Existence of a descent direction Let $d,k\in\mathbb N$ and $f:\mathbb R^d\to\mathbb R^k$. $x^\ast\in\mathbb R^d$ is called

*

*Pareto critical (for $f$) if there is no $x\in\mathbb R^d$ with $f(x)<f(x^\ast)$ (component-wisely);

*locally Pareto critical if there is a neighborhood $N$ of $x^\ast$ s.t. $x^\ast$ is Pareto critical for $\left.f\right|_N$.

Moreover, $v\in\mathbb R^d$ is called descent direction at $x\in\mathbb R^d$ if ${\rm D}f(x)v<0$ (component-wisely).

Question 1: How can we show that $x^\ast\in\mathbb R^d$ is locally Pareto critical if and only if there is no descent direction at $x^\ast$?

The "$\Rightarrow$" direction is easy to show, but I'm struggle to obtain "$\Leftarrow$", while the intuition is clear: Assuming the contrary, we could infer that $x^\ast$ is not Pareto critical for $\left.f\right|_{N_n}$, where $N_n:=B_{\frac1n}(x^\ast)$, for all $n\in\mathbb N$. Hence, there would be a $x_n\in N_n$ with $f(x_n)<f(x^\ast)$ for all $n\in\mathbb N$. We could now define $v_n:=x_n-x^\ast$ for $n\in\mathbb N$ and we should be able to find a $n_0\in\mathbb N$ such that ${\rm D}f(x^\ast)v_{n_0}<0$. But how do we conclude the existence of a descent direction rigorously?

Question 2: If $k=1$, then it's obvious that $x^\ast$ is (local) Pareto critical if and only if $x^\ast$ is a (local) minimum.
But now I've read that $x^\ast$ is locally Pareto critical if and only if $\nabla f(x^\ast)=0$. Why is that the case?

We clearly know that if $x^\ast$ is locally Pareto critical (hence a local minimum), then $\nabla f(x^\ast)=0$. But why should the other direction hold as well?

Please note that we can generally show that $x^\ast$ is Pareto critical if and only if there is a $\lambda\in[0,\infty)^k$ with $\langle\lambda,1\rangle=1$ and ${{\rm D}f(x^\ast)}^\ast\lambda=0$.
 A: Answer 1: (for $k\geq 2$)
It's obvious that $0\in\mathbb{R}^d$ can't be a descent direction. So we can substitute the hypersphere $S^{d-1}$ for $\mathbb{R}^d\setminus\{0\}$.
If there is no descent direction at $x^*$,
$$\forall v\in S^{d-1}, Df(x)v\geq 0\text{ for some component}\tag{1}$$
holds.
If there exists $v$ s.t. $Df(x)v=0$ (component-wisely), by your note at the bottom, $x^*$ is locally Pareto critical. (This note obviously doesn't hold when k=1, this is why we need $k\geq 2$.)
If such $v$ doesn't exist, we can constrcut the following map.
$$g:S^{d-1}\rightarrow\mathbb{R};v\mapsto \sup\{t\geq 0|f(x^*+\lambda v)\geq f(x^*) \text{ for some component},\forall \lambda \in [0,t]\}.$$
Because (1) holds for all $v$, and $f$ is continuous(given that it is derivable), we have $g(v)>0, \forall v$, and $g$ is a continuous function. Given that $S^{d-1}$ is a bounded closed set, i.e., a compact set in $\mathbb{R}^d$, $\delta:=\min\{g(v)|v\in S^{d-1}\}$ exists. Then we have $\delta>0$, and that $x^*$ is locally Pareto critical in $B_\delta(x^*)$.
Answer 2: I believe that the statement is wrong for $k=1$. $\nabla f(x^*)=0$ can't assure that $x^*$ is a local minimum, so it may not be locally Pareto critical.
Sorry for my poor English grammar...
