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?


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.

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    $\begingroup$ @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
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    $\begingroup$ 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


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 $0$.

  • $\begingroup$ 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

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

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    $\begingroup$ 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

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