0
$\begingroup$

Suppose that $f\colon \mathbb{R} \to \mathbb{R}$ and $f(x)$ is coercive: $\lvert x \rvert \to \infty$ implies $f(x) \to \infty$. Let $x=(x_1, x_2)^\intercal$ and $g(x)=f(x_2)$. Is $g(x)$ coercive?

$\endgroup$
1
  • $\begingroup$ @saucy-opath: see the above definition $\endgroup$
    – jsmath
    Commented Jun 8, 2018 at 17:03

1 Answer 1

1
$\begingroup$

No, because $g\left(\begin{bmatrix}t\\ 0\end{bmatrix}\right)$ is constant, whereas it ought to diverge as $t\to\infty$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .