Lower bound for $|x-y|$
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0
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For $x$,$y$ in Hilbert space $mathcal{H}$ I want a lower bound for
begin{equation}
|x-y|_{mathcal{H}}^2
end{equation}
I know
begin{equation}
| |x|_{mathcal{H}}-|y|_{mathcal{H}} |leq|x-y|_{mathcal{H}}
end{equation}
However is there a better answer than this? My Hilbert space is $L^2$. Any help is greatly appreciated.
norm upper-lower-bounds
edited Nov 16 at 7:02
Masacroso
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
|
show 3 more comments
up vote
0
down vote
favorite
For $x$,$y$ in Hilbert space $mathcal{H}$ I want a lower bound for
begin{equation}
|x-y|_{mathcal{H}}^2
end{equation}
I know
begin{equation}
| |x|_{mathcal{H}}-|y|_{mathcal{H}} |leq|x-y|_{mathcal{H}}
end{equation}
However is there a better answer than this? My Hilbert space is $L^2$. Any help is greatly appreciated.
norm upper-lower-bounds
edited Nov 16 at 7:02
Masacroso
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
1
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For $x$,$y$ in Hilbert space $mathcal{H}$ I want a lower bound for
begin{equation}
|x-y|_{mathcal{H}}^2
end{equation}
I know
begin{equation}
| |x|_{mathcal{H}}-|y|_{mathcal{H}} |leq|x-y|_{mathcal{H}}
end{equation}
However is there a better answer than this? My Hilbert space is $L^2$. Any help is greatly appreciated.
norm upper-lower-bounds
edited Nov 16 at 7:02
Masacroso
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
For $x$,$y$ in Hilbert space $mathcal{H}$ I want a lower bound for
begin{equation}
|x-y|_{mathcal{H}}^2
end{equation}
I know
begin{equation}
| |x|_{mathcal{H}}-|y|_{mathcal{H}} |leq|x-y|_{mathcal{H}}
end{equation}
However is there a better answer than this? My Hilbert space is $L^2$. Any help is greatly appreciated.
norm upper-lower-bounds
norm upper-lower-bounds
edited Nov 16 at 7:02
Masacroso
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
edited Nov 16 at 7:02
Masacroso
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
edited Nov 16 at 7:02
Masacroso
12.3k41746
edited Nov 16 at 7:02
Masacroso
12.3k41746
edited Nov 16 at 7:02
Masacroso
12.3k41746
12.3k41746
asked Nov 16 at 6:55
CuriousCat
342
asked Nov 16 at 6:55
CuriousCat
342
asked Nov 16 at 6:55
CuriousCat
342
342
1
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12
|
show 3 more comments
1
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12
1
1
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12
|
show 3 more comments
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1
Please be more specific or give us a context. What you mean by "better"? The inequality is sharp in the sense that there exist pairs $x,y$ such that equality is reached.
– maxmilgram
Nov 16 at 7:01
I would like to get rid of the absolute value. I would just like to have $|x|$ and $|y|$ on the left hand side. Is this even possible? I am trying to find an upper for bound $|u-v|^2_{L^2}$.
– CuriousCat
Nov 16 at 7:04
note that $r|x-y|le|x-y|$ for any $rin[0,1]$, choose whatever you want :)
– Masacroso
Nov 16 at 7:05
I would like to have $|x-y|^2_{L^2}geq C(|x|^2-|y|^2)$.
– CuriousCat
Nov 16 at 7:07
You can just drop the absolute value, if you like. The inequality gets weaker by doing this.
– maxmilgram
Nov 16 at 7:12