infimum of function of two distincts variables












2












$begingroup$


I am a little bit confused about the definition of infmum of a function of two variables. Let $lambda_n = 1 - e^{-n}$ and $beta = {{lambda_n}}_{n >0}$. . I would like to compute $$ inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$ I don't see what this infimum means exactly. Is it applied to $alpha$ or $mu$.
I really need you help.



Thanks.






real-analysis functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:41












  • $begingroup$
    No, I am just not used to tex. @Rebellos
    $endgroup$
    – user12313
    Dec 7 '18 at 21:43










  • $begingroup$
    Where do $alpha$ and $mu$ belong to ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:44










  • $begingroup$
    They belong to $beta$
    $endgroup$
    – user12313
    Dec 7 '18 at 21:45
















2












$begingroup$


I am a little bit confused about the definition of infmum of a function of two variables. Let $lambda_n = 1 - e^{-n}$ and $beta = {{lambda_n}}_{n >0}$. . I would like to compute $$ inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$ I don't see what this infimum means exactly. Is it applied to $alpha$ or $mu$.
I really need you help.



Thanks.






real-analysis functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:41












  • $begingroup$
    No, I am just not used to tex. @Rebellos
    $endgroup$
    – user12313
    Dec 7 '18 at 21:43










  • $begingroup$
    Where do $alpha$ and $mu$ belong to ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:44










  • $begingroup$
    They belong to $beta$
    $endgroup$
    – user12313
    Dec 7 '18 at 21:45














2












2








2





$begingroup$


I am a little bit confused about the definition of infmum of a function of two variables. Let $lambda_n = 1 - e^{-n}$ and $beta = {{lambda_n}}_{n >0}$. . I would like to compute $$ inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$ I don't see what this infimum means exactly. Is it applied to $alpha$ or $mu$.
I really need you help.



Thanks.






real-analysis functional-analysis







share|cite|improve this question











$endgroup$




I am a little bit confused about the definition of infmum of a function of two variables. Let $lambda_n = 1 - e^{-n}$ and $beta = {{lambda_n}}_{n >0}$. . I would like to compute $$ inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$ I don't see what this infimum means exactly. Is it applied to $alpha$ or $mu$.
I really need you help.



Thanks.






real-analysis functional-analysis




real-analysis functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 21:44









Rebellos

15.3k31250




15.3k31250










asked Dec 7 '18 at 21:32









user12313user12313

111




111












  • $begingroup$
    Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:41












  • $begingroup$
    No, I am just not used to tex. @Rebellos
    $endgroup$
    – user12313
    Dec 7 '18 at 21:43










  • $begingroup$
    Where do $alpha$ and $mu$ belong to ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:44










  • $begingroup$
    They belong to $beta$
    $endgroup$
    – user12313
    Dec 7 '18 at 21:45


















  • $begingroup$
    Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:41












  • $begingroup$
    No, I am just not used to tex. @Rebellos
    $endgroup$
    – user12313
    Dec 7 '18 at 21:43










  • $begingroup$
    Where do $alpha$ and $mu$ belong to ?
    $endgroup$
    – Rebellos
    Dec 7 '18 at 21:44










  • $begingroup$
    They belong to $beta$
    $endgroup$
    – user12313
    Dec 7 '18 at 21:45
















$begingroup$
Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
$endgroup$
– Rebellos
Dec 7 '18 at 21:41






$begingroup$
Is there a particular reason for the notation $alpha times mu$ and not $alpha cdot mu$ ?
$endgroup$
– Rebellos
Dec 7 '18 at 21:41














$begingroup$
No, I am just not used to tex. @Rebellos
$endgroup$
– user12313
Dec 7 '18 at 21:43




$begingroup$
No, I am just not used to tex. @Rebellos
$endgroup$
– user12313
Dec 7 '18 at 21:43












$begingroup$
Where do $alpha$ and $mu$ belong to ?
$endgroup$
– Rebellos
Dec 7 '18 at 21:44




$begingroup$
Where do $alpha$ and $mu$ belong to ?
$endgroup$
– Rebellos
Dec 7 '18 at 21:44












$begingroup$
They belong to $beta$
$endgroup$
– user12313
Dec 7 '18 at 21:45




$begingroup$
They belong to $beta$
$endgroup$
– user12313
Dec 7 '18 at 21:45










2 Answers
2






active

oldest

votes


















0












$begingroup$

The infimum is over the set of $alpha$ and $mu$ in $beta$ such that $alpha neq mu$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Let $f : beta to mathbb R$ such that :



    $$f(a) = frac{alpha - mu}{1 - alpha cdot mu}$$



    The expression



    $$inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$



    means that you want to find the infimum of the function $f(a)$ with respect to $a$ when it's not equal to $mu$ and they both belong to $beta$.



    The infimum of $f(a)$ is the greatest element in $mathbb R$ such that is less than or equal to all elements of $f^{-1}(mathbb R)$, meaning the image of $f$.



    Now, in order to find it, let $alpha, mu in beta$ with $alpha neq mu$ and :



    $$alpha = 1 - e^{-ell}, mu = 1-e^{-m}$$



    Then :



    $$inf_{aneq mu in beta} frac{alpha - mu}{1-alphacdot mu} = inf_{ell in mathbb N neq m in mathbb N} frac{e^{-ell} - e^{-m}}{1-(1-e^{-ell})(1-e^{-m})}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
      $endgroup$
      – user12313
      Dec 7 '18 at 21:57










    • $begingroup$
      @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
      $endgroup$
      – Rebellos
      Dec 7 '18 at 21:58












    • $begingroup$
      Ok, but I still don't know how to compute this infimum then? @Rebellos
      $endgroup$
      – user12313
      Dec 7 '18 at 22:06










    • $begingroup$
      @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
      $endgroup$
      – Rebellos
      Dec 7 '18 at 22:08










    • $begingroup$
      I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
      $endgroup$
      – user12313
      Dec 7 '18 at 22:09













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    2 Answers
    2






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    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    The infimum is over the set of $alpha$ and $mu$ in $beta$ such that $alpha neq mu$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The infimum is over the set of $alpha$ and $mu$ in $beta$ such that $alpha neq mu$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The infimum is over the set of $alpha$ and $mu$ in $beta$ such that $alpha neq mu$.






        share|cite|improve this answer











        $endgroup$



        The infimum is over the set of $alpha$ and $mu$ in $beta$ such that $alpha neq mu$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 21:54

























        answered Dec 7 '18 at 21:48









        zoidbergzoidberg

        1,080113




        1,080113























            0












            $begingroup$

            Let $f : beta to mathbb R$ such that :



            $$f(a) = frac{alpha - mu}{1 - alpha cdot mu}$$



            The expression



            $$inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$



            means that you want to find the infimum of the function $f(a)$ with respect to $a$ when it's not equal to $mu$ and they both belong to $beta$.



            The infimum of $f(a)$ is the greatest element in $mathbb R$ such that is less than or equal to all elements of $f^{-1}(mathbb R)$, meaning the image of $f$.



            Now, in order to find it, let $alpha, mu in beta$ with $alpha neq mu$ and :



            $$alpha = 1 - e^{-ell}, mu = 1-e^{-m}$$



            Then :



            $$inf_{aneq mu in beta} frac{alpha - mu}{1-alphacdot mu} = inf_{ell in mathbb N neq m in mathbb N} frac{e^{-ell} - e^{-m}}{1-(1-e^{-ell})(1-e^{-m})}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 21:57










            • $begingroup$
              @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
              $endgroup$
              – Rebellos
              Dec 7 '18 at 21:58












            • $begingroup$
              Ok, but I still don't know how to compute this infimum then? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:06










            • $begingroup$
              @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
              $endgroup$
              – Rebellos
              Dec 7 '18 at 22:08










            • $begingroup$
              I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:09


















            0












            $begingroup$

            Let $f : beta to mathbb R$ such that :



            $$f(a) = frac{alpha - mu}{1 - alpha cdot mu}$$



            The expression



            $$inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$



            means that you want to find the infimum of the function $f(a)$ with respect to $a$ when it's not equal to $mu$ and they both belong to $beta$.



            The infimum of $f(a)$ is the greatest element in $mathbb R$ such that is less than or equal to all elements of $f^{-1}(mathbb R)$, meaning the image of $f$.



            Now, in order to find it, let $alpha, mu in beta$ with $alpha neq mu$ and :



            $$alpha = 1 - e^{-ell}, mu = 1-e^{-m}$$



            Then :



            $$inf_{aneq mu in beta} frac{alpha - mu}{1-alphacdot mu} = inf_{ell in mathbb N neq m in mathbb N} frac{e^{-ell} - e^{-m}}{1-(1-e^{-ell})(1-e^{-m})}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 21:57










            • $begingroup$
              @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
              $endgroup$
              – Rebellos
              Dec 7 '18 at 21:58












            • $begingroup$
              Ok, but I still don't know how to compute this infimum then? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:06










            • $begingroup$
              @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
              $endgroup$
              – Rebellos
              Dec 7 '18 at 22:08










            • $begingroup$
              I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:09
















            0












            0








            0





            $begingroup$

            Let $f : beta to mathbb R$ such that :



            $$f(a) = frac{alpha - mu}{1 - alpha cdot mu}$$



            The expression



            $$inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$



            means that you want to find the infimum of the function $f(a)$ with respect to $a$ when it's not equal to $mu$ and they both belong to $beta$.



            The infimum of $f(a)$ is the greatest element in $mathbb R$ such that is less than or equal to all elements of $f^{-1}(mathbb R)$, meaning the image of $f$.



            Now, in order to find it, let $alpha, mu in beta$ with $alpha neq mu$ and :



            $$alpha = 1 - e^{-ell}, mu = 1-e^{-m}$$



            Then :



            $$inf_{aneq mu in beta} frac{alpha - mu}{1-alphacdot mu} = inf_{ell in mathbb N neq m in mathbb N} frac{e^{-ell} - e^{-m}}{1-(1-e^{-ell})(1-e^{-m})}$$






            share|cite|improve this answer











            $endgroup$



            Let $f : beta to mathbb R$ such that :



            $$f(a) = frac{alpha - mu}{1 - alpha cdot mu}$$



            The expression



            $$inf_{alpha neq mu in beta}frac{alpha - mu}{1 - alpha cdot mu}$$



            means that you want to find the infimum of the function $f(a)$ with respect to $a$ when it's not equal to $mu$ and they both belong to $beta$.



            The infimum of $f(a)$ is the greatest element in $mathbb R$ such that is less than or equal to all elements of $f^{-1}(mathbb R)$, meaning the image of $f$.



            Now, in order to find it, let $alpha, mu in beta$ with $alpha neq mu$ and :



            $$alpha = 1 - e^{-ell}, mu = 1-e^{-m}$$



            Then :



            $$inf_{aneq mu in beta} frac{alpha - mu}{1-alphacdot mu} = inf_{ell in mathbb N neq m in mathbb N} frac{e^{-ell} - e^{-m}}{1-(1-e^{-ell})(1-e^{-m})}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 21:55

























            answered Dec 7 '18 at 21:50









            RebellosRebellos

            15.3k31250




            15.3k31250












            • $begingroup$
              Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 21:57










            • $begingroup$
              @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
              $endgroup$
              – Rebellos
              Dec 7 '18 at 21:58












            • $begingroup$
              Ok, but I still don't know how to compute this infimum then? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:06










            • $begingroup$
              @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
              $endgroup$
              – Rebellos
              Dec 7 '18 at 22:08










            • $begingroup$
              I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:09




















            • $begingroup$
              Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 21:57










            • $begingroup$
              @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
              $endgroup$
              – Rebellos
              Dec 7 '18 at 21:58












            • $begingroup$
              Ok, but I still don't know how to compute this infimum then? @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:06










            • $begingroup$
              @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
              $endgroup$
              – Rebellos
              Dec 7 '18 at 22:08










            • $begingroup$
              I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
              $endgroup$
              – user12313
              Dec 7 '18 at 22:09


















            $begingroup$
            Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 21:57




            $begingroup$
            Ok, and then I think this will give us that the infimum is equal to zero? @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 21:57












            $begingroup$
            @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
            $endgroup$
            – Rebellos
            Dec 7 '18 at 21:58






            $begingroup$
            @user12313 It would be, if it could be $ell = mu$. But it's not, since we want $ell neq mu$.
            $endgroup$
            – Rebellos
            Dec 7 '18 at 21:58














            $begingroup$
            Ok, but I still don't know how to compute this infimum then? @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 22:06




            $begingroup$
            Ok, but I still don't know how to compute this infimum then? @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 22:06












            $begingroup$
            @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
            $endgroup$
            – Rebellos
            Dec 7 '18 at 22:08




            $begingroup$
            @user12313 The question was about what infimum means that over what variable is defined or how to find it ?
            $endgroup$
            – Rebellos
            Dec 7 '18 at 22:08












            $begingroup$
            I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 22:09






            $begingroup$
            I could say the two, I would like to understand it and compute it. Thanks for your explanations @Rebellos
            $endgroup$
            – user12313
            Dec 7 '18 at 22:09




















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