We prove it by induction on $\dim_{k}V$. If $\dim_{k}V=1$, then the assertion is obvious. By the hypothesis of induction, there is a $\sqrt{\alpha}$-orthogonal basis $(e_{1},\ldots,e_{r-1})$ of $V^{\prime}:=ke^{\prime}_{1}+\cdots+ke^{\prime}_{r-1}$ with respect to $\|\raisebox{1.72218pt}{.}\|$ such that

$$ke_{1}+\cdots+ke_{i}=ke^{\prime}_{1}+\cdots+ke^{\prime}_{i}$$

for $i=1,\ldots,r-1$. Choose $v\in V\setminus V^{\prime}$. As

$$\mathrm{dist}(v,V^{\prime}):=\inf\{\|v-x\|:x\in V^{\prime}\}>0,$$

there is $y\in V^{\prime}$ such that $\|v-y\|\leq(\sqrt{\alpha})^{-1}\mathrm{dist}(v,V^{\prime})$. We set $e_{r}=v-y$. Clearly $(e_{1},\ldots,e_{r-1},e_{r})$ forms a basis of $V$. It is sufficient to see that

$$\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|\geq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|,\|e_{r}\|\}$$

for all $a_{1},\ldots,a_{r-1}\in k$. Indeed, as $\|e_{r}\|\leq(\sqrt{\alpha})^{-1}\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$, we have

$$\alpha\|e_{r}\|\leq\sqrt{\alpha}\|e_{r}\|\leq\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|.$$

If $\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|\leq\|e_{r}\|$, then

	$\displaystyle\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$	$\displaystyle\geq\sqrt{\alpha}\|e_{r}\|\geq\sqrt{\alpha}\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|$
		$\displaystyle\geq\sqrt{\alpha}\left(\sqrt{\alpha}\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}\right)$
		$\displaystyle=\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}.$

Otherwise,

	$\displaystyle\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$	$\displaystyle=\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|$
		$\displaystyle\geq\sqrt{\alpha}\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}$
		$\displaystyle\geq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\},$

as required.

For the second assertion, it is sufficient to show the following lemma because it implies that the set $\{\|v-x\|\mid x\in V^{\prime}\}$ has the minimal value. ∎

Let us consider a map $\beta:V\setminus\{0\}\to{\mathbb{R}}_{>0}/|k^{\times}|$ given by

$$\beta(v)=\text{the class of $\|v\|$ in ${\mathbb{R}}_{>0}/|k^{\times}|$}.$$

It is sufficient to see that $\beta(V\setminus\{0\})$ is finite. Let $\beta_{1},\ldots,\beta_{l}$ be distinct elements of $\beta(V\setminus\{0\})$. We choose $v_{1},\ldots,v_{l}\in V\setminus\{0\}$ with $\beta(v_{i})=\beta_{i}$ for $i=1,\ldots,l$. If $i\not=j$, then $\|a_{i}v_{i}\|\not=\|a_{j}v_{j}\|$ for all $a_{i},a_{j}\in k^{\times}$. Therefore, we obtain

$$\|a_{1}v_{1}+\cdots+a_{l}v_{l}\|=\max\{\|a_{1}v_{1}\|,\ldots,\|a_{1}v_{l}\|\}$$

for all $a_{1},\ldots,a_{l}\in k$. In particular, $v_{1},\ldots,v_{l}$ are linearly independent. Therefore, we have $\#(\beta(V\setminus\{0\}))\leq\dim_{k}V$. ∎

Fix $\alpha\in(0,1)$. Let $(e_{1},\ldots,e_{r})$ be an $\alpha$-orthogonal basis of $V$ (cf. Proposition 1.3). We set

$$C_{1}=\max\{\|\phi(e_{1})\|^{\prime},\ldots,\|\phi(e_{r})\|^{\prime}\}\quad\text{and}\quad C_{2}=\min\{\|e_{1}\|,\ldots,\|e_{r}\|\}.$$

Then, for $v=a_{1}e_{1}+\cdots+a_{r}e_{r}\in V\setminus\{0\}$,

	$\displaystyle\frac{\|\phi(v)\|^{\prime}}{\|v\|}$	$\displaystyle\leq\frac{\max\{|a_{1}|\|\phi(e_{1})\|^{\prime},\ldots,|a_{r}|\|\phi(e_{r})\|^{\prime}\}}{\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r}|\|e_{r}\|\}}$
		$\displaystyle\leq\frac{\max\{|a_{1}|C_{1},\ldots,|a_{r}|C_{1}\}}{\alpha\max\{|a_{1}|C_{2},\ldots,|a_{r}|C_{2}\}}=\frac{C_{1}}{\alpha C_{2}},$

as desired. ∎

Let $(e_{1},\ldots,e_{r})$ be an $\alpha$-orthogonal basis of $V$ such that $W=ke_{1}+\cdots+ke_{l}$ (cf. Proposition 1.3). We define $\varphi\in V^{\vee}$ to be

$$\varphi(a_{1}e_{1}+\cdots+a_{r}e_{r}):=\psi(a_{1}e_{1}+\cdots+a_{l}e_{l})$$

for $a_{1},\ldots,a_{r}\in k$. Then $\left.{\varphi}\right|_{{W}}=\psi$. Moreover, note that

$$\alpha\|a_{1}e_{1}+\cdots+a_{l}e_{l}\|\leq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{l}|\|e_{l}\|\}\\ \leq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r}|\|e_{r}\|\}\leq\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|,$$

so that

$$\frac{|\varphi(a_{1}e_{1}+\cdots+a_{r}e_{r})|}{\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|}\leq\alpha^{-1}\frac{|\psi(a_{1}e_{1}+\cdots+a_{l}e_{l})|}{\|a_{1}e_{1}+\cdots+a_{l}e_{l}\|}\leq\alpha^{-1}\|\psi\|^{\vee}$$

for all $a_{1},\ldots,a_{r}\in k$ with $(a_{1},\ldots,a_{l})\not=(0,\ldots,0)$. Thus the assertion follows. ∎

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 $\alpha\to 1$. ∎

Let $(e_{1}^{\vee},\ldots,e_{r}^{\vee})$ be the dual basis of $(e_{1},\ldots,e_{r})$. For $a_{1},\ldots,a_{r}\in k$ with $a_{i}\not=0$,

$$\frac{|(e_{i}^{\vee})(a_{1}e_{1}+\cdots+a_{r}e_{r})|}{\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|}\leq\frac{|a_{i}|}{\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r}|\|e_{r}\|\}}\leq\frac{|a_{i}|}{\alpha|a_{i}|\|e_{i}\|}=\frac{1}{\alpha\|e_{i}\|},$$

and hence $\|e_{i}^{\vee}\|^{\vee}\leq(\alpha\|e_{i}\|)^{-1}$. Therefore, for $a^{\prime}_{1},\ldots,a^{\prime}_{r}\in k^{\prime}$,

	$\displaystyle\|a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}\|$	$\displaystyle\geq\frac{|(e_{i}^{\vee}\otimes 1)(a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r})|^{\prime}}{\|e_{i}^{\vee}\|^{\vee}}$
		$\displaystyle=\frac{|a^{\prime}_{i}|^{\prime}}{\|e_{i}^{\vee}\|^{\vee}}\geq\frac{|a^{\prime}_{i}|^{\prime}}{(\alpha\|e_{i}\|)^{-1}}=\alpha|a^{\prime}_{i}|^{\prime}\|e_{i}\|.$

Thus we have the assertion. ∎

For $\epsilon>0$, let $(e_{1},\ldots,e_{r})$ be an $e^{-\epsilon}$-orthogonal basis of $V$ with respect to $\|\raisebox{1.72218pt}{.}\|$. Then, by Proposition 1.9, $(e_{1},\ldots,e_{r})$ forms an $e^{-\epsilon}$-orthogonal basis of $V_{k^{\prime}}$ and $V_{k^{\prime\prime}}$ with respect to $\|\raisebox{1.72218pt}{.}\|_{k^{\prime}}$ and $\|\raisebox{1.72218pt}{.}\|_{k^{\prime\prime}}$, respectively, so that $(e_{1},\ldots,e_{r})$ is also an $e^{-\epsilon}$-orthogonal basis of $V_{k^{\prime\prime}}$ with respect to $\|\raisebox{1.72218pt}{.}\|_{k^{\prime},k^{\prime\prime}}$. Note that $\|e_{i}\|=\|e_{i}\|_{k^{\prime\prime}}=\|e_{i}\|_{k^{\prime},k^{\prime\prime}}$ for all $i=1,\ldots,r$. Thus, for $a^{\prime\prime}_{1},\ldots,a^{\prime\prime}_{r}\in k^{\prime\prime}$,

$$\|a^{\prime\prime}_{1}e_{1}+\ldots+a^{\prime\prime}_{r}e_{r}\|_{k^{\prime},k^{\prime\prime}}\leq\max\{|a^{\prime\prime}_{1}|^{\prime\prime}\|e_{1}\|,\ldots,|a^{\prime\prime}_{r}|^{\prime\prime}\|e_{r}\|\}\\ \leq e^{\epsilon}\|a^{\prime\prime}_{1}e_{1}+\ldots+a^{\prime\prime}_{r}e_{r}\|_{k^{\prime\prime}}$$

and

$$\|a^{\prime\prime}_{1}e_{1}+\ldots+a^{\prime\prime}_{r}e_{r}\|_{k^{\prime\prime}}\leq\max\{|a^{\prime\prime}_{1}|^{\prime\prime}\|e_{1}\|,\ldots,|a^{\prime\prime}_{r}|^{\prime\prime}\|e_{r}\|\}\\ \leq e^{\epsilon}\|a^{\prime\prime}_{1}e_{1}+\ldots+a^{\prime\prime}_{r}e_{r}\|_{k^{\prime},k^{\prime\prime}}.$$

Thus, we have the assertion by taking $\epsilon\to 0$. ∎

Let $\|\raisebox{1.72218pt}{.}\|^{\prime}_{W_{k^{\prime}}}$ be the quotient norm of $\|\raisebox{1.72218pt}{.}\|_{V,k^{\prime}}$ with respect to the surjection $f_{k^{\prime}}:V_{k^{\prime}}\to W_{k^{\prime}}$. Let $e$ be an non-zero element of $W$. As $\|e\|_{W,k^{\prime}}=\|e\|_{W}$, it is sufficient to show that $\|e\|^{\prime}_{W_{k^{\prime}}}=\|e\|_{W}$. Note that

$$\{v\in V\mid f(v)=e\}\subseteq\{v^{\prime}\in V_{k^{\prime}}\mid f_{k^{\prime}}(v^{\prime})=e\},$$

so that we have $\|e\|_{W}\geq\|e\|^{\prime}_{W_{k^{\prime}}}$. Let us consider an inequality $\|e\|_{W}\leq\|e\|^{\prime}_{W_{k^{\prime}}}$. For $\epsilon>0$, let $(e_{1},\ldots,e_{r})$ be an $e^{-\epsilon}$-orthogonal basis of $V$ such that $(e_{2},\ldots,e_{r})$ forms a basis of $\operatorname{Ker}(f)$. Clearly we may assume that $f(e_{1})=e$. Then

	$\displaystyle\|e\|^{\prime}_{W_{k^{\prime}}}$	$\displaystyle=\inf\{\|e_{1}+a^{\prime}_{2}e_{2}+\cdots+a^{\prime}_{r}e_{r}\|_{V,k^{\prime}}\mid a^{\prime}_{2},\ldots,a^{\prime}_{r}\in k^{\prime}\}$
		$\displaystyle\geq\inf\{e^{-\epsilon}\max\{\|e_{1}\|,|a^{\prime}_{2}|^{\prime}\|e_{2}\|_{V},\ldots,|a^{\prime}_{r}|^{\prime}\|e_{r}\|_{V}\}\mid a^{\prime}_{2},\ldots,a^{\prime}_{r}\in k^{\prime}\}$
		$\displaystyle\geq e^{-\epsilon}\|e_{1}\|\geq e^{-\epsilon}\|e\|_{W}.$

Therefore, we have $\|e\|^{\prime}_{W_{k^{\prime}}}\geq\|e\|_{W}$ by taking $\epsilon\to 0$. ∎

(1) Let $(e_{1},\ldots,e_{r})$ be an orthogonal basis of $(V,\|\raisebox{1.72218pt}{.}\|)$ (cf. Proposition 1.3). Then

$$\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|=\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r}|\|e_{r}\|\}$$

for all $a_{1},\ldots,a_{r}\in k$, so that

$$\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|\in\{0,\|e_{1}\|,\ldots,\|e_{r}\|\}.$$

(2) First we assume that

$$\|e_{1}\|=\cdots=\|e_{r}\|=c.$$

Then, for any $v\in V$,

$$\|v\|=\begin{cases}c&\text{if $v\not=0$},\\ 0&\text{if $v=0$}.\end{cases}$$

Let us see that

$$\|a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}\|^{\prime}=c\max\{|a^{\prime}_{1}|^{\prime},\ldots,|a^{\prime}_{r}|^{\prime}\}$$

for $a^{\prime}_{1},\ldots,a^{\prime}_{r}\in k^{\prime}$. Clearly we may assume that

$$(a^{\prime}_{1},\ldots,a^{\prime}_{r})\not=(0,\ldots,0).$$

We set $\gamma:=\max\{|a^{\prime}_{1}|^{\prime},\ldots,|a^{\prime}_{r}|^{\prime}\}$. We fix $\omega\in k^{\prime}$ with $|\omega|^{\prime}=\gamma$. By the assumption (i), for each $j=1,\ldots,r$, we can find $a_{j}\in k$ and $b^{\prime}_{j}\in k^{\prime}$ such that

$$a^{\prime}_{j}=a_{j}\omega+b^{\prime}_{j}\quad\text{and}\quad|b^{\prime}_{j}|^{\prime}<\gamma.$$

Note that

$$a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}=\omega\left(\sum\nolimits_{j=1}^{r}a_{j}e_{j}\right)+b^{\prime}_{1}e_{1}+\cdots+b^{\prime}_{r}e_{r}.$$

Moreover, as $\sum_{j=1}^{r}a_{j}e_{j}\not=0$, we have

	$\displaystyle\left\|\omega\left(\sum\nolimits_{j=1}^{r}a_{j}e_{j}\right)\right\|^{\prime}$	$\displaystyle=\gamma\left\|\sum\nolimits_{j=1}^{r}a_{j}e_{j}\right\|=c\gamma$
and
	$\displaystyle\|b^{\prime}_{1}e_{1}+\cdots+b^{\prime}_{r}e_{r}\|^{\prime}$	$\displaystyle\leq c\max\{|b^{\prime}_{1}|^{\prime},\ldots,|b^{\prime}_{r}|^{\prime}\}<c\gamma.$

Therefore,

$$\|a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}\|^{\prime}=c\gamma=c\max\{|a^{\prime}_{1}|^{\prime},\ldots,|a^{\prime}_{r}|^{\prime}\}.$$

In general, we take positive numbers $c_{1}<\cdots<c_{b}$ and non-empty subsets $I_{1},\ldots,I_{b}$ of $\{1,\ldots,r\}$ such that $\{\|e_{l}\|\mid l\in I_{s}\}=\{c_{s}\}$ for $s=1,\ldots,b$ and $I_{1}\cup\cdots\cup I_{b}=\{1,\ldots,r\}$. Note that $I_{s}\cap I_{s^{\prime}}=\emptyset$ for $s\not=s^{\prime}$. Let us consider

$$x=a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}=\sum_{s=1}^{b}x_{s}\in V_{k^{\prime}}\quad(a^{\prime}_{1},\ldots,a^{\prime}_{r}\in k^{\prime}),$$

where $x_{s}=\sum_{l\in I_{s}}a^{\prime}_{l}e_{l}$. Note that $(e_{l})_{l\in I_{s}}$ forms an orthogonal basis of $\bigoplus_{l\in I_{s}}ke_{l}$ and $\|e_{l}\|=c_{s}$ for all $l\in I_{s}$. Therefore, by the above observation,

$$\left\|x_{s}\right\|^{\prime}=c_{s}\max_{l\in I_{s}}\{|a^{\prime}_{l}|^{\prime}\}=\max_{l\in I_{s}}\{\|a^{\prime}_{l}e_{l}\|^{\prime}\},$$

so that it is sufficient to see that

$$\|x\|^{\prime}=\max_{s=1,\ldots,b}\left\{\left\|x_{s}\right\|^{\prime}\right\}.$$

Clearly we may assume that $x\not=0$. We set

$$\Sigma:=\left\{s\in\{1,\ldots,b\}\mid x_{s}\not=0\right\}.$$

For $s,s^{\prime}\in\Sigma$ with $s\not=s^{\prime}$, we have $\|x_{s}\|^{\prime}\not=\|x_{s^{\prime}}\|^{\prime}$. Indeed, we choose $l_{s}\in I_{s}$ and $l_{s^{\prime}}\in I_{s^{\prime}}$ with $\left\|x_{s}\right\|^{\prime}=\|a^{\prime}_{l_{s}}e_{l_{s}}\|^{\prime}$ and $\left\|x_{s^{\prime}}\right\|^{\prime}=\|a^{\prime}_{l_{s^{\prime}}}e_{l_{s^{\prime}}}\|^{\prime}$. If $\|x_{s}\|^{\prime}=\|x_{s^{\prime}}\|^{\prime}$, then

$$\left|a^{\prime}_{l_{s}}/a^{\prime}_{l_{s^{\prime}}}\right|^{\prime}=\|e_{l_{s^{\prime}}}\|/\|e_{l_{s}}\|,$$

so that, by the assumption (ii), $\|e_{l_{s^{\prime}}}\|=\|e_{l_{s}}\|$, which is a contradiction. Therefore,

$$\|x\|^{\prime}=\left\|\sum\nolimits_{s\in\Sigma}x_{s}\right\|^{\prime}=\max_{s\in\Sigma}\{\|x_{s}\|^{\prime}\}=\max_{s=1,\ldots,b}\{\|x_{s}\|^{\prime}\},$$

as required. ∎