Proving the complete monotonicity of $f$ given $(-log f)'$ is completely monotone












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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 $f$.



Prove that if $(-log f(x))'$ is completely monotonic, then $f(x)$ is also completely monotonic.




There are a few papers that use this without explicitly displaying the proof. It may help to display the function as $g(x)=(-log f(x))'$ then put $f(x)=exp(-g(x))$ but I have not yet find the pattern to induct the theorem.










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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 $f$.



    Prove that if $(-log f(x))'$ is completely monotonic, then $f(x)$ is also completely monotonic.




    There are a few papers that use this without explicitly displaying the proof. It may help to display the function as $g(x)=(-log f(x))'$ then put $f(x)=exp(-g(x))$ but I have not yet find the pattern to induct the theorem.










    share|cite|improve this question











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      2








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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 $f$.



      Prove that if $(-log f(x))'$ is completely monotonic, then $f(x)$ is also completely monotonic.




      There are a few papers that use this without explicitly displaying the proof. It may help to display the function as $g(x)=(-log f(x))'$ then put $f(x)=exp(-g(x))$ but I have not yet find the pattern to induct the theorem.










      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 $f$.



      Prove that if $(-log f(x))'$ is completely monotonic, then $f(x)$ is also completely monotonic.




      There are a few papers that use this without explicitly displaying the proof. It may help to display the function as $g(x)=(-log f(x))'$ then put $f(x)=exp(-g(x))$ but I have not yet find the pattern to induct the theorem.







      analysis monotone-functions






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      edited Dec 18 '18 at 10:34







      adli farhan

















      asked Dec 18 '18 at 7:42









      adli farhanadli farhan

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          $begingroup$

          For an open (and not necessarily connected) set $Asubseteqmathbb{R}$, let $P^infty(A)$ be the set of nonnegative infinitely differentiable functions $Atomathbb{R}$ with all the derivatives also nonnegative.



          Now if $g : Ato B$ with $A,Bsubseteqmathbb{R}$ open, $g'in P^infty(A)$, and $hin P^infty(B)$, then $hcirc gin P^infty(A)$ (this can be proven immediately or just seen from this formula).



          Your claim reduces to this (consider $g(x)=log f(-x)$ and $h(x)=e^x$).






          share|cite|improve this answer











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            $begingroup$

            For an open (and not necessarily connected) set $Asubseteqmathbb{R}$, let $P^infty(A)$ be the set of nonnegative infinitely differentiable functions $Atomathbb{R}$ with all the derivatives also nonnegative.



            Now if $g : Ato B$ with $A,Bsubseteqmathbb{R}$ open, $g'in P^infty(A)$, and $hin P^infty(B)$, then $hcirc gin P^infty(A)$ (this can be proven immediately or just seen from this formula).



            Your claim reduces to this (consider $g(x)=log f(-x)$ and $h(x)=e^x$).






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              For an open (and not necessarily connected) set $Asubseteqmathbb{R}$, let $P^infty(A)$ be the set of nonnegative infinitely differentiable functions $Atomathbb{R}$ with all the derivatives also nonnegative.



              Now if $g : Ato B$ with $A,Bsubseteqmathbb{R}$ open, $g'in P^infty(A)$, and $hin P^infty(B)$, then $hcirc gin P^infty(A)$ (this can be proven immediately or just seen from this formula).



              Your claim reduces to this (consider $g(x)=log f(-x)$ and $h(x)=e^x$).






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                For an open (and not necessarily connected) set $Asubseteqmathbb{R}$, let $P^infty(A)$ be the set of nonnegative infinitely differentiable functions $Atomathbb{R}$ with all the derivatives also nonnegative.



                Now if $g : Ato B$ with $A,Bsubseteqmathbb{R}$ open, $g'in P^infty(A)$, and $hin P^infty(B)$, then $hcirc gin P^infty(A)$ (this can be proven immediately or just seen from this formula).



                Your claim reduces to this (consider $g(x)=log f(-x)$ and $h(x)=e^x$).






                share|cite|improve this answer











                $endgroup$



                For an open (and not necessarily connected) set $Asubseteqmathbb{R}$, let $P^infty(A)$ be the set of nonnegative infinitely differentiable functions $Atomathbb{R}$ with all the derivatives also nonnegative.



                Now if $g : Ato B$ with $A,Bsubseteqmathbb{R}$ open, $g'in P^infty(A)$, and $hin P^infty(B)$, then $hcirc gin P^infty(A)$ (this can be proven immediately or just seen from this formula).



                Your claim reduces to this (consider $g(x)=log f(-x)$ and $h(x)=e^x$).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 18 '18 at 10:10

























                answered Dec 18 '18 at 9:36









                metamorphymetamorphy

                3,6821621




                3,6821621






























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