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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.