\lambda_i(x) v_i$ and $n\in \mathbb{N}$. Fix also $e=\sum_{i=1}^{n} v_i$ that is a tripotent in $\mathcal{U}$ of rank $n$. For every minimal tripotent $v$ in $\mathcal{U}_2(e)$ we have that $$\|x\|_{v}^2= \| P_2(v)\{x,x,v\|= \|P_2(v)\{\sum_{i=1}^{n} \lambda_i(x) v_i, \sum_{i=1}^{n} \lambda_i(x) v_i,v\} \| =$$ $$\sum_{i=1}^{n} \lambda_i(x)^2 \|P_2(v)\{v_i,v_i,v\}\|\geq  \lambda_n(x)^2 \sum_{i=1}^{n} \|v_i\|_{v}^2=\lambda_n(x)^2  \|\sum_{i=1}^{n} v_i\|_{v}^2=$$ $$\lambda_n(x)^2 \|e\|_{v}^2=\lambda_n(x)^2 \|P_2(v)\{e,e,v\}\|=\lambda_n(x)^2 \|v\|=\lambda_n(x)^2.$$ Thus $\displaystyle \lambda_n(x) \leq \max_e \inf \{\|x\|_{v}: v\in \mathcal{U}_2(e) \hbox{ minimal tripotent} \}  $.

The reverse inequality is a direct consequence of Lemma \[l innequality in minimax theorem\].

We will show now that $\displaystyle \lambda_n(x) =\min_f \sup\{ \|x\|_{v}: v\in \mathcal{U}_0(f) \hbox{ minimal tripotent}\}$ where $f$ runs over the set of tripotents in $\mathcal{U}$ of rank $n-1$. In case $n$ equals one, the result is trivial, since the only tripotent $f$ with rank zero is $f=0$ thus $\mathcal{U}_0(f)=\mathcal{U}$, and $\|x\|_{v}\leq \|x\|=\|x\|_{v_1}=\lambda_1(x)$ for every minimal tripotent $v$ in $\mathcal{U}$. Assume henceforth that $n\geq 2$. For $f=v_1+\ldots+v_{n-1}$ we have that, for every $v$ minimal tripotent in $\mathcal{U}_0(f)$, $$\|x\|_{v}=\|\sum_{i\geq n} \lambda_i(x) v_i \|_{v} \leq \|\sum_{i\geq n} \lambda_i(x) v_i \| = \|x\|_{v_n}=\lambda_n(x),$$ which shows $\displaystyle\lambda_n(x) \geq \min_f \sup\{ \|x\|_{v}: v\in \mathcal{U}_0(f) \hbox{ minimal tripotent}\}$.

On the other hand consider $f$ an arbitrary tripotent in $\mathcal{U}$ with $rank(f)=n-1$. We define $g=v_1+\ldots+v_n$ a tripotent in $\mathcal{U}$ 