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

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.

• @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). Jul 29, 2020 at 17:22
• 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 $d^kf(v,v,\ldots, v, w,w) > 0$. Jul 29, 2020 at 17:29

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 $$0$$.
• 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 $d^6f$? Jul 29, 2020 at 16:40
• @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 $f(x_0, y_0)$. Jul 29, 2020 at 17:11
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).
• 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. Jul 29, 2020 at 14:33