# Is a function of several variables convex near a local minimum when the derivatives are non-degenerate?

This is a cross-post.

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

Let $$dfk(x):(Rn)k→Rdf^k(x):(\mathbb R^n)^k \to \mathbb R$$ be its $$kk$$ “derivative”, i.e. the symmetric multilinear map defined by setting
$$dfk(x)(ei1,…,eik)=∂i1…∂ikf(x)df^k(x)(e_{i_1},\dots,e_{i_k})=\partial_{i_1} \dots \partial_{i_k}f(x)$$.

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

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

Question: Is $$ff$$ is strictly convex in some neighbourhood of $$xx$$?

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

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

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

Perhaps we need to understand the Taylor’s expansion of $$\text{Hess}f\text{Hess}f$$ around $$xx$$, 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 $$df^k(x)df^k(x)$$ is non-zero. Indeed, consider $$f(x,y) = x^2 y^2 + x^8 + y^8 f(x,y) = x^2 y^2 + x^8 + y^8$$.

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

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

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

\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}\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 $$ff$$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $$44$$, and hence $$d^j f(\vec 0) = 0d^j f(\vec 0) = 0$$ for $$j < 4j < 4$$ and $$d^4 f(\vec 0) > 0d^4 f(\vec 0) > 0$$ (indeed: $$d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0$$ whenever $$\vec h \ne \vec 0\vec h \ne \vec 0$$). On the other hand, $$\partial_{xx} f(0,y) = -2 y^2 < 0\partial_{xx} f(0,y) = -2 y^2 < 0$$ whenever $$y \ne 0y \ne 0$$, and so $$ff$$ is not convex near $$00$$.