Mixing
Lower bound $mathcal{K}$-class function
$begingroup$
Let $alpha_0: mathbb{R}_{geq 0} to mathbb{R}_{geq 0}$ be a $mathcal{K}$-function, i.e., a strictly increasing function such that $alpha_0(0) = 0$, and $b geq 0$ a given constant. Is there any way to lower bound what follow
$$
alpha_0(x) + b geq alpha(x + b)
$$
with a $mathcal{K}$-function $alpha$? I guess this should be a well-known result, any suggestions for some related theorems?
real-analysis functional-analysis monotone-functions
edited Jan 30 at 14:23
Scientifica
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
$endgroup$
add a comment |
$begingroup$
Let $alpha_0: mathbb{R}_{geq 0} to mathbb{R}_{geq 0}$ be a $mathcal{K}$-function, i.e., a strictly increasing function such that $alpha_0(0) = 0$, and $b geq 0$ a given constant. Is there any way to lower bound what follow
$$
alpha_0(x) + b geq alpha(x + b)
$$
with a $mathcal{K}$-function $alpha$? I guess this should be a well-known result, any suggestions for some related theorems?
real-analysis functional-analysis monotone-functions
edited Jan 30 at 14:23
Scientifica
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
$endgroup$
add a comment |
$begingroup$
Let $alpha_0: mathbb{R}_{geq 0} to mathbb{R}_{geq 0}$ be a $mathcal{K}$-function, i.e., a strictly increasing function such that $alpha_0(0) = 0$, and $b geq 0$ a given constant. Is there any way to lower bound what follow
$$
alpha_0(x) + b geq alpha(x + b)
$$
with a $mathcal{K}$-function $alpha$? I guess this should be a well-known result, any suggestions for some related theorems?
real-analysis functional-analysis monotone-functions
edited Jan 30 at 14:23
Scientifica
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
$endgroup$
Let $alpha_0: mathbb{R}_{geq 0} to mathbb{R}_{geq 0}$ be a $mathcal{K}$-function, i.e., a strictly increasing function such that $alpha_0(0) = 0$, and $b geq 0$ a given constant. Is there any way to lower bound what follow
$$
alpha_0(x) + b geq alpha(x + b)
$$
with a $mathcal{K}$-function $alpha$? I guess this should be a well-known result, any suggestions for some related theorems?
real-analysis functional-analysis monotone-functions
real-analysis functional-analysis monotone-functions
edited Jan 30 at 14:23
Scientifica
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
edited Jan 30 at 14:23
Scientifica
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
edited Jan 30 at 14:23
Scientifica
6,80941335
edited Jan 30 at 14:23
Scientifica
6,80941335
edited Jan 30 at 14:23
Scientifica
6,80941335
6,80941335
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
asked Jan 30 at 13:37
GuidoLaremiGuidoLaremi
145
145
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're looking for a $mathcal K$-function $alpha$ such that $alpha(x+b)lealpha_0(x)+b$.
Notice that $alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $alpha(x+b)=alpha_0(x)+b$. In other words, we take $alpha(x)=alpha_0(x-b)+b$ for $xge b$.
We need to define $alpha$ on $[0,b]$ as well! Well notice that $alpha(b)=b$, so any strictly increasing function such that $alpha(0)=0$ and $alpha(b)=b$ will do.
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
$endgroup$
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're looking for a $mathcal K$-function $alpha$ such that $alpha(x+b)lealpha_0(x)+b$.
Notice that $alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $alpha(x+b)=alpha_0(x)+b$. In other words, we take $alpha(x)=alpha_0(x-b)+b$ for $xge b$.
We need to define $alpha$ on $[0,b]$ as well! Well notice that $alpha(b)=b$, so any strictly increasing function such that $alpha(0)=0$ and $alpha(b)=b$ will do.
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
$endgroup$
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
add a comment |
$begingroup$
You're looking for a $mathcal K$-function $alpha$ such that $alpha(x+b)lealpha_0(x)+b$.
Notice that $alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $alpha(x+b)=alpha_0(x)+b$. In other words, we take $alpha(x)=alpha_0(x-b)+b$ for $xge b$.
We need to define $alpha$ on $[0,b]$ as well! Well notice that $alpha(b)=b$, so any strictly increasing function such that $alpha(0)=0$ and $alpha(b)=b$ will do.
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
$endgroup$
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
add a comment |
$begingroup$
You're looking for a $mathcal K$-function $alpha$ such that $alpha(x+b)lealpha_0(x)+b$.
Notice that $alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $alpha(x+b)=alpha_0(x)+b$. In other words, we take $alpha(x)=alpha_0(x-b)+b$ for $xge b$.
We need to define $alpha$ on $[0,b]$ as well! Well notice that $alpha(b)=b$, so any strictly increasing function such that $alpha(0)=0$ and $alpha(b)=b$ will do.
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
$endgroup$
You're looking for a $mathcal K$-function $alpha$ such that $alpha(x+b)lealpha_0(x)+b$.
Notice that $alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $alpha(x+b)=alpha_0(x)+b$. In other words, we take $alpha(x)=alpha_0(x-b)+b$ for $xge b$.
We need to define $alpha$ on $[0,b]$ as well! Well notice that $alpha(b)=b$, so any strictly increasing function such that $alpha(0)=0$ and $alpha(b)=b$ will do.
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
answered Jan 30 at 15:11
ScientificaScientifica
6,80941335
6,80941335
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
add a comment |
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Exactly, I was passing by a change of variables and I was getting the same result. It seems work,
$endgroup$
– GuidoLaremi
Jan 30 at 15:19
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
Yep it works :)
$endgroup$
– Scientifica
Jan 30 at 15:21
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
I'm trying to figure out if it works also for some given $b<0$.
$endgroup$
– GuidoLaremi
Jan 30 at 15:35
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Well pay attention $mathcal K$-functions are defined on the nonegative real numbers.
$endgroup$
– Scientifica
Jan 30 at 16:05
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
$begingroup$
Yes I know. Unfortunately, the same procedure does not apply.
$endgroup$
– GuidoLaremi
Jan 30 at 16:10
add a comment |
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