the product of two completely monotone functions












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$begingroup$


A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.



Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?



if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?






derivatives monotone-functions







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  • $begingroup$
    Yes, use Leibniz rule for differentiation of f(x)g(x)..
    $endgroup$
    – LordVader007
    Jan 8 at 8:42










  • $begingroup$
    Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
    $endgroup$
    – Greg Martin
    Jan 8 at 8:52
















0












$begingroup$


A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.



Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?



if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?






derivatives monotone-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, use Leibniz rule for differentiation of f(x)g(x)..
    $endgroup$
    – LordVader007
    Jan 8 at 8:42










  • $begingroup$
    Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
    $endgroup$
    – Greg Martin
    Jan 8 at 8:52














0












0








0





$begingroup$


A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.



Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?



if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?






derivatives monotone-functions







share|cite|improve this question









$endgroup$




A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.



Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?



if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?






derivatives monotone-functions




derivatives monotone-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 8:12









adli farhanadli farhan

214




214












  • $begingroup$
    Yes, use Leibniz rule for differentiation of f(x)g(x)..
    $endgroup$
    – LordVader007
    Jan 8 at 8:42










  • $begingroup$
    Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
    $endgroup$
    – Greg Martin
    Jan 8 at 8:52


















  • $begingroup$
    Yes, use Leibniz rule for differentiation of f(x)g(x)..
    $endgroup$
    – LordVader007
    Jan 8 at 8:42










  • $begingroup$
    Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
    $endgroup$
    – Greg Martin
    Jan 8 at 8:52
















$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42




$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42












$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52




$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint:



We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.



Then



$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can you help me in proving that $f$ is CM ⟺$−f′$ is CM
    $endgroup$
    – adli farhan
    2 days ago



















0












$begingroup$

This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.



CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.



It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint:



    We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.



    Then



    $$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      can you help me in proving that $f$ is CM ⟺$−f′$ is CM
      $endgroup$
      – adli farhan
      2 days ago
















    2












    $begingroup$

    Hint:



    We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.



    Then



    $$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      can you help me in proving that $f$ is CM ⟺$−f′$ is CM
      $endgroup$
      – adli farhan
      2 days ago














    2












    2








    2





    $begingroup$

    Hint:



    We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.



    Then



    $$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$






    share|cite|improve this answer









    $endgroup$



    Hint:



    We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.



    Then



    $$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 9:17









    Yves DaoustYves Daoust

    131k676229




    131k676229












    • $begingroup$
      can you help me in proving that $f$ is CM ⟺$−f′$ is CM
      $endgroup$
      – adli farhan
      2 days ago


















    • $begingroup$
      can you help me in proving that $f$ is CM ⟺$−f′$ is CM
      $endgroup$
      – adli farhan
      2 days ago
















    $begingroup$
    can you help me in proving that $f$ is CM ⟺$−f′$ is CM
    $endgroup$
    – adli farhan
    2 days ago




    $begingroup$
    can you help me in proving that $f$ is CM ⟺$−f′$ is CM
    $endgroup$
    – adli farhan
    2 days ago











    0












    $begingroup$

    This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.



    CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
    $$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
    (provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
    $$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
    i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.



    It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.



      CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
      $$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
      (provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
      $$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
      i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.



      It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.



        CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
        $$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
        (provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
        $$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
        i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.



        It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.






        share|cite|improve this answer









        $endgroup$



        This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.



        CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
        $$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
        (provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
        $$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
        i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.



        It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 9:08









        postmortespostmortes

        2,21031422




        2,21031422






























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