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$begingroup$
Let $f:mathbb{Z}rightarrow mathbb{R}$ be a quasi-homomorphism, i.e $|f(a+b)-f(a)-f(b)|leq D$ $forall$ $a$ and $b$ in $mathbb{Z}$ ($mathbb{R}$ and $mathbb{Z}$ are here considered as additive groups and so you see the plus sign). I have to prove that there exists a unique number $tau$ $epsilon $ $mathbb{R}$ such that $f(n)-ntau$ is bounded.
I have separately shown the uniqueness part and have found bounds that work in following restrictive cases :
(i) If $n$ is positive, I have bounded it above by $f(0)$ using $tau=f(1)+D$.
(ii) If $n$ is positive, I have bounded it below by $f(0)$ using $tau=f(1)-D$.
(iii) If $n$ is negative, I have bounded it above by $f(0)$ using $tau=f(-1)+D$.
(iv) If $n$ is negative, I have bounded it below by $f(0)$ using $tau=f(-1)-D$. I understand that I have always used $f(0)$ to bound them, but that is only what I can see since I am splitting $n$ as $n$ times the generator $1$ $epsilon$ $mathbb{Z}$. Please help me bound this universally using just one unique $tau$.
group-theory analysis upper-lower-bounds
asked Feb 13 at 11:32
CircleCircle
112
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{Z}rightarrow mathbb{R}$ be a quasi-homomorphism, i.e $|f(a+b)-f(a)-f(b)|leq D$ $forall$ $a$ and $b$ in $mathbb{Z}$ ($mathbb{R}$ and $mathbb{Z}$ are here considered as additive groups and so you see the plus sign). I have to prove that there exists a unique number $tau$ $epsilon $ $mathbb{R}$ such that $f(n)-ntau$ is bounded.
I have separately shown the uniqueness part and have found bounds that work in following restrictive cases :
(i) If $n$ is positive, I have bounded it above by $f(0)$ using $tau=f(1)+D$.
(ii) If $n$ is positive, I have bounded it below by $f(0)$ using $tau=f(1)-D$.
(iii) If $n$ is negative, I have bounded it above by $f(0)$ using $tau=f(-1)+D$.
(iv) If $n$ is negative, I have bounded it below by $f(0)$ using $tau=f(-1)-D$. I understand that I have always used $f(0)$ to bound them, but that is only what I can see since I am splitting $n$ as $n$ times the generator $1$ $epsilon$ $mathbb{Z}$. Please help me bound this universally using just one unique $tau$.
group-theory analysis upper-lower-bounds
asked Feb 13 at 11:32
CircleCircle
112
$endgroup$
$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago
add a comment |
$begingroup$
Let $f:mathbb{Z}rightarrow mathbb{R}$ be a quasi-homomorphism, i.e $|f(a+b)-f(a)-f(b)|leq D$ $forall$ $a$ and $b$ in $mathbb{Z}$ ($mathbb{R}$ and $mathbb{Z}$ are here considered as additive groups and so you see the plus sign). I have to prove that there exists a unique number $tau$ $epsilon $ $mathbb{R}$ such that $f(n)-ntau$ is bounded.
I have separately shown the uniqueness part and have found bounds that work in following restrictive cases :
(i) If $n$ is positive, I have bounded it above by $f(0)$ using $tau=f(1)+D$.
(ii) If $n$ is positive, I have bounded it below by $f(0)$ using $tau=f(1)-D$.
(iii) If $n$ is negative, I have bounded it above by $f(0)$ using $tau=f(-1)+D$.
(iv) If $n$ is negative, I have bounded it below by $f(0)$ using $tau=f(-1)-D$. I understand that I have always used $f(0)$ to bound them, but that is only what I can see since I am splitting $n$ as $n$ times the generator $1$ $epsilon$ $mathbb{Z}$. Please help me bound this universally using just one unique $tau$.
group-theory analysis upper-lower-bounds
asked Feb 13 at 11:32
CircleCircle
112
$endgroup$
Let $f:mathbb{Z}rightarrow mathbb{R}$ be a quasi-homomorphism, i.e $|f(a+b)-f(a)-f(b)|leq D$ $forall$ $a$ and $b$ in $mathbb{Z}$ ($mathbb{R}$ and $mathbb{Z}$ are here considered as additive groups and so you see the plus sign). I have to prove that there exists a unique number $tau$ $epsilon $ $mathbb{R}$ such that $f(n)-ntau$ is bounded.
I have separately shown the uniqueness part and have found bounds that work in following restrictive cases :
(i) If $n$ is positive, I have bounded it above by $f(0)$ using $tau=f(1)+D$.
(ii) If $n$ is positive, I have bounded it below by $f(0)$ using $tau=f(1)-D$.
(iii) If $n$ is negative, I have bounded it above by $f(0)$ using $tau=f(-1)+D$.
(iv) If $n$ is negative, I have bounded it below by $f(0)$ using $tau=f(-1)-D$. I understand that I have always used $f(0)$ to bound them, but that is only what I can see since I am splitting $n$ as $n$ times the generator $1$ $epsilon$ $mathbb{Z}$. Please help me bound this universally using just one unique $tau$.
group-theory analysis upper-lower-bounds
group-theory analysis upper-lower-bounds
asked Feb 13 at 11:32
CircleCircle
112
asked Feb 13 at 11:32
CircleCircle
112
asked Feb 13 at 11:32
CircleCircle
112
asked Feb 13 at 11:32
CircleCircle
112
asked Feb 13 at 11:32
CircleCircle
112
112
$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago
add a comment |
$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago
$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Define the function $f^*:mathbb Ztomathbb R$ by:
$$f^*(a)=lim_{nto +infty}frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= lim_{nto+infty}frac{f(nka)}{n}= klim_{nto+infty}frac{f(nka)}{nk}= klim_{nto+infty}frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|leqslant D+(n-1)D=nD.$$
Therefore:
$$frac{|f(na)-nf(a)|}{n}leqslant frac{n-1}{n}D<D,$$
and so:
$$|f^*(a)-f(a)|= |lim_{nto+infty}frac{f(na)}{n}-f(a)|= lim_{nto+infty}|frac{f(na)}{n}-f(a)|= lim_{nto+infty}frac{|f(na)-nf(a)|}{n}leqslant D.$$
Put $tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-atau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|leqslant D.$$
It remains to prove that $|f(-a)+atau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nDleqslant f(0)-nDleqslant f(-na)+nf(a)leqslant f(0)+nDleqslant |f(0)|+nD,$$
so:
$$frac{|f(-na)+nf(a)|}{n}leqslant frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |lim_{ntoinfty}frac{f(-na)}{n}+f(a)|= lim_{nto+infty}frac{|f(-na)+nf(a)|}{n}leqslant lim_{nto+infty}frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+atau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|leqslant D+D+D=3D.$$
So $|f(a)-atau|$ is bounded by $3D$ for every $a$.
answered 18 hours ago
SMMSMM
2,513510
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Define the function $f^*:mathbb Ztomathbb R$ by:
$$f^*(a)=lim_{nto +infty}frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= lim_{nto+infty}frac{f(nka)}{n}= klim_{nto+infty}frac{f(nka)}{nk}= klim_{nto+infty}frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|leqslant D+(n-1)D=nD.$$
Therefore:
$$frac{|f(na)-nf(a)|}{n}leqslant frac{n-1}{n}D<D,$$
and so:
$$|f^*(a)-f(a)|= |lim_{nto+infty}frac{f(na)}{n}-f(a)|= lim_{nto+infty}|frac{f(na)}{n}-f(a)|= lim_{nto+infty}frac{|f(na)-nf(a)|}{n}leqslant D.$$
Put $tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-atau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|leqslant D.$$
It remains to prove that $|f(-a)+atau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nDleqslant f(0)-nDleqslant f(-na)+nf(a)leqslant f(0)+nDleqslant |f(0)|+nD,$$
so:
$$frac{|f(-na)+nf(a)|}{n}leqslant frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |lim_{ntoinfty}frac{f(-na)}{n}+f(a)|= lim_{nto+infty}frac{|f(-na)+nf(a)|}{n}leqslant lim_{nto+infty}frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+atau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|leqslant D+D+D=3D.$$
So $|f(a)-atau|$ is bounded by $3D$ for every $a$.
answered 18 hours ago
SMMSMM
2,513510
$endgroup$
add a comment |
$begingroup$
Define the function $f^*:mathbb Ztomathbb R$ by:
$$f^*(a)=lim_{nto +infty}frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= lim_{nto+infty}frac{f(nka)}{n}= klim_{nto+infty}frac{f(nka)}{nk}= klim_{nto+infty}frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|leqslant D+(n-1)D=nD.$$
Therefore:
$$frac{|f(na)-nf(a)|}{n}leqslant frac{n-1}{n}D<D,$$
and so:
$$|f^*(a)-f(a)|= |lim_{nto+infty}frac{f(na)}{n}-f(a)|= lim_{nto+infty}|frac{f(na)}{n}-f(a)|= lim_{nto+infty}frac{|f(na)-nf(a)|}{n}leqslant D.$$
Put $tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-atau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|leqslant D.$$
It remains to prove that $|f(-a)+atau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nDleqslant f(0)-nDleqslant f(-na)+nf(a)leqslant f(0)+nDleqslant |f(0)|+nD,$$
so:
$$frac{|f(-na)+nf(a)|}{n}leqslant frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |lim_{ntoinfty}frac{f(-na)}{n}+f(a)|= lim_{nto+infty}frac{|f(-na)+nf(a)|}{n}leqslant lim_{nto+infty}frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+atau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|leqslant D+D+D=3D.$$
So $|f(a)-atau|$ is bounded by $3D$ for every $a$.
answered 18 hours ago
SMMSMM
2,513510
$endgroup$
add a comment |
$begingroup$
Define the function $f^*:mathbb Ztomathbb R$ by:
$$f^*(a)=lim_{nto +infty}frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= lim_{nto+infty}frac{f(nka)}{n}= klim_{nto+infty}frac{f(nka)}{nk}= klim_{nto+infty}frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|leqslant D+(n-1)D=nD.$$
Therefore:
$$frac{|f(na)-nf(a)|}{n}leqslant frac{n-1}{n}D<D,$$
and so:
$$|f^*(a)-f(a)|= |lim_{nto+infty}frac{f(na)}{n}-f(a)|= lim_{nto+infty}|frac{f(na)}{n}-f(a)|= lim_{nto+infty}frac{|f(na)-nf(a)|}{n}leqslant D.$$
Put $tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-atau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|leqslant D.$$
It remains to prove that $|f(-a)+atau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nDleqslant f(0)-nDleqslant f(-na)+nf(a)leqslant f(0)+nDleqslant |f(0)|+nD,$$
so:
$$frac{|f(-na)+nf(a)|}{n}leqslant frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |lim_{ntoinfty}frac{f(-na)}{n}+f(a)|= lim_{nto+infty}frac{|f(-na)+nf(a)|}{n}leqslant lim_{nto+infty}frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+atau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|leqslant D+D+D=3D.$$
So $|f(a)-atau|$ is bounded by $3D$ for every $a$.
answered 18 hours ago
SMMSMM
2,513510
$endgroup$
Define the function $f^*:mathbb Ztomathbb R$ by:
$$f^*(a)=lim_{nto +infty}frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= lim_{nto+infty}frac{f(nka)}{n}= klim_{nto+infty}frac{f(nka)}{nk}= klim_{nto+infty}frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|leqslant D+(n-1)D=nD.$$
Therefore:
$$frac{|f(na)-nf(a)|}{n}leqslant frac{n-1}{n}D<D,$$
and so:
$$|f^*(a)-f(a)|= |lim_{nto+infty}frac{f(na)}{n}-f(a)|= lim_{nto+infty}|frac{f(na)}{n}-f(a)|= lim_{nto+infty}frac{|f(na)-nf(a)|}{n}leqslant D.$$
Put $tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-atau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|leqslant D.$$
It remains to prove that $|f(-a)+atau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nDleqslant f(0)-nDleqslant f(-na)+nf(a)leqslant f(0)+nDleqslant |f(0)|+nD,$$
so:
$$frac{|f(-na)+nf(a)|}{n}leqslant frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |lim_{ntoinfty}frac{f(-na)}{n}+f(a)|= lim_{nto+infty}frac{|f(-na)+nf(a)|}{n}leqslant lim_{nto+infty}frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+atau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|leqslant D+D+D=3D.$$
So $|f(a)-atau|$ is bounded by $3D$ for every $a$.
answered 18 hours ago
SMMSMM
2,513510
answered 18 hours ago
SMMSMM
2,513510
answered 18 hours ago
SMMSMM
2,513510
answered 18 hours ago
SMMSMM
2,513510
2,513510
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$begingroup$
if you have separately shown uniqueness, why bound it? why not just show there must exist one?
$endgroup$
– Rylee Lyman
Feb 13 at 12:01
$begingroup$
Because uniqueness will just tell me that if at all universal bounds exist (not merely restrictive cases as I have been able to do as yet), all of them will be the same. I tried 'there must exist one' by contradiction but no conclusion yet.
$endgroup$
– Circle
Feb 13 at 12:13
$begingroup$
Probably $tau=lim_{ntoinfty}frac{f(n)}{n}$.
$endgroup$
– SMM
Feb 13 at 12:13
$begingroup$
@SMM $n$ can be negative as well. In that case, I will get two values for $tau$ (one in each direction). And from what I am getting, they will be negative of each other, which would imply it to be zero in case we expect the (directional) limits to be equal. Can you check it once?
$endgroup$
– Circle
yesterday
$begingroup$
@Circle I wrote an explanation in the answer.
$endgroup$
– SMM
18 hours ago