On the spectrum of the sum of two commuting elements in a Banach algebra Original: Soit A une algèbre de Banach unitaire et a et b deux éléments tels que $a*b=b*a$.
Pourquoi  $σ (a+b) \subset σ(a)+σ(b)$.
Et  qu’elle est la relation entre σ (a*b)  et σ(a) et σ(b)?

Translation: Let $A$ be a unitary Banach algebra and the elements $a$ and $b$ such that $ab = ba$. Why is the spectrum of $a + b$ contained in the sum of the spectra of $a$ and $b$? What relations do we have between the spectrum of $ab$ and the spectra of $a$ and $b$?

 A: Two noncommuting quasinilpotent elements $a,b$ can give a non quasinilpotent sum $a+b$ and product $ab$. E.g., in $M_2(\mathbb{C})$:
$$
a=\pmatrix{0&1\\0&0}\quad b=\pmatrix{0&0\\1&0}\qquad \sigma(a)=\sigma(b)=\{0\}\quad\sigma(a+b)=\{\pm 1\}\quad \sigma(ab)=\{0,1\}.
$$
But, indeed:

In a unital Banach algebra, for every commuting elements $ab=ba$, the spectra of $a+b$ and $ab$ satisfy
  $$
\sigma(a+b)\subseteq \sigma(a)+\sigma(b)\qquad \sigma(ab)\subseteq \sigma(a)\sigma(b).
$$

Sketch: 1) We do it in the commutative case 2) We restrict to the commutative case by considering the bicommutant of $\mathbb{C}[x,y]$.
Proof: assume that $A$ is commutative first. Then by Gelfand, for every $x\in A$, we have
$$
\sigma(x)=\{\phi(x)\;;\;\phi\in \hat{A}\}
$$
where $\hat{A}$ denotes the set of characters (nonzero algebra homomorphisms from $A$ to $\mathbb{C}$). 
It follows readily that for all $x,y\in A$:
$$
\sigma(x+y)\subseteq\sigma(x)+\sigma(y)\qquad \mbox{and}\qquad\sigma(xy)\subseteq\sigma(x)\sigma(y).
$$
Now back to the general case where $A$ is not assumed to be commutative, take $x,y\in A$ such that $xy=yx$.
Let $B=\mathbb{C}[x,y]$ be the unital algebra generated by $x,y$. And denote $B'$ its commutant, that is the set of all $z$ in $A$ which commute with every element of $B$ (equivalently, with $x$ and $y$). Note that for any sets $S\subseteq T$, we have $T'\subseteq S'$. Also $S$ is commutative if and only if $S\subseteq S'$.
Since $B$ is commutative, we have:
$$
B\subseteq B'\qquad \Rightarrow \qquad B''\subseteq B'\qquad\Rightarrow\qquad B''\subseteq (B'')'.
$$
So $B''$ is commutative. Moreover, one checks easily that it is closed (hence a unital Banach algebra) and contains $B$.
We need to compare the spectra relative to $B''$ and to $A$ for elements of $B''$. So take $z$ in $B''$. Of course, if $z$ is invertible in $B''$, it is invertible in $A$. Now conversely, assume that $z$ is invertible in $A$. Now for all $u$ in $B'$, we have $zu=uz$ so $uz^{-1}=z^{-1}u$. Hence $z^{-1}$ belongs to $B''$. Therefore
$$
z\;\mbox{invertible in } A\quad\Leftrightarrow\quad z\;\mbox{invertible in}\; B''
$$
for all $z\in B''$. Applying this to $z=b-\lambda 1$, we see that
$$
\sigma_A(b)=\sigma_{B''}(b)\qquad\forall b\in B''.
$$
This shows that we can restrict to the commutative unital Banach algebra $B''$ and apply the above commutative case to $x,y,x+y,xy$ which all belong to $B$, hence to $B''$. $\Box$
Note: to see why this result is natural, and why the inclusions are not equalities in general, consider the case of $A=M_n(\mathbb{C})$. If $x$ and $y$ commute, they can be triangularized simulatneously. Then the result is clear. You see also why you don't necessarily have equalities.
