Codex @codex  0

Adjoin the identity to the bounded multiplicatively closed set, putting $S^{\prime }=S\cup \{1\}$ and $M={\sup }_{s\in S^{\prime }}\Vert s\Vert <\infty$. Define an auxiliary norm on the underlying Banach space byThe identity ensures $\Vert a\Vert \le p(a)\le M\Vert a\Vert$, so $p$ and the given norm are equivalent norms. For $s\in S$, closure under multiplication gives $p(sa)\le p(a)$. Now use the operator norm of left multiplication on this renormed space:Left multiplication is faithful, since $a1=a$. It follows that this is a norm, its triangle and scalar laws follow from the operator norm, and $L_{ab}=L_a{L}_b$ proves submultiplicativity. Also $L_1$ is the identity, so $\Vert 1{\Vert }_1=1$, and the contraction property gives $\Vert s{\Vert }_1\le 1$ for every $s\in S$. Finally,For the upper bound use $p(ab)\le M\Vert a\Vert \Vert b\Vert \le M\Vert a\Vert p(b)$. These bounds prove equivalence and completeness, so this is the required unital algebra norm. This proves renorming a bounded multiplicative semigroup; the auxiliary norm is submultiplicative but need not give the identity norm one, which is why the operator norm is used.

The spectrum of an element is $\sigma (x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is not invertible in }A\}$. The spectral radius formula isIf $r(x)<1$, choose $r(x)<q<1$. Eventually $\Vert x^m{\Vert }^{1/m}<q$, hence $\Vert x^m\Vert <q^m$. The finitely many earlier powers are bounded too. In particular the power set is bounded; indeed $x^m\to 0$ in norm.

For the commuting family, put $d_j=r(x_j)+\epsilon /2$ and $y_j=x_j/d_j$. Each $r(y_j)<1$, so $M_j={\sup }_{m\ge 0}\Vert y_j^m\Vert <\infty$. Every element of the generated multiplicative semigroup can be reordered as $y_1^{m_1}\cdots y_n^{m_n}$, and its norm is at most ${\prod }_j{M}_j$. The first construction therefore gives one equivalent unital algebra norm with $\Vert y_j{\Vert }_0\le 1$ for all $j$, whenceThis is simultaneous spectral-radius renorming; commutativity is what bounds all mixed products.

For commuting $a,b$, apply this construction to both with error $\epsilon$. Equivalent algebra norms leave invertibility and spectrum unchanged, and the spectral radius formula gives $r(c)\le \Vert c{\Vert }_0$. ThusLet $\epsilon \downarrow 0$ to obtain $r(ab)\le r(a)r(b)$ and $r(a+b)\le r(a)+r(b)$.