7
$\begingroup$

This is a cross-post.

Let $U \subseteq \mathbb R^n$ be an open subset, and let $f:U \to \mathbb R$ be smooth. Suppose that $x \in U$ is a strict local minimum point of $f$.

Let $df^k(x):(\mathbb R^n)^k \to \mathbb R$ be its $k$ "derivative", i.e. the symmetric multilinear map defined by setting $df^k(x)(e_{i_1},\dots,e_{i_k})=\partial_{i_1} \dots \partial_{i_k}f(x)$.

Assume that $df^j(x) \neq 0$ for some natural $j$. Let $k$ be the minimal such that $df^k(x) \neq 0$. Since $x$ is a local minimum, $k$ must be even.

Suppose now that $df^k(x)$ is non-degenerate, i.e. $df^k(x)(h,\dots,h) \neq 0$ for any non-zero $h \in \mathbb R^n$. (Since $x$ is a minimum, this is equivalent to $df^k(x)$ being positive-definite, i.e. $df^k(x)(h,\dots,h) > 0$ for any non-zero $h \in \mathbb R^n$).

Question: Is $f$ is strictly convex in some neighbourhood of $x$?

In the one-dimensional case, when $f$ is a map $\mathbb R \to \mathbb R$, the answer is positive:

We have $f^k(x)>0$, and the Taylor expansion of $f''$ near $x$ is $$ f''(y) = {1 \over (k-2)!} f^{(k)}(x)(y - x)^{k-2} + O((y - x)^{k-1}). $$ Thus, $f''(y)>0$ for $y \ne x$ sufficiently close to $x$, so $f$ is strictly convex around $x$.


Returning back to the high-dimensional case, if $k>2$, we have $\text{Hess}f(x)=df^2(x)=0$, and I guess that we should somehow prove that $\text{Hess}f(y)$ becomes positive-definite for $y$ sufficiently close to $x$.

Perhaps we need to understand the Taylor's expansion of $\text{Hess}f$ around $x$, similarly to the one-dimensional case, but I am not sure how to do that.

Is there a nice way?


Comment:

It is certainly not enough to assume that $df^k(x)$ is non-zero. Indeed, consider $ f(x,y) = x^2 y^2 + x^8 + y^8$.

$f$ has a strict global minimum at $(0,0)$.
$$\det(\text{Hess}f(x,y))=3136 x^6 y^6 + 112 x^8 + 112 y^8 - 12 x^2 y^2,$$ which is negative when $x=y$ is small and nonzero. Thus, $f$ is not convex at a neighbourhood of zero.

Note that $\text{Hess}f(0,0)=0$; The first non-zero derivative at $(0,0)$ is the fourth-order derivative $df^4(0)$. It is degenerate, however, since $df^4(0)(h^1e_1+h^2e_2,h^1e_1+h^2e_2)=4(h^1)^2(h^2)^2$ vanishes when either $h_i$ is zero.

So, non-vanishing of some derivatives does not ensure convexity.

$\endgroup$
2
  • 1
    $\begingroup$ @Mateusz already gave you a counter example, but I think that $d^kf > 0$ is definitely too weak. Morally you want $Hess(f)$ to be positive semi definite in a neighborhood, and this suggests that since you are in the case $Hess(f)(0) = 0$, you want a suitable number of higher derivatives of the matrix valued function $Hess(f)$ to take values in the symmetric positive semidefinite matrices. Assuming $d^3f > 0$ does not control $\partial_x \partial^2_{yy} f$ (which is essentially what Mateusz used in his example). $\endgroup$ Jul 29, 2020 at 17:22
  • 1
    $\begingroup$ So something like $d^k Hess(f)$ being positive would probably work. (So if $k = 4$ is the smallest $k$ for which $d^k f \neq 0$ you want $\partial_v \partial_v Hess(f)$ to be a PD matrix.) This can be written as for any $v, w\neq 0$ that $d^kf(v,v,\ldots, v, w,w) > 0$. $\endgroup$ Jul 29, 2020 at 17:29

2 Answers 2

12
$\begingroup$

Let $$\begin{aligned} f(x,y) & = x^4 - x^2 y^2 + y^4 \\ & = \tfrac{1}{2} x^4 + \tfrac{1}{2} y^4 + \tfrac{1}{2} (x^2 - y^2)^2 . \end{aligned}$$ Then $f$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $4$, and hence $d^j f(\vec 0) = 0$ for $j < 4$ and $d^4 f(\vec 0) > 0$ (indeed: $d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0$ whenever $\vec h \ne \vec 0$). On the other hand, $$\partial_{xx} f(0,y) = -2 y^2 < 0$$ whenever $y \ne 0$, and so $f$ is not convex near $0$.

$\endgroup$
3
  • $\begingroup$ Thanks. Just one question: Why does the fact that $f$ is a homogeneous polynomial of degree $6$ imply that $d^6f$ is non-degenerate? Can we deduce that without explicitly computing $d^6f$? $\endgroup$ Jul 29, 2020 at 16:40
  • $\begingroup$ @AsafShachar: if $f$ is a homogeneous polynomial, it is equal to its taylor polynomial of that degree, and hence if $v = (x_0, y_0)$ up to some numerical constant $d^6f(v,v,\ldots, v)$ is the same as $f(x_0, y_0)$. $\endgroup$ Jul 29, 2020 at 17:11
  • $\begingroup$ @WillieWong and Asaf Schachar: Thanks! I edited this into the answer (and also reduced the degree from 6 to 4). $\endgroup$ Jul 29, 2020 at 21:20
5
$\begingroup$

Let $n=1$, $f(t)=t^2 + |t|^{7/2}\sin(1/|t|)$ for $t\ne0$, $f(0):=0$. Then $f'(0)=0$ and $f''(0)=2>0$, so that $0$ is a strict local minimum of $f$. However, $f''(t)\sim-|t|^{-1/2}\sin(1/|t|)$ as $t\to0$, and so, $f$ is not convex (let alone strictly convex) in any neighborhood of $0$.


Here are the graphs $\{(t,f(t))\colon|t|<0.1\}$ (left) and $\{(t,f''(t))\colon|t|<0.1\}$ (right).

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ Thanks, this is a nice example. However, I assumed that $f$ is smooth. I think you should keep this answer, to show what can happen without smoothness. $\endgroup$ Jul 29, 2020 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.