Let $E$ be a TVS and let $(V_n)_{n \in \mathbb{N}}$ be a fundamental system of neighborhoods of $0$ in $E$ such that $\bigcap_{n \in \mathbb{N}} V_n=\{0\}$. Let us set $W_1=V_1$ and define by induction the sequence $(W_n)_{n \in \mathbb{N}}$ os balanced neighborhoods of $0$ which satisfy the relation $$W_{n+1}+W_{n+1}+W_{n+1} \subset V_n \cap W_n.$$ Let us define $\gamma$ on $E$ as follows:
- $\gamma(0)=0$
- $\gamma(x)=2^{-k}$ if $x \in W_k$ but $x \not\in W_{k+1}$
- $\gamma(x)=1$ if $x \not\in W_1$.
My question: How to prove that $\gamma(\lambda x) \leq \gamma(x)$ for all $x \in E$ and $|\lambda|\leq 1?$
If $\lambda=0$ then $\lambda x=0$ for all $x \in E$. Then, $\gamma(\lambda x)=\gamma(0)=0\leq \gamma(x)$ for all $x \in E$. If $|\lambda|=1$ then $\lambda W_n=W_n$ for all $n \in \mathbb{N}$ since $W_n$ is balanced. From the definition of $\gamma$ we easily concluded that $\gamma(\lambda x)=\gamma(x)$ for all $x \in E$ provided that $|\lambda|=1$.
How to prove this inequality in the case $0<|\lambda|<1$?
Suppose $x \in W_k\setminus W_{k+1} $ and $\lambda x \in W_j\setminus W_{j+1} $. Then $\gamma (x)=2^{-k}$ and $\gamma (\lambda x)=2^{-j}$. We have to show that $2^{-j} \leq 2^{-k}$ or $j \geq k$. Suppose $j <k$. Then $j+1 \leq k$. Now $\lambda x \in W_k$ because $W_k$ is balanced and $|\lambda | \leq 1$. Hence $\lambda x \in W_k \subseteq W_{j+1}$ which is a contradiction. I will let you finish the proof by considering the cases $x \notin W_1$ and $\lambda x \notin W_1$.