We denote the norm of $(V^{\vee})^{\vee}$ by $\|\raisebox{1.72218pt}{.}\|^{\prime}$, that is,

$$\|v\|^{\prime}=\sup\left\{\frac{|\phi(v)|}{\|\phi\|^{\vee}}\mid\phi\in V^{\vee}\setminus\{0\}\right\}.$$

Note that $|\phi(v)|\leq\|v\|\|\phi\|^{\vee}$ for all $v\in V$ and $\phi\in V^{\vee}$. In particular, $\|v\|^{\prime}\leq\|v\|$. For $v\in V\setminus\{0\}$, we set $W:=kv$ and choose $\psi\in W^{\vee}$ with $\psi(v)=1$. Then $\|\psi\|^{\vee}=1/\|v\|$. For any $\alpha\in(0,1)$, by Lemma 1.6, there is $\varphi\in V^{\vee}$ such that $\left.{\varphi}\right|_{{W}}=\psi$ and $\|\varphi\|^{\vee}\leq\alpha^{-1}\|\psi\|^{\vee}$. As $|\varphi(v)|/\|\varphi\|^{\vee}\leq\|v\|^{\prime}$, we have $\alpha\|v\|\leq\|v\|^{\prime}$. Thus we obtain $\|v\|\leq\|v\|^{\prime}$ by taking $\alpha\to 1$. ∎