let A be a banach algebra with identity and $a \in A$.
the spectral radius $a$: $ r(a)= sup \{ \lambda : \lambda \in \sigma(a)\}$
a) if $A$ is a abelian banach algebra, $a,b \in A$, will below terms be correct? $r(ab) \leq r(a) r(b) \\ r(a+b) \leq r(a) +r(b)$
b) say that the map $ r:A\longrightarrow\mathbb{R}\\a\mapsto r(a) $ is upper semicountinuous?
On
$\newcommand{fs}[1]{\lVert #1\rVert}$ Without loss of generality, suppose $A$ is a unital abelian Banach algebra (that is $\lVert 1\rVert=1$). Theorem of Gelfand Representation states that $r(a)=\lVert \hat{a}\rVert_{\infty}$(see Theorem 1.3.6 in Murphy's book $C^*$-algebra and Operator Theory), Hence, \begin{align*} &r(a+b)=\lVert\widehat{a+b}\rVert_\infty=\fs{\hat{a}+\hat{b}}_\infty\leq\fs{\hat{a}}_\infty+\fs{\hat{b}}_\infty=r(a)+r(b),\\ &r(ab)=\lVert\widehat{ab}\rVert_\infty=\fs{\hat{a}\hat{b}}_\infty\leq\fs{\hat{a}}_\infty\fs{\hat{b}}_\infty=r(a)r(b) \end{align*} My answer to (b) is as same as the other.
From this answer we know that $$ r(x)=\inf\{\Vert x^n\Vert^{1/n}:n\in\mathbb{N}\} $$ So from this post we get the answer for a). The functions $f_n(x)=\Vert x^n\Vert^{1/n}$ are continuous and a fortiori upper semi-continuous. Hence from this post we know that their infimum (which is actually $r(x)$) is upper semi-continuous too. This answers b).