A function have minimum or maximum












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Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?



I could not even construct such an example satisfying the assumptions.






calculus convex-analysis







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    Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?



    I could not even construct such an example satisfying the assumptions.






    calculus convex-analysis







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      Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?



      I could not even construct such an example satisfying the assumptions.






      calculus convex-analysis







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      Let $f$ be strictly convex or concave on $Bbb R$. $lim_{xto -infty}f'(x)=A<0, lim_{xto+infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?



      I could not even construct such an example satisfying the assumptions.






      calculus convex-analysis




      calculus convex-analysis






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      edited Nov 21 '18 at 11:47

























      asked Nov 21 '18 at 11:16









      xldd

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          If $$f(x)=-ln(1+e^{-x})-2x$$



          then
          $$f'(x)=-1-frac{1}{1+e^{-x}}$$



          and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because



          $$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$



          for all $xinmathbb{R}$.



          The function $f$ has no global maximum or minimum (its derivative is always negative).




          Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).



          (You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)






          share|cite|improve this answer































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            Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.






            share|cite|improve this answer





























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              Take a generic function $f(x)$ that meets the required criteria.
              Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.



              As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.






              share|cite|improve this answer





















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                3 Answers
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                0














                If $$f(x)=-ln(1+e^{-x})-2x$$



                then
                $$f'(x)=-1-frac{1}{1+e^{-x}}$$



                and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because



                $$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$



                for all $xinmathbb{R}$.



                The function $f$ has no global maximum or minimum (its derivative is always negative).




                Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).



                (You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)






                share|cite|improve this answer




























                  0














                  If $$f(x)=-ln(1+e^{-x})-2x$$



                  then
                  $$f'(x)=-1-frac{1}{1+e^{-x}}$$



                  and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because



                  $$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$



                  for all $xinmathbb{R}$.



                  The function $f$ has no global maximum or minimum (its derivative is always negative).




                  Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).



                  (You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)






                  share|cite|improve this answer


























                    0












                    0








                    0






                    If $$f(x)=-ln(1+e^{-x})-2x$$



                    then
                    $$f'(x)=-1-frac{1}{1+e^{-x}}$$



                    and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because



                    $$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$



                    for all $xinmathbb{R}$.



                    The function $f$ has no global maximum or minimum (its derivative is always negative).




                    Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).



                    (You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)






                    share|cite|improve this answer














                    If $$f(x)=-ln(1+e^{-x})-2x$$



                    then
                    $$f'(x)=-1-frac{1}{1+e^{-x}}$$



                    and we have $lim_{xto-infty}f'(x)=-1<0$ and $lim_{xtoinfty}f'(x)=-2<0$. The function $f$ is strictly concave because



                    $$f''(x)=-frac{e^{-x}}{(1+e^{-x})^2}<0$$



                    for all $xinmathbb{R}$.



                    The function $f$ has no global maximum or minimum (its derivative is always negative).




                    Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-infty$ and $infty$) and subtracting one (to make both the limits negative).



                    (You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 21 '18 at 12:44

























                    answered Nov 21 '18 at 12:38









                    smcc

                    4,297517




                    4,297517























                        1














                        Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.






                        share|cite|improve this answer


























                          1














                          Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.






                            share|cite|improve this answer












                            Suppose $f'$ exists for every $xinmathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 21 '18 at 12:06









                            P De Donato

                            4147




                            4147























                                0














                                Take a generic function $f(x)$ that meets the required criteria.
                                Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.



                                As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.






                                share|cite|improve this answer


























                                  0














                                  Take a generic function $f(x)$ that meets the required criteria.
                                  Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.



                                  As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.






                                  share|cite|improve this answer
























                                    0












                                    0








                                    0






                                    Take a generic function $f(x)$ that meets the required criteria.
                                    Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.



                                    As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.






                                    share|cite|improve this answer












                                    Take a generic function $f(x)$ that meets the required criteria.
                                    Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-infty,a)$, where $|f(x)-(C_1+Ax)|<epsilon$; $(b,infty)$, where $|f(x)-(C_2+Bx)|<epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $forall x<a', f(x)>M+epsilon>M$ and $forall x>b', f(x)<N-epsilon<N$.



                                    As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 21 '18 at 12:06









                                    MoKo19

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