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Decreasing $f:mathbbRtomathbbR$ tending to $0$ at $∞$ not convex beyond any point?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Convex function which has a limit in $-infty$ is non decreasingLocal monotonicity and differentiability a.e.Strictly decreasing function with a horizontal asymptote is convex?For which values of $x_0inmathbbR$ is $f$ differentiable$f$ convex strictly decreasing function , is $f'(x+delta)-f'(x)$ convexComputing the limit of a convex, decreasing functionSufficient condition for inflection pointStrictly convex except in a single pointSufficient condition for absolute maximumExample of convex and injective $f:I to mathbbR$ such that $f^-1$ is not concave










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Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.










    share|cite|improve this question











    $endgroup$














      1












      1








      1


      1



      $begingroup$


      Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.










      share|cite|improve this question











      $endgroup$




      Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.







      real-analysis limits monotone-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 16:50









      user21820

      40.4k544163




      40.4k544163










      asked Mar 24 at 13:08









      Andrew VAndrew V

      342111




      342111




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Edit: added additional term to make the function actually decreasing. The idea remains the same.



          Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is this decreasing for all reals?
            $endgroup$
            – marty cohen
            Mar 24 at 15:21










          • $begingroup$
            @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
            $endgroup$
            – RiaD
            Mar 24 at 16:37










          • $begingroup$
            @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
            $endgroup$
            – WimC
            Mar 24 at 18:29











          Your Answer








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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Edit: added additional term to make the function actually decreasing. The idea remains the same.



          Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is this decreasing for all reals?
            $endgroup$
            – marty cohen
            Mar 24 at 15:21










          • $begingroup$
            @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
            $endgroup$
            – RiaD
            Mar 24 at 16:37










          • $begingroup$
            @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
            $endgroup$
            – WimC
            Mar 24 at 18:29















          5












          $begingroup$

          Edit: added additional term to make the function actually decreasing. The idea remains the same.



          Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is this decreasing for all reals?
            $endgroup$
            – marty cohen
            Mar 24 at 15:21










          • $begingroup$
            @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
            $endgroup$
            – RiaD
            Mar 24 at 16:37










          • $begingroup$
            @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
            $endgroup$
            – WimC
            Mar 24 at 18:29













          5












          5








          5





          $begingroup$

          Edit: added additional term to make the function actually decreasing. The idea remains the same.



          Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.






          share|cite|improve this answer











          $endgroup$



          Edit: added additional term to make the function actually decreasing. The idea remains the same.



          Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 24 at 18:47

























          answered Mar 24 at 13:30









          WimCWimC

          24.4k23063




          24.4k23063







          • 1




            $begingroup$
            Is this decreasing for all reals?
            $endgroup$
            – marty cohen
            Mar 24 at 15:21










          • $begingroup$
            @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
            $endgroup$
            – RiaD
            Mar 24 at 16:37










          • $begingroup$
            @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
            $endgroup$
            – WimC
            Mar 24 at 18:29












          • 1




            $begingroup$
            Is this decreasing for all reals?
            $endgroup$
            – marty cohen
            Mar 24 at 15:21










          • $begingroup$
            @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
            $endgroup$
            – RiaD
            Mar 24 at 16:37










          • $begingroup$
            @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
            $endgroup$
            – WimC
            Mar 24 at 18:29







          1




          1




          $begingroup$
          Is this decreasing for all reals?
          $endgroup$
          – marty cohen
          Mar 24 at 15:21




          $begingroup$
          Is this decreasing for all reals?
          $endgroup$
          – marty cohen
          Mar 24 at 15:21












          $begingroup$
          @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
          $endgroup$
          – RiaD
          Mar 24 at 16:37




          $begingroup$
          @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
          $endgroup$
          – RiaD
          Mar 24 at 16:37












          $begingroup$
          @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
          $endgroup$
          – WimC
          Mar 24 at 18:29




          $begingroup$
          @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
          $endgroup$
          – WimC
          Mar 24 at 18:29

















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