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The unit criterion for the Jacobson radical implies that every radical element is a quasinilpotent element, because $1-x/\lambda$ is invertible for every nonzero $\lambda$. Conversely, if a two-sided ideal consists of quasinilpotent elements, then $yx$ is quasinilpotent for every member $x$ and every $y\in A$. Hence $1+yx$ is invertible and $x$ lies in the Jacobson radical. The maximality statement concerns ideals contained in the quasinilpotent set, not the whole set itself.

If $x\in J$, part (ii) makes $1-x/\lambda$ invertible for every nonzero scalar $\lambda$. Thus $\sigma (x)\subseteq \{0\}$ and $r(x)=0$: it is a quasinilpotent element. Conversely, let $I$ be a two-sided ideal consisting of quasinilpotent elements. For $x\in I$ and $y\in A$, $yx\in I$ has spectrum $\{0\}$, so $1+yx$ is invertible. Part (ii) implies $x\in J$. This proves $J$ is the greatest ideal contained in the quasinilpotent set, the Jacobson radical as greatest quasinilpotent ideal characterization.

A semisimple Banach algebra is one with $J=\{0\}$. For $B=\mathcal{B}(X)$, its natural action on $X$ is faithful and algebraically irreducible. Indeed, given $v\ne 0$ and any $w\in X$, the Hahn-Banach theorem gives a bounded functional $g$ with $g(v)=1$. The rank-one operator $z\mapsto g(z)w$ then maps $v$ to $w$. Any invariant subspace containing $v$ consequently contains every $w$. The faithful representation has kernel zero, so zero is primitive and $B$ is semisimple.

Nevertheless, choose a nonzero bounded functional $g$ and $0\ne x_0\in \ker g$, possible when $\dim X>1$. The nonzero rank-one operator $Tz=g(z)x_0$ satisfies $T^{2}z=g(z)g(x_0)x_0=0$. Hence $\sigma (T)=\{0\}$ and $r(T)=0$, although $T\ne 0$. This proves the quasinilpotent set of $B$ is not $\{0\}$; it is not the same as the radical.

For infinite-dimensional $X$, the finite-rank operators form a nonzero proper two-sided ideal: products with bounded operators are finite-rank, and the identity has infinite rank. Thus zero is not maximal. Even if maximal ideals are required to be closed, the compact operators give a nonzero proper closed ideal containing the finite-rank operators. The identity is not compact on an infinite-dimensional Banach space, since the Riesz lemma supplies a bounded sequence with pairwise distances bounded below. Therefore zero is primitive but not maximal in infinite dimension.

The first algebra exists. Let $A=L^1(0,1)$ with truncated convolutionThe Tonelli theorem givesCommutativity follows by replacing $s$ with $t-s$; associativity follows from the Fubini theorem applied to the simplex $s,u\ge 0$, $s+u\le t$. Together with completeness of $L^1$, this makes $A$ a commutative Banach algebra, not necessarily unital. Adjoin an identity using unitization:This norm is complete and submultiplicative by the preceding bound, and $e=(1,0)$ is the identity. This is the Volterra convolution algebra on a finite interval with an identity adjoined.

Take $x=(0,1)$, where $1(t)=1$. Direct induction under convolution givesThus no power vanishes. For every $\lambda \ne 0$ the seriesconverges absolutely, since its norms have factorial denominators. Multiplication by $\lambda e-x$ telescopes to $e$, so it is an inverse. On the other hand $x$ cannot be invertible: its scalar coordinate is zero, and scalar coordinates multiply, so $xy$ can never equal $e$. HenceThis explicitly exhibits a nonnilpotent quasinilpotent element without assuming a spectral-radius formula.

The second algebra does not exist. Suppose it did, and choose a nonscalar $y$. Nonemptiness of its spectrum of an element, together with $\rho (y)=0$, gives ${\sigma }_B(y)=\{0\}$. But $y+e$ is still nonscalar: if $y+e=\alpha e$, then $y=(\alpha -1)e$. By translation of the spectrum by a scalar,This contradicts the required zero spectral radius for every nonscalar element. The contradiction already uses the spectral condition alone; the nonvanishing-power condition cannot remedy it.

A Banach algebra is semisimple when its Jacobson radical is zero. This does not assert that every quasinilpotent element is zero. For example, the algebra of bounded operators on a Banach space of dimension greater than one is semisimple, but contains nonzero square-zero rank-one operators. This notion also does not require the finite-dimensional direct-sum decomposition associated with a semisimple algebra.

The identity $(x+\alpha e)-\lambda e=x-(\lambda -\alpha )e$ proves the formula directly from invertibility. In a nonzero complex unital Banach algebra, the nonemptiness of the Banach-algebra spectrum implies that if $x$ is a quasinilpotent element, then $\rho (x+e)=1$. Consequently every nonscalar element cannot simultaneously be quasinilpotent: adding the identity preserves nonscalarity while translating its spectrum.

Truncated convolution makes $L^1(0,1)$ a commutative nonunital Banach algebra. The Tonelli theorem gives $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$, and the Fubini theorem gives associativity over the integration simplex. Its unitization is $\mathbb{C}\oplus L^1(0,1)$ with norm $|\alpha |+\Vert f{\Vert }_1$ and product $(\alpha ,f)(\beta ,g)=(\alpha \beta ,\alpha g+\beta f+f\ast g)$. The constant function one has convolution powers $t^{n-1}/(n-1)!$, of norm $1/n!$. Thus it is a nonnilpotent quasinilpotent element in this unitization.