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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, I think 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?

Edit:

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$. (I thank Robert Israel for this nice example).

$f$ 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 not non-degenerate, however, since if $h=h^1e_1+h^2e_2$ then $df^4(0)(h,h,h,h)=4(h^1)^2(h^2)^2$ vanishes when either $h_i$ is zero.

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

Asaf Shachar
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    I think you could show this works for a strongly convex function, since the "strictness" of a strongly convex function has a tangible magnitude, which you can argue will eventually dominate the Taylor remainder for a small enough neighborhood. There is some interesting interplay between strict convexity and strong convexity; you may be able to use the fact that $f$ is differentiable to suss out local strong convexity from its strict convexity, but I'm not positive. – Zim Jul 20 '20 at 18:01
  • Hi, nice to hear from you again:) Are you saying that for a strongly convex function, the first non-zero derivative at a point of local minimum should be non-degenerate? – Asaf Shachar Jul 20 '20 at 18:47
  • You too! That was not exactly what I meant (although if $f$ is $\beta$-strongly convex, then for every $(x,y)$ in its domain, $\langle x-y, |, \nabla f(x)-\nabla f(y)\rangle\geq\beta|x-y|^2$). Actually I think I misread the question, oops! – Zim Jul 20 '20 at 19:01
  • actually I might miss something, but I don't think it's true (even in the) one-dimensional case. See this counterexample to a slightly more general question https://math.stackexchange.com/questions/3303409/find-monotonic-interval-for-real-c-infty-function. – Targon Jul 21 '20 at 12:14
  • @Targon I don't think this is related. The one-dimensional case is certainly true here. This other question is interesting, but I don't think it is directly connected. Indeed, here I assume that some derivative do not vanish. To get the 'pathological' counterexamples, you need all the derivatives to vanish. You may see here: https://math.stackexchange.com/questions/3677358/is-a-smooth-function-convex-near-a-strict-minimum – Asaf Shachar Jul 22 '20 at 12:57
  • Maybe in Robert Israel’s example $x^8+y^8$ can be replaced by an other high-degree polynomial, which provides a counterexample with non-degenerate $f^4(0)$ (or $f^6(0)$)? – Alex Ravsky Jul 22 '20 at 17:02
  • I guess in the Taylor expansion of $f′′$ near $x$ should be $O((y - x)^{k-3})$ instead of $O((y - x)^{k-1})$. – Alex Ravsky Jul 29 '20 at 11:41
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    The answer is still "No". Take $(x^2-y^2)^2+\varepsilon(x^{4}+y^{4})$. The first part vanishes on 2 lines but not between them, so it cannot be convex, and the second part does not matter if $\varepsilon$ is small enough. – fedja Jul 29 '20 at 14:03
  • @fedja Thanks, that's a nice example! If you want you can turn this into an answer, or I can do that (as a wiki...). – Asaf Shachar Jul 29 '20 at 14:19

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