Lower bound for $|x-y|$











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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.










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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















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.










share|cite|improve this question




















  • 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













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.










share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 16 at 7:02









Masacroso

12.3k41746




12.3k41746










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














  • 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















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