# Minimum of $z:\mathbb{R}^n \to \mathbb{R}$ along paths implies local minimum of $z$

Suppose we are given a smooth function $$z: \mathbb{R}^n \to \mathbb{R}$$, a point $$x_0 \in \mathbb{R}^n$$ and a set $$\mathcal{F}$$ consisting of certain paths in $$\mathbb{R}^n$$, i.e. $$f: [0,1] \to \mathbb{R}^n$$ for $$f \in \mathcal{F}$$. Assume that for each $$f \in \mathcal{F}$$, there exists a $$t \in [0,1]$$ such that $$f(t) = x_0$$, and that the function $$z \circ f: [0,1] \to \mathbb{R}$$ has a local minimum in $$t$$. Under what conditions on $$\mathcal{F}$$ does it follow that $$z$$ has a local minimum at $$x_0$$?

Is is easily seen that it is enough if $$\mathcal{F}$$ consists of all continuous paths going through $$x_0$$. However, the set of straight lines going through $$x_0$$ is not enough (consider $$x_0 = (0,0)$$ and $$z(x,y) = (y^2 - 2x)(y^2 - x)$$, by Peano).

The standard argument to see that continuous paths are sufficient, is to assume not; i.e. there are points $$(x_k)_{k\in\mathbb{N}}$$ such that $$x_k \to x_0$$ and $$z(x_k) < z(x_0)$$. Now construct a continuous path along going through $$(x_k)$$ and you are done. So one may also consider which $$\mathcal{F}$$ are such that for each converging sequence of points $$(x_k)$$ one can find a subsequence $$(x_{k_l})$$ and an $$f \in \mathcal{F}$$ such that $$f$$ goes through $$(x_{k_l})$$.

Can we classify sets $$\mathcal{F}$$ for which the local minimality of $$z$$ holds? More specifically, I am interested in paths for which each component is twice differentiable.

• Have you considered permutations $\sigma$ of $\mathcal{F}$ commuting to $z$? – Sylvain JULIEN Aug 26 at 21:11
• Also your local minimality assumption on $z$ should mean there is some abstract derivative $\partial_{x_0}$ such that $\partial_{x_0}(z\circ f)=0\Rightarrow\partial_{x_0}(z)=0$. – Sylvain JULIEN Aug 26 at 21:36
• So writing $\Phi_{f}(z)$ for $z\circ f$, one should get $\Phi_{f}\circ\partial_{x_0}=\partial_{x_0}\circ\Phi_{f}$ and $\Phi_{f}(0)=0$. – Sylvain JULIEN Aug 26 at 21:44
• The equalities above holding replacing $f$ by $\sigma(f)$ for $\sigma$ any permutation of $\mathcal{F}$. – Sylvain JULIEN Aug 26 at 21:56

Answer: $$C^\infty$$ curves suffice for arbitrary functions $$z: \mathbb{R}^n \to \mathbb{R}$$.

Suppose we are given any function $$z: \mathbb{R}^n \to \mathbb{R}$$, a point $$x_0 \in \mathbb{R}^n$$ and let $$\mathcal{F}$$ consisting of $$C^\infty$$ (i.e. infinitely differentiable) $$f: [0,1] \to \mathbb{R}^n$$ such that $$f(0)=x_0$$ and all derivatives of $$f$$ vanish at $$x_0$$ (these functions are all restrictions of $$C^\infty$$ maps from $$\mathbb{R}$$ to $$\mathbb{R}^n$$.)

Claim: If for every $$f \in \mathcal{F}$$ the function $$z \circ f: [0,1] \to \mathbb{R}$$ has a local minimum at $$0$$, then $$z$$ has a local minimum at $$x_0$$.

Proof: We may assume that $$x_0=0$$. If $$z$$ does not have a local minimum at $$x_0=0$$, then as noted by the OP, there are points $$(x_k)_{k\in\mathbb{N}}$$ such that $$x_k \to 0$$ and $$z(x_k) < z(0)$$ for all $$k \ge 1$$. Passing to a subsequence, we may assume that the Euclidean norm $$| \cdot|$$ satisfies $$|x_k|<2^{-k}$$.

We will construct $$f \in \mathcal{F}$$ such that $$f(1/k)=x_k$$ for all $$k>1$$ and this will prove the claim.

Let $$\psi_0$$ be a $$C^\infty$$ bump function on $$[0,1]$$ such that $$\psi_0$$ and all its derivatives vanish at 0 and 1 yet $$\psi_0$$ is positive on $$(0,1)$$. e.g., $$\psi_0(t)=\exp(-1/[t(1-t)])$$. Define $$\psi_1(t)=\int_0^t \psi_0(s) ds$$ and $$\psi(t)=\psi_1(t)/\psi_1(1)$$.

Then $$\psi(0)=0$$ and $$\psi(1)=1$$ and all the derivatives of $$\psi$$ vanish at 0 and 1. For $$k \ge 2$$, write $$d_k= 1/(k-1)-1/k$$ and $$y_k=x_{k-1}-x_{k}$$. Define $$f: [0,1] \to \mathbb{R}^n$$ as follows:

First $$f(0)=0$$. Second, if $$k\ge 2$$ and $$1/k then $$f(t):=x_{k}+\psi((t-1/k)/d_k) y_k \,.$$ In particular $$f(1/k)=x_k$$ for $$k\ge 1$$. Clearly $$f$$ is $$C^\infty$$ in $$(0,1]$$. For $$1/k and $$j \ge 0$$, the $$j$$'th derivative of $$f$$ satisfies $$|f^{(j)}(t)| \le nC_j \delta_k^{-j} 2^{-k} \le nC_j2^j t^{-2j} 2^{-t/2} \, ,$$ where $$C_j$$ depend only on $$j$$. This implies that $$f$$ is differentiable at 0 and all its derivatives vanish at 0, so $$f \in \mathcal{F}$$.

• It would be interesting to get the result for a class $\mathcal{F}$ with quantified regularity (i.e. with a bound on the $C^k$ norm for some $k\ge 1$), but I would guess that this is not possible. – Benoît Kloeckner Aug 27 at 15:44
• Let me point out that if you select the sequence of $x_k$'s generically (in the sense of Baire category), with the requirement that $x_k$ is within $2^{-k}$ of $z$, then infinitely many of the $x_k$ will still have $f(x_k) < f(z)$, and this is enough to show that $z$ is not a local minimum. By combining this idea with some standard techniques from set theory, one could prove that it is consistent to have a family $\mathcal F$ as requested by the OP with $|\mathcal F| < \mathfrak{c}$. – Will Brian Aug 29 at 14:00