Prove $p(\alpha_n x_n)\to 0$ provided $\{\alpha_n\}_{n\in\mathbb{N}}$ converges and $p(x_n)\to 0$

Asked Dec 31 '19 at 02:47

Active Dec 31 '19 at 23:51

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Let $E$ be a vector space. A map $p : E \to \mathbb{R}$ satisfies :

$p(x+y)\leq p(x)+p(y)$ ;
For a fixed $x\in E$, $\lambda \mapsto p(\lambda x)$ is continuous ;
For all $\lambda\in\mathbb{R}$, if $p(x_n)\to 0$ as $n\to+\infty$, then $p(\lambda x_n)\to 0$ as $n\to +\infty$.

Prove that

$p(\mathbf{0})=0$, where $\mathbf{0}$ is the zero vector in $E$ ;
If $\{\alpha_n\}_{n\in\mathbb N}\subset \mathbb{R}$ converges, and $p(x_n)\to 0$ as $n\to\infty$, then $p(\alpha_n x_n)\to 0$ as $n\to\infty$.

I can answer the first question, but the second one is confused to me. Maybe there are some conditions lost, I have no idea. Looking forward to good ideas, thank you for your help!

edited Dec 31 '19 at 03:05

Sangchul Lee

167,468

asked Dec 31 '19 at 02:47

ling

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I changed the formatting a bit so that it reads a bit better. Hope this is to your liking. – Sangchul Lee Dec 31 '19 at 02:58
4

The first claim is false. Take the function $p :\equiv 1$. Note that it satisfies all the conditions and $p(\textbf{0}) \neq 0$. – acreativename Dec 31 '19 at 03:28

Prove $p(\alpha_n x_n)\to 0$ provided $\{\alpha_n\}_{n\in\mathbb{N}}$ converges and $p(x_n)\to 0$

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