This is a cross-post.

Let U⊆Rn be an open subset, and let f:U→R be smooth. Suppose that x∈U is a

strict local minimumpoint of f.Let dfk(x):(Rn)k→R be its k “derivative”, i.e. the symmetric multilinear map defined by setting

dfk(x)(ei1,…,eik)=∂i1…∂ikf(x).

Assume that dfj(x)≠0 forLet k be the minimal such that dfk(x)≠0. Since x is a local minimum, k must be even.somenatural j.

Suppose now that dfk(x) is non-degenerate, i.e. dfk(x)(h,…,h)≠0 for any non-zero h∈Rn. (Since x is a minimum, this is equivalent to dfk(x) being positive-definite, i.e. dfk(x)(h,…,h)>0 for any non-zero h∈Rn).

Question:Is f is strictly convex in some neighbourhood of x?In the one-dimensional case, when f is a map R→R, the answer is positive:

We have fk(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.

**Answer**

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.

**Attribution***Source : Link , Question Author : Asaf Shachar , Answer Author : Mateusz Kwaśnicki*