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If stationary points and minima are equivalent, then is the function convex?


Counterexample to show that the set of global minima of a function $f$ is a strict subset of the set of minima of the convex envelope of $f$Is every monotone map the gradient of a convex function?A convex function is differentiable at all but countably many pointsAre the stationary points of a strongly convex function unique in each dimension?Strictly increasing, strictly convex function: is the second derivative positive?What does it mean for the Hessian of a convex function to be positive semidefinite?Question on equivalent definitions of a convex functionGeneralization of properties of the subgradient of a convex function $f$Showing a function is convex by looking at the hessian.How to show two different definitions of $alpha$-strongly convex are equivalent?













4












$begingroup$


Let $f : mathbb R^n to mathbb R$ be a differentiable function for which a minimum exists. If $f$ is convex, then



$$x in mathbb R^n : nabla f(x) = 0 = x in mathbb R^n : f(x) leq f(y), ; forall y in mathbb R^n.$$



However, is the converse statement true? That is, if the above equation holds (and the two sets are non-empty), then is $f$ necessarily convex? Furthermore, would compactness of these sets be relevant?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
    $endgroup$
    – Andreas Blass
    May 2 at 1:28










  • $begingroup$
    @AndreasBlass Thank you for the observation. I edited the question.
    $endgroup$
    – Justin Le
    May 2 at 1:33






  • 1




    $begingroup$
    To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
    $endgroup$
    – Andreas Blass
    May 2 at 1:42






  • 1




    $begingroup$
    This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
    $endgroup$
    – Theo Bendit
    May 2 at 2:29















4












$begingroup$


Let $f : mathbb R^n to mathbb R$ be a differentiable function for which a minimum exists. If $f$ is convex, then



$$x in mathbb R^n : nabla f(x) = 0 = x in mathbb R^n : f(x) leq f(y), ; forall y in mathbb R^n.$$



However, is the converse statement true? That is, if the above equation holds (and the two sets are non-empty), then is $f$ necessarily convex? Furthermore, would compactness of these sets be relevant?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
    $endgroup$
    – Andreas Blass
    May 2 at 1:28










  • $begingroup$
    @AndreasBlass Thank you for the observation. I edited the question.
    $endgroup$
    – Justin Le
    May 2 at 1:33






  • 1




    $begingroup$
    To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
    $endgroup$
    – Andreas Blass
    May 2 at 1:42






  • 1




    $begingroup$
    This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
    $endgroup$
    – Theo Bendit
    May 2 at 2:29













4












4








4


1



$begingroup$


Let $f : mathbb R^n to mathbb R$ be a differentiable function for which a minimum exists. If $f$ is convex, then



$$x in mathbb R^n : nabla f(x) = 0 = x in mathbb R^n : f(x) leq f(y), ; forall y in mathbb R^n.$$



However, is the converse statement true? That is, if the above equation holds (and the two sets are non-empty), then is $f$ necessarily convex? Furthermore, would compactness of these sets be relevant?










share|cite|improve this question











$endgroup$




Let $f : mathbb R^n to mathbb R$ be a differentiable function for which a minimum exists. If $f$ is convex, then



$$x in mathbb R^n : nabla f(x) = 0 = x in mathbb R^n : f(x) leq f(y), ; forall y in mathbb R^n.$$



However, is the converse statement true? That is, if the above equation holds (and the two sets are non-empty), then is $f$ necessarily convex? Furthermore, would compactness of these sets be relevant?







convex-analysis convex-optimization maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 3 at 7:19









Rodrigo de Azevedo

13.3k41964




13.3k41964










asked May 2 at 0:53









Justin LeJustin Le

1378




1378







  • 2




    $begingroup$
    Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
    $endgroup$
    – Andreas Blass
    May 2 at 1:28










  • $begingroup$
    @AndreasBlass Thank you for the observation. I edited the question.
    $endgroup$
    – Justin Le
    May 2 at 1:33






  • 1




    $begingroup$
    To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
    $endgroup$
    – Andreas Blass
    May 2 at 1:42






  • 1




    $begingroup$
    This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
    $endgroup$
    – Theo Bendit
    May 2 at 2:29












  • 2




    $begingroup$
    Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
    $endgroup$
    – Andreas Blass
    May 2 at 1:28










  • $begingroup$
    @AndreasBlass Thank you for the observation. I edited the question.
    $endgroup$
    – Justin Le
    May 2 at 1:33






  • 1




    $begingroup$
    To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
    $endgroup$
    – Andreas Blass
    May 2 at 1:42






  • 1




    $begingroup$
    This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
    $endgroup$
    – Theo Bendit
    May 2 at 2:29







2




2




$begingroup$
Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
$endgroup$
– Andreas Blass
May 2 at 1:28




$begingroup$
Take $n=1$ and $f(x)=-e^x$. The sets in your question are equal because both are empty.
$endgroup$
– Andreas Blass
May 2 at 1:28












$begingroup$
@AndreasBlass Thank you for the observation. I edited the question.
$endgroup$
– Justin Le
May 2 at 1:33




$begingroup$
@AndreasBlass Thank you for the observation. I edited the question.
$endgroup$
– Justin Le
May 2 at 1:33




1




1




$begingroup$
To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
$endgroup$
– Andreas Blass
May 2 at 1:42




$begingroup$
To get the sets nonempty, try letting (still with $n=1$) $f(x)=x^2-1$ for $xleq1$ and $f(x)=2ln x$ for $xgeq 1$. (I hope I did the arithmetic right so that the values and derivatives match up at $x=1$ which makes $f$ differentiable. With more work, you could make an infinitely differentiable $f$ with the same general shape.) Both sets in your question are $0$, so they're equal, nonempty, and compact. But $f$ isn't convex to the right of $x=1$.
$endgroup$
– Andreas Blass
May 2 at 1:42




1




1




$begingroup$
This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
$endgroup$
– Theo Bendit
May 2 at 2:29




$begingroup$
This question reminds me of a paper by Saint Raymond. He shows that that in a general Banach space $X$, a lower-semicontinuous function $f$ is convex if, given any $l in X^*$, the function $f + l$ achieves its minimum on a non-empty, convex set.
$endgroup$
– Theo Bendit
May 2 at 2:29










1 Answer
1






active

oldest

votes


















9












$begingroup$

No, the converse is not correct. Here is a counterexample:



nonconvex counter-example



This function is smooth, nonconvex, yet it has a unique global minimiser which satisfies Fermat's condition.



By the way, a convex function does not necessarily satisfy the condition you mentioned - you need additional conditions. Take for example $f(x) = e^x$.



If a (convex or nonconvex) function $f:mathbbR^ntomathbbR$ is lower semicontinuous and level bounded, then $inf f$ is finite and its set of minimisers is nonempty and compact. A function $f$ is said to be level bounded if its level sets (the sets $xinmathbbR^n: f(x) leq a$) are bounded for every $ain mathbbR$ (they might be empty for some $a$).



Update: Another counter-example is the following function



$$
f(x) = fracxe^-2x + 1
$$



Its graph looks a little like the one above.






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    No, the converse is not correct. Here is a counterexample:



    nonconvex counter-example



    This function is smooth, nonconvex, yet it has a unique global minimiser which satisfies Fermat's condition.



    By the way, a convex function does not necessarily satisfy the condition you mentioned - you need additional conditions. Take for example $f(x) = e^x$.



    If a (convex or nonconvex) function $f:mathbbR^ntomathbbR$ is lower semicontinuous and level bounded, then $inf f$ is finite and its set of minimisers is nonempty and compact. A function $f$ is said to be level bounded if its level sets (the sets $xinmathbbR^n: f(x) leq a$) are bounded for every $ain mathbbR$ (they might be empty for some $a$).



    Update: Another counter-example is the following function



    $$
    f(x) = fracxe^-2x + 1
    $$



    Its graph looks a little like the one above.






    share|cite|improve this answer











    $endgroup$

















      9












      $begingroup$

      No, the converse is not correct. Here is a counterexample:



      nonconvex counter-example



      This function is smooth, nonconvex, yet it has a unique global minimiser which satisfies Fermat's condition.



      By the way, a convex function does not necessarily satisfy the condition you mentioned - you need additional conditions. Take for example $f(x) = e^x$.



      If a (convex or nonconvex) function $f:mathbbR^ntomathbbR$ is lower semicontinuous and level bounded, then $inf f$ is finite and its set of minimisers is nonempty and compact. A function $f$ is said to be level bounded if its level sets (the sets $xinmathbbR^n: f(x) leq a$) are bounded for every $ain mathbbR$ (they might be empty for some $a$).



      Update: Another counter-example is the following function



      $$
      f(x) = fracxe^-2x + 1
      $$



      Its graph looks a little like the one above.






      share|cite|improve this answer











      $endgroup$















        9












        9








        9





        $begingroup$

        No, the converse is not correct. Here is a counterexample:



        nonconvex counter-example



        This function is smooth, nonconvex, yet it has a unique global minimiser which satisfies Fermat's condition.



        By the way, a convex function does not necessarily satisfy the condition you mentioned - you need additional conditions. Take for example $f(x) = e^x$.



        If a (convex or nonconvex) function $f:mathbbR^ntomathbbR$ is lower semicontinuous and level bounded, then $inf f$ is finite and its set of minimisers is nonempty and compact. A function $f$ is said to be level bounded if its level sets (the sets $xinmathbbR^n: f(x) leq a$) are bounded for every $ain mathbbR$ (they might be empty for some $a$).



        Update: Another counter-example is the following function



        $$
        f(x) = fracxe^-2x + 1
        $$



        Its graph looks a little like the one above.






        share|cite|improve this answer











        $endgroup$



        No, the converse is not correct. Here is a counterexample:



        nonconvex counter-example



        This function is smooth, nonconvex, yet it has a unique global minimiser which satisfies Fermat's condition.



        By the way, a convex function does not necessarily satisfy the condition you mentioned - you need additional conditions. Take for example $f(x) = e^x$.



        If a (convex or nonconvex) function $f:mathbbR^ntomathbbR$ is lower semicontinuous and level bounded, then $inf f$ is finite and its set of minimisers is nonempty and compact. A function $f$ is said to be level bounded if its level sets (the sets $xinmathbbR^n: f(x) leq a$) are bounded for every $ain mathbbR$ (they might be empty for some $a$).



        Update: Another counter-example is the following function



        $$
        f(x) = fracxe^-2x + 1
        $$



        Its graph looks a little like the one above.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 2 at 2:01

























        answered May 2 at 1:37









        Pantelis SopasakisPantelis Sopasakis

        2,5221040




        2,5221040



























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