Commutative unital Banach algebra with nilpotent elements

Question

What would be a concrete example of a commutative unital Banach algebra that contains nilpotent elements?

Answer 1

Upper triangular Toeplitz matrices give another example, i.e., the unital subalgebra of the $n$-by-$n$ matrices generated by the matrix with $1$s on the superdiagonal and $0$s elsewhere.

More generally, take any (nonzero) nilpotent element $a$ of a unital Banach algebra $B$, then take the unital Banach subalgebra of $B$ generated by $a$.

Answer 2

If you consider a closed (two-sided) ideal $I$ of the Banach algebra $A$, then you can endow $A/I$ with the quotient norm

$$ \|a+I\|=\inf_{x\in I}\|a+x\| $$

and this norm makes $A/I$ into a Banach algebra (see proof on PlanetMath).

If you now consider $a\in A$ and the closure $I$ of the ideal generated by $a^n$, then $a+I$ will be nilpotent in $A/I$.

Another example. Let $A$ be a commutative Banach algebra and consider $B=A\oplus A$ endowed with the norm $\|(a,v)\|=\|a\|+\|v\|$. Then $B$ is a Banach space. Define an operation on $B$ by

$$(a,v)(b,w)=(ab,aw+bv)$$

It's easy to check that this is a good multiplication turning $B$ into an algebra. For the norm we have \begin{align} \|(a,v)(b,w)\|&=\|(ab,aw+bv)\|\\ &=\|ab\|+\|aw+bv\|\\ &\le\|ab\|+\|aw\|+\|bv\| &&\text{(triangle inequality)}\\ &\le\|a\|\,\|b\|+\|a\|\,\|w\|+\|b\|\,\|v\| &&\text{($A$ is a Banach algebra)}\\ &\le\|a\|\,\|b\|+\|a\|\,\|w\|+\|b\|\,\|v\|+\|v\|\,\|w\|\\ &=(\|a\|+\|v\|)(\|b\|+\|w\|)\\ &=\|(a,v)\|\,\|(b,w)\| \end{align} Thus $B$ is a Banach algebra and every element of the form $(0,v)$ is nilpotent, because $(0,v)^2=(0,0)$.