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?
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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?
$begingroup$
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?
convex-analysis convex-optimization maxima-minima
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
$endgroup$
add a comment |
$begingroup$
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?
convex-analysis convex-optimization maxima-minima
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
$endgroup$
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
add a comment |
$begingroup$
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?
convex-analysis convex-optimization maxima-minima
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
$endgroup$
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
convex-analysis convex-optimization maxima-minima
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
edited May 3 at 7:19
Rodrigo de Azevedo
13.3k41964
13.3k41964
asked May 2 at 0:53
Justin LeJustin Le
1378
asked May 2 at 0:53
Justin LeJustin Le
1378
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, the converse is not correct. Here is a counterexample:
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.
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
$endgroup$
add a comment |
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$begingroup$
No, the converse is not correct. Here is a counterexample:
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.
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
$endgroup$
add a comment |
$begingroup$
No, the converse is not correct. Here is a counterexample:
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.
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
$endgroup$
add a comment |
$begingroup$
No, the converse is not correct. Here is a counterexample:
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.
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
$endgroup$
No, the converse is not correct. Here is a counterexample:
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.
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
edited May 2 at 2:01
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
answered May 2 at 1:37
Pantelis SopasakisPantelis Sopasakis
2,5221040
2,5221040
add a comment |
add a comment |
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XXI0DVQdvN I,jGxoO,QYt21cfsuj tCjElVoF,7,mYsfZ
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