A function $f:\mathbb{R}^n\to\mathbb{R}$ is called coercive if
$$
\lim_{\|x\|\to\infty}f(x)=\infty.
$$
To show that $f$ is coercive, we need to prove that for every sequence $\{x_n\}$ with $\|x_n\|\to\infty$, it holds that $f(x_n)\to\infty$.
My question: Is it sufficient to check that $f$ is coercive along every line? That is, does the condition
$$
\lim_{t\to\infty}f(x+td)=\infty,\quad\forall\,x\in\mathbb{R}^n,\quad\forall\,d\in\mathbb{R}^n\backslash\{0\}
$$
imply that $f$ is coercive? If not, what additional assumptions (e.g., convexity and/or $C^1$) can we impose on $f$ to make this true?
Best Answer
In general, this is not true. However, if we assume $f$ to be continuous and quasi-convex, then it is correct.
Suppose $f$ is not coercive, then there must exist an unbounded sublevel set $S_a=\{x:f(x)\leq a\}$ of $f$ (see this link). Since $f$ is quasi-convex, $S_a$ is convex. For an unbounded convex set, its recession cone $R_{S_a}$ must contain a nonzero element $d\ne 0$ (see Proposition 1.4.2 of this book). That's to say, there is $x\in S_a$ such that $x+td\in S_a$ for all $t\geq 0$. Recall that $f$ is coercive along every line, so $f(x+td)\to \infty$ as $t\to\infty$. However, that $x+td\in S_a$ implies $f(x+td)\leq a$ for all $t\geq 0$. This leads to a contradiction. Therefore, $f$ must be coercive.