0
$\begingroup$

Suppose that for a function $h(x)$, there is a sequence $x_n$ such that $\lim_{n \to \infty}h(x_n)=h(c)=\min_x h(x)$. My understanding is $\lim_{ n\to \infty} x_n \neq c$ in general. But can somebody provide examples to verify the inequality?

$\endgroup$

2 Answers 2

2
$\begingroup$

Let $h(x)=x^2e^{-x^2}$, then $\min_{x\in\mathbb{R}} h(x)=0$. If we take $x_n=n$ we get $\lim h(x_n)=0=h(0)$, but $x_n\not\to 0$.

$\endgroup$
1
$\begingroup$

Simplest counterexample I could come up with: take $h\colon\mathbb{R}\to\mathbb{R}$ be the identically zero function, i.e., $h(x)=0$ for all $x$.

Take any $\ell\in\mathbb{R}$, and any sequence $(x_n)_n$ converging to $\ell$. Then $$ \lim_{n\to\infty} h(x_n) = h(\ell) = 0 = h(\ell+1) = \min_x h(x) $$ but clearly $\ell\neq \ell+1$.

$\endgroup$
1
  • $\begingroup$ You can also use any non-injective function. $\endgroup$
    – Clement C.
    Commented Jun 6, 2018 at 1:29

You must log in to answer this question.

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