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 $infty/infty$
$begingroup$
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
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
$endgroup$
add a comment |
$begingroup$
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
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
$endgroup$
$begingroup$
I updated my answer to reflect the edit.
$endgroup$
– user159517
Mar 17 at 14:01
add a comment |
$begingroup$
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
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
$endgroup$
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
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
edited Mar 17 at 12:59
Lorenzo Quarisa
3,775623
3,775623
asked Mar 17 at 8:15
Chris S.Chris S.
356
asked Mar 17 at 8:15
Chris S.Chris S.
356
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $M$ be a diagonal binary matrix. There are two possible cases to consider
$M$ is singular
$M$ 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.
answered Mar 17 at 10:01
user159517user159517
4,585931
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
Let $M$ be a diagonal binary matrix. There are two possible cases to consider
$M$ is singular
$M$ 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.
answered Mar 17 at 10:01
user159517user159517
4,585931
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Let $M$ be a diagonal binary matrix. There are two possible cases to consider
$M$ is singular
$M$ 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.
answered Mar 17 at 10:01
user159517user159517
4,585931
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Let $M$ be a diagonal binary matrix. There are two possible cases to consider
$M$ is singular
$M$ 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.
answered Mar 17 at 10:01
user159517user159517
4,585931
$endgroup$
Let $M$ be a diagonal binary matrix. There are two possible cases to consider
$M$ is singular
$M$ 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.
answered Mar 17 at 10:01
user159517user159517
4,585931
edited Mar 17 at 13:58
answered Mar 17 at 10:01
user159517user159517
4,585931
answered Mar 17 at 10:01
user159517user159517
4,585931
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
I updated my answer to reflect the edit.
$endgroup$
– user159517
Mar 17 at 14:01