Examples of functions characterized by sequence












0












$begingroup$


Let $ Psi$ be a set of functions $chi :mathbb{R}^+rightarrow [0,1)$ satisfying
$$chi(t_n)rightarrow 1Rightarrow t_nrightarrow 0$$



I want to find some functions that belongs to $ Psi$, other than



$chi_1 (t)=left{begin{matrix}
e^{-2t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$



$chi_2 (t)=left{begin{matrix}
frac{1}{1+t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$




I'm working on this kind of functions, and i want to know more about it.






functional-analysis functions examples-counterexamples







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
    $endgroup$
    – Yanko
    Dec 7 '18 at 14:45








  • 1




    $begingroup$
    Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 0:47












  • $begingroup$
    @ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
    $endgroup$
    – Motaka
    Dec 8 '18 at 10:16










  • $begingroup$
    @Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
    $endgroup$
    – Motaka
    Dec 10 '18 at 9:13






  • 1




    $begingroup$
    It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
    $endgroup$
    – Paul Sinclair
    Dec 11 '18 at 1:20
















0












$begingroup$


Let $ Psi$ be a set of functions $chi :mathbb{R}^+rightarrow [0,1)$ satisfying
$$chi(t_n)rightarrow 1Rightarrow t_nrightarrow 0$$



I want to find some functions that belongs to $ Psi$, other than



$chi_1 (t)=left{begin{matrix}
e^{-2t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$



$chi_2 (t)=left{begin{matrix}
frac{1}{1+t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$




I'm working on this kind of functions, and i want to know more about it.






functional-analysis functions examples-counterexamples







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
    $endgroup$
    – Yanko
    Dec 7 '18 at 14:45








  • 1




    $begingroup$
    Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 0:47












  • $begingroup$
    @ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
    $endgroup$
    – Motaka
    Dec 8 '18 at 10:16










  • $begingroup$
    @Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
    $endgroup$
    – Motaka
    Dec 10 '18 at 9:13






  • 1




    $begingroup$
    It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
    $endgroup$
    – Paul Sinclair
    Dec 11 '18 at 1:20














0












0








0





$begingroup$


Let $ Psi$ be a set of functions $chi :mathbb{R}^+rightarrow [0,1)$ satisfying
$$chi(t_n)rightarrow 1Rightarrow t_nrightarrow 0$$



I want to find some functions that belongs to $ Psi$, other than



$chi_1 (t)=left{begin{matrix}
e^{-2t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$



$chi_2 (t)=left{begin{matrix}
frac{1}{1+t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$




I'm working on this kind of functions, and i want to know more about it.






functional-analysis functions examples-counterexamples







share|cite|improve this question











$endgroup$




Let $ Psi$ be a set of functions $chi :mathbb{R}^+rightarrow [0,1)$ satisfying
$$chi(t_n)rightarrow 1Rightarrow t_nrightarrow 0$$



I want to find some functions that belongs to $ Psi$, other than



$chi_1 (t)=left{begin{matrix}
e^{-2t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$



$chi_2 (t)=left{begin{matrix}
frac{1}{1+t} & text{if } t>0\
1 & text{if } t=0
end{matrix}right.$




I'm working on this kind of functions, and i want to know more about it.






functional-analysis functions examples-counterexamples




functional-analysis functions examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:09







Motaka

















asked Dec 7 '18 at 14:39









MotakaMotaka

239111




239111








  • 1




    $begingroup$
    Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
    $endgroup$
    – Yanko
    Dec 7 '18 at 14:45








  • 1




    $begingroup$
    Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 0:47












  • $begingroup$
    @ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
    $endgroup$
    – Motaka
    Dec 8 '18 at 10:16










  • $begingroup$
    @Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
    $endgroup$
    – Motaka
    Dec 10 '18 at 9:13






  • 1




    $begingroup$
    It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
    $endgroup$
    – Paul Sinclair
    Dec 11 '18 at 1:20














  • 1




    $begingroup$
    Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
    $endgroup$
    – Yanko
    Dec 7 '18 at 14:45








  • 1




    $begingroup$
    Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 0:47












  • $begingroup$
    @ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
    $endgroup$
    – Motaka
    Dec 8 '18 at 10:16










  • $begingroup$
    @Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
    $endgroup$
    – Motaka
    Dec 10 '18 at 9:13






  • 1




    $begingroup$
    It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
    $endgroup$
    – Paul Sinclair
    Dec 11 '18 at 1:20








1




1




$begingroup$
Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
$endgroup$
– Yanko
Dec 7 '18 at 14:45






$begingroup$
Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $tnot =1$ for any $ain(0,1)$. In this case $f(t_n)rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero.
$endgroup$
– Yanko
Dec 7 '18 at 14:45






1




1




$begingroup$
Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 0:47






$begingroup$
Neither of your sample functions has the property you describe. Both satisfy $chi(t_n) to 0 implies t_n to infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2cdot 0} = frac 1{1 + 0} = 1$$Did you mean something else?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 0:47














$begingroup$
@ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
$endgroup$
– Motaka
Dec 8 '18 at 10:16




$begingroup$
@ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition
$endgroup$
– Motaka
Dec 8 '18 at 10:16












$begingroup$
@Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
$endgroup$
– Motaka
Dec 10 '18 at 9:13




$begingroup$
@Paul Sinclair: I'm so sorry I made an error in the definition of $chi$.. It's correct now
$endgroup$
– Motaka
Dec 10 '18 at 9:13




1




1




$begingroup$
It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
$endgroup$
– Paul Sinclair
Dec 11 '18 at 1:20




$begingroup$
It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$.
$endgroup$
– Paul Sinclair
Dec 11 '18 at 1:20










1 Answer
1






active

oldest

votes


















1












$begingroup$

Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.



The condition $chi(t_n) to 1 implies t_n to 0$ is equivalent to (for functions $[0,infty) to [0,1)$):




For every $epsilon > 0$ there is some $m_epsilon > 0$ such that $chi(t) le 1- m_epsilon$ for all $t in [epsilon, infty)$.




I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.



Your condition does not require any behavior for $chi$ near $0$. It could be that $lim_{tto 0} chi(t) = 1$, but it could also be true that $chi$ is bounded away from $1$ everywhere. In that case, there are no sequences ${t_n}$ such that $chi(t_n) to 1$, so your condition is vacuously true.



Any function bounded away from $1$ is an example:





  • $chi(t) = c$ for any constant $c < 1$.


  • $chi(t) = a + bsin t$, with $1 > a + b$ and $ a ge b$.

  • $chi(t) = begin{cases} frac{arctan t}pi&t text{ rational}\ 0.5& ttext{ irrational}end{cases}$


Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and




  • $chi(t) = begin{cases} 1-t & t < 1\0& t > 1end{cases}$

  • $chi(t) = e^{-(t-1)^2}$

  • $chi(t) = frac{2arctan (1/t)}pi$


The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n in (p - 1/n, p + 1/n)$ with $chi(t_n) > 1 - 1/n$, and therefore $chi(t_n) to 1$ while $t_n to p$. A similar argument holds for $p = infty$. Thus every point in $[epsilon, infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_epsilon$.






share|cite|improve this answer











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    1












    $begingroup$

    Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.



    The condition $chi(t_n) to 1 implies t_n to 0$ is equivalent to (for functions $[0,infty) to [0,1)$):




    For every $epsilon > 0$ there is some $m_epsilon > 0$ such that $chi(t) le 1- m_epsilon$ for all $t in [epsilon, infty)$.




    I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.



    Your condition does not require any behavior for $chi$ near $0$. It could be that $lim_{tto 0} chi(t) = 1$, but it could also be true that $chi$ is bounded away from $1$ everywhere. In that case, there are no sequences ${t_n}$ such that $chi(t_n) to 1$, so your condition is vacuously true.



    Any function bounded away from $1$ is an example:





    • $chi(t) = c$ for any constant $c < 1$.


    • $chi(t) = a + bsin t$, with $1 > a + b$ and $ a ge b$.

    • $chi(t) = begin{cases} frac{arctan t}pi&t text{ rational}\ 0.5& ttext{ irrational}end{cases}$


    Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and




    • $chi(t) = begin{cases} 1-t & t < 1\0& t > 1end{cases}$

    • $chi(t) = e^{-(t-1)^2}$

    • $chi(t) = frac{2arctan (1/t)}pi$


    The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n in (p - 1/n, p + 1/n)$ with $chi(t_n) > 1 - 1/n$, and therefore $chi(t_n) to 1$ while $t_n to p$. A similar argument holds for $p = infty$. Thus every point in $[epsilon, infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_epsilon$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.



      The condition $chi(t_n) to 1 implies t_n to 0$ is equivalent to (for functions $[0,infty) to [0,1)$):




      For every $epsilon > 0$ there is some $m_epsilon > 0$ such that $chi(t) le 1- m_epsilon$ for all $t in [epsilon, infty)$.




      I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.



      Your condition does not require any behavior for $chi$ near $0$. It could be that $lim_{tto 0} chi(t) = 1$, but it could also be true that $chi$ is bounded away from $1$ everywhere. In that case, there are no sequences ${t_n}$ such that $chi(t_n) to 1$, so your condition is vacuously true.



      Any function bounded away from $1$ is an example:





      • $chi(t) = c$ for any constant $c < 1$.


      • $chi(t) = a + bsin t$, with $1 > a + b$ and $ a ge b$.

      • $chi(t) = begin{cases} frac{arctan t}pi&t text{ rational}\ 0.5& ttext{ irrational}end{cases}$


      Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and




      • $chi(t) = begin{cases} 1-t & t < 1\0& t > 1end{cases}$

      • $chi(t) = e^{-(t-1)^2}$

      • $chi(t) = frac{2arctan (1/t)}pi$


      The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n in (p - 1/n, p + 1/n)$ with $chi(t_n) > 1 - 1/n$, and therefore $chi(t_n) to 1$ while $t_n to p$. A similar argument holds for $p = infty$. Thus every point in $[epsilon, infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_epsilon$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.



        The condition $chi(t_n) to 1 implies t_n to 0$ is equivalent to (for functions $[0,infty) to [0,1)$):




        For every $epsilon > 0$ there is some $m_epsilon > 0$ such that $chi(t) le 1- m_epsilon$ for all $t in [epsilon, infty)$.




        I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.



        Your condition does not require any behavior for $chi$ near $0$. It could be that $lim_{tto 0} chi(t) = 1$, but it could also be true that $chi$ is bounded away from $1$ everywhere. In that case, there are no sequences ${t_n}$ such that $chi(t_n) to 1$, so your condition is vacuously true.



        Any function bounded away from $1$ is an example:





        • $chi(t) = c$ for any constant $c < 1$.


        • $chi(t) = a + bsin t$, with $1 > a + b$ and $ a ge b$.

        • $chi(t) = begin{cases} frac{arctan t}pi&t text{ rational}\ 0.5& ttext{ irrational}end{cases}$


        Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and




        • $chi(t) = begin{cases} 1-t & t < 1\0& t > 1end{cases}$

        • $chi(t) = e^{-(t-1)^2}$

        • $chi(t) = frac{2arctan (1/t)}pi$


        The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n in (p - 1/n, p + 1/n)$ with $chi(t_n) > 1 - 1/n$, and therefore $chi(t_n) to 1$ while $t_n to p$. A similar argument holds for $p = infty$. Thus every point in $[epsilon, infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_epsilon$.






        share|cite|improve this answer











        $endgroup$



        Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.



        The condition $chi(t_n) to 1 implies t_n to 0$ is equivalent to (for functions $[0,infty) to [0,1)$):




        For every $epsilon > 0$ there is some $m_epsilon > 0$ such that $chi(t) le 1- m_epsilon$ for all $t in [epsilon, infty)$.




        I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.



        Your condition does not require any behavior for $chi$ near $0$. It could be that $lim_{tto 0} chi(t) = 1$, but it could also be true that $chi$ is bounded away from $1$ everywhere. In that case, there are no sequences ${t_n}$ such that $chi(t_n) to 1$, so your condition is vacuously true.



        Any function bounded away from $1$ is an example:





        • $chi(t) = c$ for any constant $c < 1$.


        • $chi(t) = a + bsin t$, with $1 > a + b$ and $ a ge b$.

        • $chi(t) = begin{cases} frac{arctan t}pi&t text{ rational}\ 0.5& ttext{ irrational}end{cases}$


        Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and




        • $chi(t) = begin{cases} 1-t & t < 1\0& t > 1end{cases}$

        • $chi(t) = e^{-(t-1)^2}$

        • $chi(t) = frac{2arctan (1/t)}pi$


        The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n in (p - 1/n, p + 1/n)$ with $chi(t_n) > 1 - 1/n$, and therefore $chi(t_n) to 1$ while $t_n to p$. A similar argument holds for $p = infty$. Thus every point in $[epsilon, infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_epsilon$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 1:59

























        answered Dec 8 '18 at 2:02









        Paul SinclairPaul Sinclair

        19.6k21442




        19.6k21442






























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