$.
This looks very simple and maybe it is, but unfornately I couldn't find such a function.



Edit: We assume $Mneq0$. If it helps we can also assume that $M$ is a diagonal binary matrix.



Thank you in advance,



Chris






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    I updated my answer to reflect the edit.
    $endgroup$
    – user159517
    Mar 17 at 14:01














1












1








1





$begingroup$


Let $Minmathbb{R}^{ntimes n}$ and $yinmathbb{R}^n$. We define $l:,mathbb{R}^ntomathbb{R}$ by $xmapsto ||Mx-y||^2_2$. Now I´m looking for a a function $f:,mathbb{R}_{geq 0}tomathbb{R}$, which satisfies $f(||x||_2)leq l(x)$. In addition $f$ should be monotone increasing with $lim_{alphatoinfty}frac{f(alpha)}{alpha}=infty$.
This looks very simple and maybe it is, but unfornately I couldn't find such a function.



Edit: We assume $Mneq0$. If it helps we can also assume that $M$ is a diagonal binary matrix.



Thank you in advance,



Chris






functions







share|cite|improve this question











$endgroup$




Let $Minmathbb{R}^{ntimes n}$ and $yinmathbb{R}^n$. We define $l:,mathbb{R}^ntomathbb{R}$ by $xmapsto ||Mx-y||^2_2$. Now I´m looking for a a function $f:,mathbb{R}_{geq 0}tomathbb{R}$, which satisfies $f(||x||_2)leq l(x)$. In addition $f$ should be monotone increasing with $lim_{alphatoinfty}frac{f(alpha)}{alpha}=infty$.
This looks very simple and maybe it is, but unfornately I couldn't find such a function.



Edit: We assume $Mneq0$. If it helps we can also assume that $M$ is a diagonal binary matrix.



Thank you in advance,



Chris






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 12:59









Lorenzo Quarisa

3,775623




3,775623










asked Mar 17 at 8:15









Chris S.Chris S.

356




356












  • $begingroup$
    I updated my answer to reflect the edit.
    $endgroup$
    – user159517
    Mar 17 at 14:01


















  • $begingroup$
    I updated my answer to reflect the edit.
    $endgroup$
    – user159517
    Mar 17 at 14:01
















$begingroup$
I updated my answer to reflect the edit.
$endgroup$
– user159517
Mar 17 at 14:01




$begingroup$
I updated my answer to reflect the edit.
$endgroup$
– user159517
Mar 17 at 14:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $M$ be a diagonal binary matrix. There are two possible cases to consider





  1. $M$ is singular


  2. $M$ is the identity matrix.


First, lets assume that 1) holds. Let $x_0 in text{ker} M$ with $|x_0| = 1$. Let $alpha$ be big enough such that $f(alpha) > |y|_2^2$. Then for all $c > alpha$, we find



$$f(|cx_0|_2) > l(cx_0).$$
So the statement is wrong.



In the case $2)$, where $M$ is the identity matrix, take $$g(|x|_2) := (|x|_2- (1+|y|_2))^{+} $$
where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $|x-y|_2 geq 1$ and $$g(|x|_2) leq |x|_2 - |y|_2 leq |x-y|_2.$$ Now for any $p in (1,2)$, the function $$f(|x|_2) := g(|x|_2)^p$$ does the job.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
    $endgroup$
    – Chris S.
    Mar 17 at 12:48












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1 Answer
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active

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1












$begingroup$

Let $M$ be a diagonal binary matrix. There are two possible cases to consider





  1. $M$ is singular


  2. $M$ is the identity matrix.


First, lets assume that 1) holds. Let $x_0 in text{ker} M$ with $|x_0| = 1$. Let $alpha$ be big enough such that $f(alpha) > |y|_2^2$. Then for all $c > alpha$, we find



$$f(|cx_0|_2) > l(cx_0).$$
So the statement is wrong.



In the case $2)$, where $M$ is the identity matrix, take $$g(|x|_2) := (|x|_2- (1+|y|_2))^{+} $$
where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $|x-y|_2 geq 1$ and $$g(|x|_2) leq |x|_2 - |y|_2 leq |x-y|_2.$$ Now for any $p in (1,2)$, the function $$f(|x|_2) := g(|x|_2)^p$$ does the job.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
    $endgroup$
    – Chris S.
    Mar 17 at 12:48
















1












$begingroup$

Let $M$ be a diagonal binary matrix. There are two possible cases to consider





  1. $M$ is singular


  2. $M$ is the identity matrix.


First, lets assume that 1) holds. Let $x_0 in text{ker} M$ with $|x_0| = 1$. Let $alpha$ be big enough such that $f(alpha) > |y|_2^2$. Then for all $c > alpha$, we find



$$f(|cx_0|_2) > l(cx_0).$$
So the statement is wrong.



In the case $2)$, where $M$ is the identity matrix, take $$g(|x|_2) := (|x|_2- (1+|y|_2))^{+} $$
where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $|x-y|_2 geq 1$ and $$g(|x|_2) leq |x|_2 - |y|_2 leq |x-y|_2.$$ Now for any $p in (1,2)$, the function $$f(|x|_2) := g(|x|_2)^p$$ does the job.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
    $endgroup$
    – Chris S.
    Mar 17 at 12:48














1












1








1





$begingroup$

Let $M$ be a diagonal binary matrix. There are two possible cases to consider





  1. $M$ is singular


  2. $M$ is the identity matrix.


First, lets assume that 1) holds. Let $x_0 in text{ker} M$ with $|x_0| = 1$. Let $alpha$ be big enough such that $f(alpha) > |y|_2^2$. Then for all $c > alpha$, we find



$$f(|cx_0|_2) > l(cx_0).$$
So the statement is wrong.



In the case $2)$, where $M$ is the identity matrix, take $$g(|x|_2) := (|x|_2- (1+|y|_2))^{+} $$
where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $|x-y|_2 geq 1$ and $$g(|x|_2) leq |x|_2 - |y|_2 leq |x-y|_2.$$ Now for any $p in (1,2)$, the function $$f(|x|_2) := g(|x|_2)^p$$ does the job.






share|cite|improve this answer











$endgroup$



Let $M$ be a diagonal binary matrix. There are two possible cases to consider





  1. $M$ is singular


  2. $M$ is the identity matrix.


First, lets assume that 1) holds. Let $x_0 in text{ker} M$ with $|x_0| = 1$. Let $alpha$ be big enough such that $f(alpha) > |y|_2^2$. Then for all $c > alpha$, we find



$$f(|cx_0|_2) > l(cx_0).$$
So the statement is wrong.



In the case $2)$, where $M$ is the identity matrix, take $$g(|x|_2) := (|x|_2- (1+|y|_2))^{+} $$
where $+$ denotes the positive part. Whenever $g$ is nonzero, due to the reverse triangle inequality we have $|x-y|_2 geq 1$ and $$g(|x|_2) leq |x|_2 - |y|_2 leq |x-y|_2.$$ Now for any $p in (1,2)$, the function $$f(|x|_2) := g(|x|_2)^p$$ does the job.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 17 at 13:58

























answered Mar 17 at 10:01









user159517user159517

4,585931




4,585931












  • $begingroup$
    That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
    $endgroup$
    – Chris S.
    Mar 17 at 12:48


















  • $begingroup$
    That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
    $endgroup$
    – Chris S.
    Mar 17 at 12:48
















$begingroup$
That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
$endgroup$
– Chris S.
Mar 17 at 12:48




$begingroup$
That's right. I did not think about that corner case. I assume that $Mneq 0$. I will edit the question. Thank you!
$endgroup$
– Chris S.
Mar 17 at 12:48


















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