Question

(Lael and Nourouzi 2008) Let \((V, \mu, \nu)\) be an \(I F\)-normed space. Assume further that \(\mu(x, t)>0\) for all \(t>0\) implies \(x=0\). Define
\[
\|x\|_\alpha=\inf \{t>0: \mu(x, t)>\alpha \text { and } v(x, t)
\]
where \(\alpha \in(0,1)\). Then \(\left\{\|x\|_\alpha: \alpha \in(0,1)\right\}\) is an ascending family of norms on \(V\).

Answer 1

To show that \(\{\|x\|_\alpha: \alpha \in (0,1)\}\) is an ascending family of norms, we need to show that for any \(0\lt \alpha_1\lt;\alpha_2\lt 1\) and any \(x \in V\), we have \(\|x\|_{\alpha_1} \leq \|x\|_{\alpha_2}\).Let \(x \in V\) be arbitrary. By definition, we have
\[
\|x\|_{\alpha_1} = \inf\{t \gt 0: \mu(x,t) \gt \alpha_1 \text{ and } \nu(x,t) \leq 1\}
\]
and
\[
\|x\|_{\alpha_2} = \inf\{t \gt;0: \mu(x,t) \gt \alpha_2 \text{ and } \nu(x,t) \leq 1\}.
\]
Since \(\alpha_1 \lt \alpha_2\), we have
\[
\{t>0: \mu(x,t) \gt \alpha_1 \text{ and } \nu(x,t) \leq 1